Differential Calculus for Engineers 9811925526, 9789811925528

Table of contents :
Preface
Contents
1 Sequences of Real Numbers
1.1 Real Numbers
1.2 Real Number Sequences
1.3 Extended Real Number Line
2 Real Number Series
2.1 Convergent and Divergent Series
2.2 Series with Positive Terms
2.3 Series with Arbitrary Terms
2.4 Approximating the Sum of a Leibniz’s Series
2.5 Absolutely and Conditionally Convergent Series
2.6 Operations on Convergent Series
2.7 Sequences and Series of Complex Numbers
3 Sequences of Functions (Functional Sequences)
3.1 Simple and Uniformly Convergence
3.2 The Properties of the Uniformly Convergent Functional Sequences
4 Series of Functions (Functional Series)
4.1 Simple and Uniform Convergence
4.2 Properties of the Uniformly Convergent Series of Functions
4.3 Power Series
4.4 Taylor’s Formula
4.5 Taylor’s and Maclaurin’s Series
4.6 Elementary Functions. Euler’s Formulas. Hyperbolic Trigonometric Functions
5 Functions of Several Variables
5.1 Vector Space mathbbRn. Basic Notions and Notations
5.2 Convergent Sequences of Vectors in mathbbRn
5.3 Topology Elements on mathbbRn
5.4 Limits of Functions of Several Variables
5.5 Continuous Functions of Several Variables
5.6 Properties of Continuous Functions Defined on Compact or Connected Sets
5.7 Linear Continuous Maps from mathbbRn to mathbbRm
6 Differential Calculus of Functions of Several Variables
6.1 Partial Derivatives. Differentiability of a Function of Several Variables
6.2 Differentiability of Vector Functions. Jacobian Matrix
6.3 Differentiability of Composite Functions
6.4 The First Order Differenential and Its Invariance Form
6.5 The Directional Derivative. The Differential Operators: Gradient, Divergence, Curl and Laplacian
6.6 Partial Derivatives and Differentials of Higher Orders
6.7 Second-Order Partial Derivatives of Functions Composed of Two Variables
6.8 Change of Variables
6.9 Taylor’s Formula for Functions of Several Variables
6.10 Local Extrema of a Function of Several Variables
6.11 Local Inversion Theorem
6.12 Regular Transformations
6.13 Implicit Functions
6.14 Local Conditional Extremum
6.15 Dependent and Independent Functions
References
Index

Citation preview

Gavriil Paltineanu Ileana Bucur Mariana Zamfir

Differential Calculus for Engineers

Differential Calculus for Engineers

Gavriil Paltineanu · Ileana Bucur · Mariana Zamfir

Differential Calculus for Engineers

Gavriil Paltineanu Department of Mathematics and Computer Science Technical University of Civil Engineering Bucharest Bucharest, Romania

Ileana Bucur Department of Mathematics and Computer Science Technical University of Civil Engineering Bucharest Bucharest, Romania

Mariana Zamfir Department of Mathematics and Computer Science Technical University of Civil Engineering Bucharest Bucharest, Romania

ISBN 978-981-19-2552-8 ISBN 978-981-19-2553-5 (eBook) https://doi.org/10.1007/978-981-19-2553-5 © The Editor(s) (if applicable) and The Author(s), under exclusive license to Springer Nature Singapore Pte Ltd. 2022 This work is subject to copyright. All rights are solely and exclusively licensed by the Publisher, whether the whole or part of the material is concerned, specifically the rights of translation, reprinting, reuse of illustrations, recitation, broadcasting, reproduction on microfilms or in any other physical way, and transmission or information storage and retrieval, electronic adaptation, computer software, or by similar or dissimilar methodology now known or hereafter developed. The use of general descriptive names, registered names, trademarks, service marks, etc. in this publication does not imply, even in the absence of a specific statement, that such names are exempt from the relevant protective laws and regulations and therefore free for general use. The publisher, the authors, and the editors are safe to assume that the advice and information in this book are believed to be true and accurate at the date of publication. Neither the publisher nor the authors or the editors give a warranty, expressed or implied, with respect to the material contained herein or for any errors or omissions that may have been made. The publisher remains neutral with regard to jurisdictional claims in published maps and institutional affiliations. This Springer imprint is published by the registered company Springer Nature Singapore Pte Ltd. The registered company address is: 152 Beach Road, #21-01/04 Gateway East, Singapore 189721, Singapore

Preface

This book is based on lectures given by the authors at the Faculty of Geodesy and the Faculty of Civil Engineering from Technical University of Civil Engineering of Bucharest. The material presented in this book exceeds the content of the spoken lessons and so it is also useful for other engineering specialties and even for students in mathematics. The book provides the engineering disciplines with necessary information of differential calculus of functions with one and several variables. The paper includes all the basic notions and results of differential calculus that are taught today in higher technical education, namely: sequences and series of numbers, sequences, and series of functions, power series, elements of topology on n-dimensional space, limits of functions, continuous functions, partial derivatives of functions of several variables, Taylor’s formula, extrema of a function of several variables (free or with constrains), change of variables, dependent functions. The style of the paper is direct and few references are made to previous knowledge. The selection of the material made by the authors, the simplest, often original, versions chosen of the demonstrations presented as well as the numerous examples and exercises completely solved constitute the secret of the success of this work. We tried to offer the fundamental material concisely and without distracting details. We focused on the presentation of basic ideas in order to make in detailed and a comprehensible as possible. Besides students in technical faculties and those starting a mathematical course, the book may be useful to engineers and scientists who wish to refresh their knowledge about some aspects of differential calculus. Bucharest, Romania

Gavriil Paltineanu Ileana Bucur Mariana Zamfir

v

Contents

1 Sequences of Real Numbers . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1.1 Real Numbers . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1.2 Real Number Sequences . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1.3 Extended Real Number Line . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

1 1 4 10

2 Real Number Series . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.1 Convergent and Divergent Series . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.2 Series with Positive Terms . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.3 Series with Arbitrary Terms . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.4 Approximating the Sum of a Leibniz’s Series . . . . . . . . . . . . . . . . . . 2.5 Absolutely and Conditionally Convergent Series . . . . . . . . . . . . . . . 2.6 Operations on Convergent Series . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.7 Sequences and Series of Complex Numbers . . . . . . . . . . . . . . . . . . .

13 13 17 24 27 28 31 32

3 Sequences of Functions (Functional Sequences) . . . . . . . . . . . . . . . . . . . 3.1 Simple and Uniformly Convergence . . . . . . . . . . . . . . . . . . . . . . . . . . 3.2 The Properties of the Uniformly Convergent Functional Sequences . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

37 37

4 Series of Functions (Functional Series) . . . . . . . . . . . . . . . . . . . . . . . . . . . 4.1 Simple and Uniform Convergence . . . . . . . . . . . . . . . . . . . . . . . . . . . 4.2 Properties of the Uniformly Convergent Series of Functions . . . . . 4.3 Power Series . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4.4 Taylor’s Formula . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4.5 Taylor’s and Maclaurin’s Series . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4.6 Elementary Functions. Euler’s Formulas. Hyperbolic Trigonometric Functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

47 47 49 51 59 65

5 Functions of Several Variables . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5.1 Vector Space Rn . Basic Notions and Notations . . . . . . . . . . . . . . . . . 5.2 Convergent Sequences of Vectors in Rn . . . . . . . . . . . . . . . . . . . . . . . 5.3 Topology Elements on Rn . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5.4 Limits of Functions of Several Variables . . . . . . . . . . . . . . . . . . . . . .

73 73 75 76 87

42

70

vii

viii

Contents

5.5 5.6 5.7

Continuous Functions of Several Variables . . . . . . . . . . . . . . . . . . . . 93 Properties of Continuous Functions Defined on Compact or Connected Sets . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 97 Linear Continuous Maps from Rn to Rm . . . . . . . . . . . . . . . . . . . . . . 103

6 Differential Calculus of Functions of Several Variables . . . . . . . . . . . . . 6.1 Partial Derivatives. Differentiability of a Function of Several Variables . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6.2 Differentiability of Vector Functions. Jacobian Matrix . . . . . . . . . . 6.3 Differentiability of Composite Functions . . . . . . . . . . . . . . . . . . . . . . 6.4 The First Order Differenential and Its Invariance Form . . . . . . . . . . 6.5 The Directional Derivative. The Differential Operators: Gradient, Divergence, Curl and Laplacian . . . . . . . . . . . . . . . . . . . . . 6.6 Partial Derivatives and Differentials of Higher Orders . . . . . . . . . . . 6.7 Second-Order Partial Derivatives of Functions Composed of Two Variables . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6.8 Change of Variables . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6.9 Taylor’s Formula for Functions of Several Variables . . . . . . . . . . . . 6.10 Local Extrema of a Function of Several Variables . . . . . . . . . . . . . . 6.11 Local Inversion Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6.12 Regular Transformations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6.13 Implicit Functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6.14 Local Conditional Extremum . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6.15 Dependent and Independent Functions . . . . . . . . . . . . . . . . . . . . . . . .

107 107 117 119 125 128 134 142 145 150 154 159 161 162 171 180

References . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 183 Index . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 185

Chapter 1

Sequences of Real Numbers

1.1 Real Numbers Further we shall denote by N the set of natural numbers i.e.: N = {0, 1, 2, . . . , n, . . .} and by N∗ = N\{0}. On the set of natural numbers N are defined two arithmetic operations: the addition noted by “+” and the multiplication noted by “·”. The other two basic arithmetic operations i.e. subtraction and division are not possible in N. To make possible the subtraction operation the negative numbers are added to N, thus obtaining the set of integers: Z = {. . . , −n, . . . , −2, −1, 0, 1, 2, . . . , n, . . .}. The next extension of numbers, which makes possible also the division operation, is the set of rational numbers denoted by Q, namely the numbers of the form qp , where p, q ∈ Z, q = 0, p and q relatively prime (i.e. only integer factor divides both of them is 1). All rational numbers can be expressed as repeating decimal fractions. Also we remark that (Q, +, ·) is an Abelian (commutative) field. From ancient times it has observed that the set of rational numbers is not sufficient of any size in nature. rich to allow the express of the measure √ √ For example, the / Q. Thus, it was diagonal of a square of side 1 is 2, and it is known that 2 ∈ necessary to extend the set of rational numbers and it was created the set of real numbers denoted by R. Real numbers other than rational are called irrational. Unlike rational numbers, irrational numbers can be represented by infinite non-repeating decimal fractions. We do not intend to present here the construction of real numbers. We will just say that we can construct a set R which contains the rational numbers, on which two operations are definited: the addition noted by “+” and the multiplication noted by “·”, and an order relation, denoted by “≤”, so that the real number system (R, +, ·, ≤) © The Author(s), under exclusive license to Springer Nature Singapore Pte Ltd. 2022 G. Paltineanu et al., Differential Calculus for Engineers, https://doi.org/10.1007/978-981-19-2553-5_1

1

2

1 Sequences of Real Numbers

is a total ordered commutative field, which additionally satisfies the following properties: (A) (B)

(Archimedes’ axiom) For any x ∈ R and any y ∈ R, y > 0, there exists n ∈ N such that ny > x. (Cantor’s axiom) Every non-empty subset of R with an upper bound in R, has a least upper bound in R (see Definition 1.1.1 below).

The last property is what differentiates the real numbers from the rational numbers. For example, the set of rational numbers with square less than 2 has a rational √ upper bound in Q, namely 23 , but no rational least upper bound, because this is 2, and √ 2∈ / Q. Therefore from algebraic point of view, (R, +) is a commutative group with identity element 0, and (R\{0}, ·) is a commutative group with identity element 1.In addition we have the distributive law: x · (y + z) = x · y + x · z, ∀x, y, z ∈ R. The order relation “≤” is total, namely for any x, y ∈ R we have either x ≤ y or y ≤ x. Also the order relation is compatible with algebraic srtucture i.e.: 1. 2.

if x  ≤ y  and x  ≤ y  then x  + x  ≤ y  + y  , if x ≤ y and α > 0 then α · x ≤ α · y.

From the fact that (R, +, ·, ≤) is a total ordered commutative field results all the known rules of calculation with real numbers. Remark 1.1.1 It can show that the Archimedes’ axiom is equivalent with the following property: ∀x ∈ R, there is [x] ∈ Z such that [x] ≤ x < [x] + 1. ([x] is called the integer part of x). Proposition 1.1.1 For any x, y ∈ R, x < y, there exists r ∈ Q such that x < r < y. The proof results from Archimedes’ axiom. From Proposition 1.1.1 we deduce that between two real numbers there are an infinitude of ration numbers. The follwing proposition following from Cantor’s axiom. Proposition 1.1.2 For any x, y ∈ R, x < y, there exists at last irrational number z such that x < z < y. From Proposition 1.1.2 it results that between two real numbers there are an infinitude of irrational numbers.

1.1 Real Numbers

3

Definition 1.1.1 A set X of real numbers is said to be bounded above (below) if there exists a real number b (respectively a) such that each element x ∈ X satisfies the inequality: x ≤ b (respectively a ≤ x) In this case, the number b (respectively a) is called the upper bound (lower bound). Obviously, if b is an upper bound of X, then any number b ≥ b is also an upper bound, therefore every bounded above set has an infinitude of upper bounds. The argument concerning the lower bounds of a set bounded below is similar. Definition 1.1.2 A set X of real numbers is said to be bounded if has both an upper bound and a lower bound, that is, there are two real numbers a and b such that: a ≤ x ≤ b, ∀x ∈ X The least of all the upper bounds of a the set X bounded above is called the supremum of that set and is denoted by M = sup X . The greatest of all the lower bounds of a set X bounded below is called the infimum of that set and is denoted by m = inf X . Remark 1.1.2 A real number M is the supremum of the subset X of R if and only if the following two requirements are satisfied: 1. 2.

x ≤ M, ∀x ∈ X ; ∀ε > 0, ∃xε ∈ X such that M − ε < xε .

Proof “⇒” Indeed, if M = sup X , then M is an upper bound of X, whence it results (1). Since M is the least upper bound of X , it follows that for any ε > 0, M − ε is not an upper bound of X , hence there exists xε ∈ X such that M − ε < xε . “⇐” Let M ∈ R be with the properties (1) and (2). From (1) it results that M is an upper bound of X. Let M  < M and ε = M − M  > 0. From (2) we deduce that there exists xε ∈ X such that xε > M − ε = M  . Therefore M  is not an upper bound of X, so M = sup X . Remark 1.1.3 A real number m is the infimum of the subset X of R, if and only if the following two requirements are satisfied: 1. 2.

m ≤ x, ∀x ∈ X ; ∀ε > 0, ∃xε ∈ X such that xε < m + ε.

Remark 1.1.4 It can show that the Cantor’s axiom is equivalent with the following property: If {an } and {bn } are two rational number sequences with the properties: 1.

a1 ≤ a2 ≤ . . . ≤ an ≤ . . . ≤ bn ≤ . . . ≤ b2 ≤ b1 ,

4

1 Sequences of Real Numbers

2.

limn→∞ (bn − an ) = 0∗) ,

then there is an unique element c ∈ R such that an ≤ c ≤ bn , ∀n ∈ N. ∗)

∀ε > 0, ∃n ε ∈ N∗ such that bn − an < ε, ∀n ≥ n ε .

1.2 Real Number Sequences Definition 1.2.1 A real number sequence {xn } is said to be increasing (decreasing) if xn ≤ xn+1 , ∀n ≥ 1 (respectively xn+1 ≤ xn , ∀n ≥ 1). Both increasing and decreasing sequences are referred to as monotonic (monotone) sequences. Definition 1.2.2 A sequence {xn } is said to bounded above (below), if there exists a real numbers M (respectively m) suh that: xn ≤ M(m ≤ xn ), ∀n ≥ 1. The sequence {xn } is said to be bounded if it is bounded both above and below i.e. there exist the numbers m and M such that m ≤ xn ≤ M, ∀n ∈ N∗ . Definition 1.2.3 The sequence {xn } is called convergent if there exits a number l ∈ R with the property: for any ε > 0, there is a number n ε ∈ N, such that |xn − l| < ε, ∀n ≥ n ε . The number l is called the limit of the sequence {xn } and we shall write: R

lim xn = l or xn −→ l.

n→∞

n→∞

Remark 1.2.1 A convergent sequence has an unique limit. Indeed, let l and l  be the limits of the convergence sequence (xn ). If we denote by yn = xn − l and by yn = xn − l  , then lim x→∞ yn = lim x→∞ yn = 0. On the other hand we have:   0 = lim yn − yn = l  − l, n→∞

hence l  = l. Definition 1.2.4 Let {xn } be a sequence and let {kn } be an arbitrary increasing   sequence of positive integers (k1 < k2 < . . . < kn < . . .). The sequence xkn will be referred to as a subsequence of the sequence {xn }. In particular, the sequence {xn } itself can be regarded as a subsequence of the sequence {xn } (for kn = n).

1.2 Real Number Sequences

5

Obviously, if a sequence is convergent and has the limit l, then every subsequence is also convergent and has a same limit l. But if a given sequence {xn } has a convergent subsequence this does not imply that given sequence is itself convergent. Lemma 1.2.1 Every bounded sequence {xn } contains a convergent   (Cesàro). subsequence xkn . Proof Let {xn } be a bounded real number sequence. Then there exists a, b ∈ Q such that a < xn < b, ∀n ∈ N. If we denote by c the middle of the interval [a, b], then at least of the intervals [a, c] or [c, b] contains an infinite number of elements xn . If [a, c] has this property, then we shall denote a1 = a and b1 = c and choose xk1 ∈ [a1 , b1 ]. If c1 is the middle of the interval [a1 , b1 ] then at least of the intervals [a1 , c1 ] respectively [c1 , b1 ] contains an infinite terms of the sequence {xn }. Let’s suppose that [c1 , b1 ] has this property. Then we denote by a2 = c1 , b2 = b1 . Clearly we can choose among the elements of sequence {xn } an element xk2 ∈ [a2 , b2 ], such that k2 > k1 . Continuing in this way are obtained two sequences of rational numbers {an }, {bn } with the properties: 1. 2. 3.

a1 ≤ a2 ≤ . . . ≤ an ≤ . . . ≤ bn ≤ . . . ≤ b2 ≤ b1 . limn→∞ (bn − an ) = limn→∞ b−a = 0. 2n there is a strictly increasing sequence of natural numbers k1 < k2 < . . . < kn < . . . such that xkn ∈ [an , bn ], ∀n ∈ N.

On the other hand, from Remark 1.1.4 it follows that there exists an unique element l ∈ R with the property an ≤ l ≤ bn , ∀n ∈ N. Since   xk − l  ≤ bn − an = b − a and lim b − a = 0, n n→∞ 2n 2n   we deduce that the subsequence xkn is convergent and has the limit l. Definition 1.2.5 A real number sequence {xn } is said to be fundamental (or Cauchy), if for any ∀ε > 0, ∃n ε ∈ N∗ such that: |xm − xn | < ε, ∀m ≥ n ε , ∀n ≥ n ε If we denote by p = m − n (if m > n), respectively p = n − m (if m < n) we obtain the following equivalent definition: if ∀ε > 0, ∃n ε ∈ N∗ such that ∀n ≥ n ε and The sequence {x   n } is fundamental ∗ ∀ p ∈ N we have xn+ p − xn  < ε. Lemma 1.2.2 Every fundamental real numbers sequence is bounded. Proof Let {xn } be a fundamental real number sequence. For ε = 1, there is n 1 ∈ N∗    such that xn+ p − xn < 1, ∀n ≥ n 1 , ∀ p ∈ N∗ .

6

1 Sequences of Real Numbers

In particular case for n = n 1 it results that xn 1 − 1 < xn 1 + p < xn 1 + 1, ∀ p ∈ N∗ . If we denote by:     a = min x1 , . . . , xn 1 −1 , xn 1 − 1 and by b = max x1 , . . . , xn 1 −1 , xn 1 + 1 , then a ≤ xn ≤ b, ∀n ∈ N, hence {xn } is bounded. Theorem 1.2.1 (Cauchy’s test for convergence of a sequence). For the real number sequence {xn } to be convergent it is necessary and sufficient that it should be fundamental. Proof Necessity: Let {xn } be a convergent sequence with the limit l. Then ∀ε > 0, ∃n ε ∈ N∗ such that ∀n ≥ n ε , |xn − l| < 2ε . Of course, for m ≥ n ε , we have also |xm − l| < 2ε and further: |xm − xn | = |(xm − l) + (l − xn )| ≤ |xm − l| + |xn − l|
0, ∃n ε ∈ N∗ such that: |xn − xm |
0, ∃n ε ∈ N such that M − ε < xn ε . Since the sequence {xn } is increasing we have xn ≥ xn ε , ∀n ≥ n ε . Consequently: M − ε < xn ≤ M ≤ M + ε, ∀n ≥ n ε , that is: |xn − M| < ε, ∀n ≥ n ε , hence {xn } is convergent and limn→∞ xn = M. A reasoning for decreasing sequence is quite analogous. Example 1.2.2 (Number e). The best known application ofTheorem n 1.2.2 is the one consisting to prove the convergence of the sequence xn = 1 + n1 , n ≥ 1. We shall prove that this sequence is increasing and bounded above. Applying Newton’s binomial formula we get: n · (n − 1) 1 1 n · (n − 1) · (n − 2) · . . . · [n − (n − 1)] 1 + · 2 + ··· + · n n 1·2 n 1 · 2 · 3 · ... · n n         . 1 2 n−1 1 1 1 + ··· + · 1− · ... · 1 − =2+ · 1− · 1− 2! n n! n n n

xn = 1 + n ·

and, similarly for the term xn+1 : xn+1 = 2 +

        1 2 n 1 1 1 + ··· + · 1− · ... · 1 − · 1− · 1− 2! n+1 (n + 1)! n+1 n+1 n+1

8

1 Sequences of Real Numbers

We observe that the terms in xn are less than the corresponding terms in xn+1 , and, beside, xn+1 includes an extra positive term (the last one). Therefore: xn < xn+1 , ∀n ∈ N∗ , hence the sequence (xn ) is increasing. On the other hand we have:         1 1 1 1 2 n−1 xn = 2 + · 1 − + ··· + · 1− · 1− · ... · 1 − 2! n n! n n n 1 1 1 1 1 1 < 2 + + 2 + · · · + n−1 < 2 + + + ··· + 2! 3!  n! 2 2 2 1 1 1 1 − 2n =2+ 2 < 2 + 21 = 3 1 − 21 2 Thus we see that the sequence is bounded above. Consequently the sequence is incresing and bounded above, hence it is convergent according to Theorem 1.2.2. The limit of this sequence is refered to as the number e. Thus, 

1 lim 1 + n→∞ n

n = e.

We shall see later that the number e plays an extremely important role in mathematics. For the moment we only mention that e is an irrational number, 2 < e < 3 and e ≈ 2.718281. Remark 1.2.2 The statement of Theorem 1.2.2 it is not true for rational number sequences. n  For example, the rational numbers sequence xn = 1 + n1 , n ≥ 1 is fundamental, because it is convergent in R, and any convergent sequence is fundamental (as it results from the first part of the proof of the Theorem 1.2.2), but this sequence / Q. is not convergent in Q, since according to Remark 1.2.1 its limit is e, and e ∈ Example 1.2.3 Let us use Weierstrass’s test (Theorem 1.2.2) to establish the convergence of the sequence with the general term: xn = 1 +

1 1 1 + + · · · + − ln n, n ≥ 1. 2 3 n

We shall prove that this sequence is decreasing and bounded below. Indeed, from Lagrange’s Theorem for function f (x) = ln(x), x ∈ [k, k + 1], it results that there exists ck ∈ (k, k + 1) such that ln(k + 1) − ln(k) = c1k . On the other hand, from inequalities:

1.2 Real Number Sequences

9

1 1 1 < , ∀k ∈ N∗ < k+1 ck k we deduce: 1 1 < ln(k + 1) − ln(k) < , (∀)k ∈ N∗ k+1 k For k = 1, n we get successively: 1 < ln(2) − ln(1) < 1 2 1 1 < ln(3) − ln(2) < 3 2 ........................ 1 1 < ln(n + 1) − ln(n) < (∗) n+1 n Summing the above inequalites it follows: ln(n + 1) < 1 +

1 1 1 + + ··· + . 2 3 n

Since ln(n) < ln(n + 1), we deduce that xn = 1 + 21 + 13 + · · · + On the other hand, taking into account of inequality (*) it results: xn − xn+1 = ln(n + 1) − ln(n) −

1 n

− ln n > 0.

1 > 0. n+1

Therefore the sequence is decreasing and bounded below, hence it is convergent. The limit will be denoted by C (Euler’s constant). Thus,   1 1 1 C = lim 1 + + + · · · + − ln n n→∞ 2 3 n On can prove that C is an irrational number and C ≈ 0.57721. If we denote by εn = 1 + 21 + 13 + · · · + n1 − ln n − C, then: εn > 0 and lim εn = 0. n→∞

It results the following interesting identity: 1+

1 1 1 + + · · · + = ln n + C + εn , ∀n ≥ 1. 2 3 n

From this identity we deduce that:

10

1 Sequences of Real Numbers

  1 1 1 lim 1 + + + · · · + = ∞. n→∞ 2 3 n

1.3 Extended Real Number Line The extended real number line is denoted by R and is obtained from the real number line R by adding two elements +∞ and − ∞, i.e. R = R ∪ {−∞, +∞}. The order relation on R is obtained by extension of the order relation from R so: −∞ < +∞, −∞ < x, x < +∞, ∀x ∈ R.

In this way, R, ≤ is a totally ordered set, and every subset of R has a supremum and an infimum. If A ⊂ R is bounded, then the assertion follows from Cantor’s axiom; if A ⊂ R is unbounded above, then sup A = +∞ and if A ⊂ R is unbouded below, then inf A = −∞. The algebraic operations from R extended on R so: x + ∞ = ∞ + x = ∞, ∀x ∈ R, x  = −∞ x + (−∞) = (−∞) + x = −∞, ∀x ∈ R, x = ∞   ∞ dac a x > 0 x ·∞=∞·x = , ∀x ∈ R  −∞ dac a x < 0   −∞ dac a x > 0 , ∀x ∈ R. x · (−∞) = (−∞) · x =  ∞ dac a x < 0 The expressions: ∞−∞, −∞+∞, 0·(±∞), (±∞)·0, are usually left undefined. Definition 1.3.1 A real number sequence {x n } has the limit + ∞ (− ∞) if: ∀ε > 0, ∃n ε ∈ N∗ such that xn > ε(xn < −ε), ∀n ≥ n ε . We shall write: limn→∞ xn = +∞ (respectively limn→∞ xn = −∞). Proposition 1.3.1 Any monotone real number sequence has a limit in R and any real number sequence contains a subsequence which has a limit in R. Proof Let {xn } be a monotone increasing real number sequence. If {xn } is bounded above then, according to Theorem 1.2.2 it is convergent, hence has a finite limit. If {xn } is unbounded above then limn→∞ xn = +∞. For decreasing sequence the reasoning is quite analogous.

1.3 Extended Real Number Line

11

Let now {xn } be an arbitrary real number sequence.  If {xn } is bounded, then from Lemma 1.2.1 it results that it contains a subsequnce xkn convergent. Let us suppose that {xn } is unbounded (for example the sequence  is unbounded above). In this case we will prove that there exists a subsequence xkn such that limk→∞ xn k = +∞. Indeed, since there exist an infinity of terms xn >1, we can choose xk1 > 1. Also, since there exist an infinity of terms xn > 2 we can choose k2 > k1 with the property xk2 > 2, and so on. Thus we can construct a subsequence xkn such that xkn > n, hence limn→∞ xkn = +∞. Definition 1.3.2 Let {xn } be a real number sequence. An element a ∈ R is called the limit point of the sequence {xn } if there exists a subsequence xkn such that a = limn→∞ xkn . Remark 1.3.1 Every sequence which has a limit (finite or infinite), has only a limit point coinciding with the limit of the sequence. Example 1.3.1 1. 2. 3. 4.

The sequence xn The sequence xn The sequence xn The sequence xn

= (−1)n+1 has two limit points, namely − 1 and 1. n+1 = n√(−1) has two limit points, namely 0 and ∞. = n nhas an unique limit point, namely ∞. √ = (−1) has an unique limit point, namely 0. n

Definition 1.3.3 The largest limit point of the sequence {xn } is called the limit superior of {xn } and is denoted by limn→∞ xn , and the lowest limit point of the sequence {xn } is called the limit inferior of {xn } and is denoted by limn→∞ xn . Theorem 1.3.1 Any sequence of real numbers {xn } has the limit superior (the limit inferior) equal to a finite number or +∞ or to −∞. If the sequence is bounded its limit superior and limit inferior are finite numbers. Proof Let us denoted by A the set of all limit points of the sequence {xn }. If {xn } is unbounded above then +∞ ∈ A and limn→∞ xn = +∞. If {xn } is bounded above and A ∩R = φ, then A = {−∞} and limn→∞ xn = −∞. If the sequence is bounded, then according to Lemma 1.2.1, A ∩ R = φ. Since A ∩ R is bounded above it follows ∗ from Cantor’s axiom that there exists a = sup A ∩ R.  any n ∈ N there is an  For 1 element an ∈ A ∩ R such that a − n < an ≤ a. Let xkn be a subsequence of the sequence {xn } such that xkn → an . Obviously, for n sufficient large we can suppose that a − n1 < xn k < a + n1 , hence xkn → a. Therefore a ∈ A ∩ R and limn→∞ xn = a. An analogous reasoning is possible for limn→∞ xn . Remark 1.3.2 Let {xn } be a bounded real number sequence, L = limn→∞ xn and l = limn→∞ xn . Then: (a) (b)

For any a < L, there exist an infinite number of terms xn > a and for any b > L there are not more than a finite number of terms xn > b. For any a < l < b, then there exist an infinite number of terms xn < b and not more than a finite number of terms xn < a.

12

1 Sequences of Real Numbers

Indeed, we justify the statement for L = limn→∞ xn . From (a) it results  that for  any ∀n ∈ N∗ , there exist an infinite number of terms xn ∈ L − n1 , L + n1 . Then, by mathematical induction, we can construct a strictly increasing sequence {kn } such that xkn ∈ L − n1 , L + n1 , from which we deduce that of natural   numbers 2 xk − L  < , and further that xk → L. Therefore L is a limit point of {xn }. On the n n n other hand, if L  > L, then there exists n ∈ N∗ such that L  − n1 > L. Since there are not more than a finite number of terms xn > L  − n1 it follows that L  can not be a limit point of {xn }. Further, if we denote by m = inf{xn ; n ∈ N} and by M = sup{xn ; n ∈ N} then we have: −∞ ≤ m ≤ l ≤ L ≤ M ≤ +∞ Example 1.3.2 Find m, M, and l, L for the sequence with the general term: xn =

1 + (−1)n (−1)n + . n 2

We notice that: xn =

− n1 for n odd, 1 + 1 for n even. n

Therefore the sequence contains two convergent subsequences i.e.: − n1 → 0 and   1 + 1 → 1, hence l = 0 and L = 1. The subsequence − n1 is increasing and the n 1  subsequence n + 1 is decreasing. It follows that m = −1 and M = 2. Therefore we have: m = −1 < l = 0 < L = 1 < M = 2. Proposition 1.3.2 The necessary and sufficient condition for the sequence {xn } to have a limit (finite or infinite) is that L = limn→∞ xn = l = limn→∞ xn . Proof Neccesity: According to Remark 1.3.1, if there exists limn→∞ xn = a, then the sequence has an unique limit point coinciding with a, hence L = l = a. Sufficiency: First, let us suppose that L = l = a ∈ R. From Remark 1.3.2 it results that for any ε > 0, the interval (a − ε, a + ε) contains an infinite number of terms of the sequence {xn } and outside this interval there is a finite number of terms of the sequence. It results that a = limn→∞ xn . If L = l = +∞, then limn→∞ xn = +∞, and if L = l = a = −∞, then limn→∞ xn = −∞.

Chapter 2

Real Number Series

2.1 Convergent and Divergent Series Let us consider a real number sequence {u n }. A real number series is an expression of de form: u 1 + u 2 + ... + u n + ... =

∞ 

un

(2.1)

n=1

The “infinite sum” (2.1) is purely formal because the addition of an infinite number of summands does not make sense. The real numbers u 1 , u 2 , ..., u n , .... are called the terms of the series and the number u n is called the general term of the series. The number: Sn = u 1 + u 2 + ... + u n is called the nth partial sum of the series (1). Definition 2.1.1 Series (1) is called convergent if the sequence {Sn } of the partial sums of that series is convergent and the limit S of the sequence {Sn } is called the sum of the series. In this case we write:

S = u 1 + u 2 + ... + u n + ... =

∞ 

un

n=1

In the case where lim Sn = ±∞ or does not exist, the series is called divergent. n→∞

Example 2.1.1 Let us decide which is nature (convergent or divergent) of the following series: © The Author(s), under exclusive license to Springer Nature Singapore Pte Ltd. 2022 G. Paltineanu et al., Differential Calculus for Engineers, https://doi.org/10.1007/978-981-19-2553-5_2

13

14

1.

2 Real Number Series

∞

1 n=1 n(n+1)

=

1 1·2

+

1 2·3

+ ··· +

We notice that the partial sum Sn = Indeed, if in the identity:

1 n(n+1) 1 1·2

+

+ ··· 1 2·3

+ ··· +

1 n(n+1)

can be calculated.

1 1 1 = − , k (k + 1) k k+1 we succesively give to k the values 1, 2, . . . , n, we get: 1 1 1 + + ··· + 1·2 2·3 n(n + 1) 1 1 1 1 1 1 = 1 − + − + ··· + − =1− . 2 2 3 n n+1 n+1

Sn =

Since lim Sn = 1, we deduce that the series is convergent and its sum is n→∞  1 S = 1. Therefore we have ∞ n = 1 n (n + 1) = 1. ∞ 2. n = 1 n = 1 + 2 + ··· + n + ··· . In this case we have Sn = 1 + 2 + · · · + n = n(n+1) 2 Since lim Sn = ∞, the series is divergent. n→∞

3.

∞

n=1

(−1)n−1 = 1 − 1 + 1 − 1 + · · · + (−1)n−1 + · · ·

In this case we observe that:  Sn = 1 − 1 + 1 − 1 + · · · + (−1)

n−1

=

1 if n is odd 0 if n is even

Obviously, the limit of the sequence {Sn } does not exist, hence the series is divergent. Example 2.1.2 Geometric series. The general term of the geometric series is u n = a ·q n−1 . Therefore the geometric series has the following form: ∞ 

a · q n−1 = a + a · q + a · q 2 + · · · + a · q n−1 + · · ·

n=1 n

, for q = 1. The partial sum is Sn = a + a · q + a · q 2 + · · · + a · q n−1 = a · 1−q 1−q If |q| < 1, then lim q n = 0, hence lim Sn = 1−a q . n→∞ n→∞ Therefore, if |q| < 1, the geometric series is convergent and has the sum S = a . 1− q

2.1 Convergent and Divergent Series

15

If q = 1, then Sn = a + a + a+ · · · + a = a · n and lim Sn = ± ∞. n→∞ n times  a if n is odd . If q = − 1, then Sn = a − a + a − · · · + (−1)n−1 · a = 0 if n is even Clearly, in this case the limit of the sequence {Sn } does not exist. If q > 1, then lim q n = + ∞, hence lim Sn = ± ∞. n→∞ n→∞ Finally, if q < − 1, then the sequence {q n } does not have the limit, hence neither the sequence {Sn } does not have the limit. In conclusion, the geometric series is convergent if |q| < 1 and has the sum S = 1−a q and it is divergent if |q| ≥ 1. Further, as an application of the convergent geometric series, we shall present the conversion of a decimal fraction into a vulgar fraction. 2 ...bk . Case 1. The terminating decimal: 0 · b1 b2 . . . bk = b1 b10k 573 For example 0.57 3 = 1000 . b1 b2 ...bk Case 2. The pure recurring decimal: 0 · (b1 b2 . . . bk ) = 99 . . . 9 .   k times

Indeed, b1 b2 . . . bk b1 b2 . . . bk + + ··· 10k 102 k

b1 b2 . . . bk 1 1 = + + · · · 1 + 10k 10k 102 k

0 · (b1 b2 . . . bk ) =

We observe that in parentheses we have a convergent geometric series with the ratio q = 101k ∈ (− 1, 1). Therefore we have: 0.(b1 b2 . . . bk ) =

b1 b2 . . . bk b1 b2 . . . bk 1 · = 1 k 10 9 . . . 9 1 − 10k  9 k times

For example 0, (573) = 573 999 Case 3. The mixed recurring decimal:  b1 b2 . . . bk c1 c2 . . . c p − b1 b2 . . . bk 0.b1 b2 . . . bk c1 c2 . . . c p = 9 9 . . . 9 0 . . . 0  0 p times

k times

For proof we consider the particular case k = 1 , p = 2: c1 c2 c1 c2 b1 + + ··· + 10 103 105

b1 c1 c2 1 1 = + 1 + 2 + 4 + ··· 10 103 10 10

0 · b1 (c1 c2 ) =

16

2 Real Number Series

c1 c2 c1 c2 1 b1 1 b1 · = + + · 1 3 10 10 10 10 99 1 − 102  2 10 − 1 b1 + 10 c1 + c2 99 b1 + c1 c2 = = 990 990 b1 c1 c2 − b1 102 b1 + 10 c1 + c2 − b1 = = . 990 990 =

For example 0, 5(7 3) =

573−5 990

=

568 . 990

Example 2.1.3 Harmonic series. The harmonic series has the general term u n =

1 , n

hence the harmonic series is:

∞  1 1 1 1 = 1 + + + ··· + + ··· n 2 3 n n=1

The partial sum is Sn = 1 + 21 + 13 + · · · + n1 = ln n + C − εn , where lim εn = 0 n →∞ (Example 1.2.3, Chap. 1). It results that lim Sn = + ∞, hence the harmonic n→∞ series is divergent.  Proposition 2.1.1 If the series ∞ n = 1 u n is convergent, then lim u n = 0. n →∞

Proof Let S = lim Sn . Since u n = Sn − Sn − 1 , it results that: n→∞

lim u n = S − S = 0.

n →∞

 1 The reverse statement is not true. For example the harmonic series ∞ n = 1 n is divergent although lim u n = lim n1 = 0. n →∞ n →∞ From Proposition 2.1.1 it results the following practical criterion of divergence. Corollary 2.1.1 If lim u n is not equal to zero or does not exist, then the series n →∞ ∞ n = 1 u n is divergent.  Thus, when we investigate the convergence of the series ∞ n = 1 u n , you must first check if lim u n = 0. n →∞

Example 2.1.4 The series

∞

ln(3 + en ) n = 1 ln(1 + e3 n )

is divergent, because

 ln en 1 + 3 e−n  lim u n = lim n→∞ n→∞ ln e3n 1 + e−3n  n + ln 1 + 3 e−n 1 = = 0  = lim n→∞ 3n + ln 1 + e−3n 3

2.2 Series with Positive Terms

17

Theorem  2.1.1 (Cauchy criterion) A necessry and sufficient condition for a number ∗ series ∞ n = 1 u n to be convergent is that for any ε > 0, there exists n ε ∈ N such ∗ that for any n ≥ n ε and any p ∈ N there holds:

u n + 1 + u n + 2 + ... + u n + p < ε Proof The series

∞

n=1

u n is convergent if the sequence {Sn } of its.

partial sums is convergent. On the other hand, according to Theorem 1.2.1, the sequence {Sn } is convergent if it is fundamental i.e.: ∀ ε > 0, ∃ n ε ∈ N∗ such that ∀ n ≥ n ε and ∀ p ∈ N∗ there holds:



Sn+ p − Sn = u n+1 + u n+2 + · · · + u n+ p < ε Remark 2.1.1 Deletion of a finite number of terms of a series (or addition of a finite number of terms of a series) does not affect its convergence or divergence. Indeed, if {Sn } is the sequence of partial sums of given series then the sequence of partial sums of modified series is of form {Sn ± c},where c is the sum of terms add or delete.

2.2 Series with Positive Terms In this section we shall deal with series with positive terms i.e. the series of de form: ∞ 

u n , u n > 0∀ n ∈ N∗

n=1

The special place that these series occupy among the numerical series is highlighted by the following theorem. Theorem 2.2.1 A necessry and sufficient condition for a series with positive terms to be convergent is that its sequence of partial sums to be bounded above. Proof Necessity follows from the fact that if series is convergent then the sequence of partial sums is convergent, hence bounded. The sufficiency follows from the fact that the sequence of the partial sums of a series with positive terms is increasing and, being also bounded above it is convergent, according to Theorem 1.2.1. ∞  Theorem 2.2.2 (The first comparison test). Let ∞ n = 1 u n and n = 1 vn be two series with positive terms. Also we suppose that there exists k ∈ N∗ such that u n ≤ vn , ∀ n ≥ k. Then:

18

(1) (2)

2 Real Number Series

 ∞ If the series ∞ n = 1 vn is convergent, then the series n = 1 u n is convergent. ∞ If the series ∞ n = 1 u n is divergent, then the series n = 1 vn is divergent.

Proof From Remark 2.1.1 it results that deleting possible a finite number of terms of given two series we can suppose that u n ≤ vn , ∀ n ∈ N∗ . If we shall denote by Sn = u 1 + u 2 + · · · + u n and by Tn = v1 + v2 + · · ·+ vn , then we deduce that Sn ≤ Tn , ∀ n ∈ N∗ . The last inequality implies that the boundedness of the sequence {Tn } entails the boundedness of the sequence {Sn } and, conversely, the unboundedness of the sequence {Sn } entails the unboundedness of the sequence {Tn }. The Theorem follows now from Theorem 2.2.1.  1 Example 2.2.1 Find the nature of the series ∞ n = 2 ln n . Since 0 < ln n < n , ∀n ≥ 2, it results that n1 < ln1n , ∀ n ≥ 2.  1 As the harmonic series ∞ n = 1 n is divergent (Example 2.1.3), we deduce from ∞ Theorem 2.2.2 that the series n = 2 ln1n is also divergent. Example 2.2.2 The generalized harmonic series The generalized harmonic series is: ∞  1 1 1 1 = 1 + α + α + ··· + α + ··· ,α > 0 α n 2 3 n n=1

We shall prove that this series is convergent if α > 1 and divergent if α ≤ 1. Indeed, if α ≤ 1 then n α ≤ n, whence it results n1 ≤ n1α .  1 As the harmonic series ∞ n = 1 n is divergent, we deduce, from Theorem 2.2.2, ∞ 1 that the series n = 1 n α is divergent. If α > 1, then there exits β > 0 such that α = 1 + β. Further we have: 1 + 2α 1 + 4α ... 1



2k−1

1 1 1 2 1 < α + α = α = β 3α 2 2 2 2 1 1 1 1 1 1 1 4 1 + α + α < α + α + α + α = α = β 5α 6 7 4 4 4 4 4 4 α + 

1

1

α + · · · + 

2k−1 + 1

2k−1 1 α =  β . k−1 2 2k−1

2k − 1

α < 

Suming the previous inequalities it results: S2k − 1 = 1 +

1 1 + α 2α 3



+

1 1 1 1 + α + α + α 4α 5 6 7

+ ···

2.2 Series with Positive Terms

 + +

19



1 2k − 1

1

α + 

α + · · · + 

2k−1 + 1



1

α

2k − 1

0 there exists ε > 0 such that l − ε > 0 and let n ε ∈ N∗ be with the property l − ε < uvnn < l + ε, ∀ n ≥ n ε .

(1)

As vn > 0 , n ≥ 1 , it results (l − ε) · vn < u n < ( l + ε) · vn , ∀ n ≥ n ε . The assertion follows now from Theorem 2.2.2. l = 0, then, ∀ ε > 0, ∃ n ε ∈ N∗ such that:

(2)

− ε · vn < u n < ε · vn , ∀ n ≥ n ε The assertion follows now from Theorem 2.2.2. If l = + ∞, then, ∀ ε > 0, ∃ n ε ∈ N∗ such that u n > ε vn , ∀ n ≥ n ε .

(3)

The assertion follows now from Theorem 2.2.2.  Example 2.2.3 Find the nature of the series ∞ n=1

1√ . n2 · n n

√ . Since lim n n = 1 it results n→∞  1 lim uvnn = 1 ∈ (0, ∞). Since the series ∞ 2 is convergent (Example n = 1 n n→∞  1√ 2.2.2, α = 2 > 1) it results, from Theorem 2.2.4, that the series ∞ n = 1 n 2 · n n is also convergent.  Theorem 2.2.5 (Cauchy’s root test). Let ∞ n = 1 u n be a series with positive terms and let. √ L = lim n u n .Then: Let us denote by u n =

n2 ·

1√ n n

and by vn =

1 n2

n→∞

(1) (2)

If L < 1, then the series is convergent. If L > 1, then the series is divergent.

Proof Let q ∈ R such that L < q < 1. From Remark 1.3.2 it results that there exists √ only a finite number of terms n u n > q. In other words, there exists k ∈ N∗ √ such that n u n < q , ∀ n ≥ k. Therefore we have u n < q n , ∀ n ≥ k.  n As the series ∞ n = 1 q is convergent, being a geometric  series with the ratio 0 < q < 1, from Theorem 2.2.2 it follows that the series ∞ n = 1 u n is convergent. √ (2) If L > 1, then there exists an infinite  term n u n > 1. Thus u n > 1 for an infinity of indices, hence the series ∞ n = 1 u n is divergent according to Proposition 2.1.1. (1)

2.2 Series with Positive Terms

Corollary 2.2.1 Let there exists l = lim

21

∞ n = 1 u n be a series with positive terms, and we suppose that √ n u n . Then:

n →∞

(1) (2)

If l < 1 then the series is convergent. If l > 1 then the series is divergent.

Proof The assertion follows from Theorem 2.2.5.    n n Example 2.2.4 Find the nature of the series ∞ · n = 1 3 + (−1)   n If we denote by u n = 3 + (−1)n · a1n > 0 , n ≥ 1, then L = lim

n→∞

√ n

u n = lim

n→∞

1 , for a an

> 0.

  1 4 · 3 + (−1)n = a a

From Theorem 2.2.5 it follows that the series is convergent if a > 4 and divergent if a < 4.  1 if n is odd . If a = 4, then u n = 2n 1 if n is even Since lim u n = 0, according to Proposition 2.1.1, it results that the series is n →∞ divergent.  n3 Example 2.2.5 Find the nature of the series ∞ n = 1 (e + 1 )n . n √ √ n 3 √ n n 3 = 1 , it results that lim n u n = lim e+n1 = 1e < 1. Since lim n→∞ n→∞ n→∞ n From Corollary 2.2.1 we deduce that the series is convergent.  Theorem 2.2.6 (D’Alembert’s test). Let ∞ n = 1 u n be a series with positive terms, and we suppose that there exists lim u nu+n 1 = l. Then: n→∞

(1) (2)

If l < 1 the series is convergent. If l > 1 the series is divergent.

Proof Since lim

n→∞

un + 1 un

= l, for any ε > 0 there exists n ε ∈ N∗ such that:

l −ε < (1)

u n+1 0 with the property q = l + ε < 1. From (2.2) it results that:

22

2 Real Number Series

un + 1 < q · un , ∀ n ≥ nε

(2.3)

According to the Remark 2.1.2 we can suppose that n ε = 1. If in the inequality (2.3) we successively give to n the values 1 , 2 , ..., n − 1, ... we get: u2 ≤ q · u1 u3 ≤ q · u2 ≤ q 2 · u1 ... u n ≤ q · u n−1 ≤ q n−1 · u 1 ... Therefore we have u n ≤ u 1  · q n−1 , ∀n ∈ N∗ . The assertion follows now from n −1 is convergent, being a geometric Theorem 2.2.2, since the series ∞ n = 1 u1 · q series with the ratio q ∈ (0, 1). (2)

As l > 1, we can choose ε > 0 such that l − ε > 1. From (2.2) it follows that: u n+1 > l − ε > 1, ∀n ≥ n ε . un From Remark 2.1.1 we can suppose that n ε = 1. Therefore we have: un < un + 1 , ∀ n ≥ 1

Thus, the sequence {u n } is strictly increasing and so does not have the limit 0. The assertion follows now from Corollary 2.1.1.  n! Example 2.2.6 Find the nature of the series ∞ n = 1 nn .  n n Since lim u nu+n 1 = lim (n(n ++ 1)1)n n+ 1 = lim n +n 1 = 1e < 1, it results, n →∞ n →∞ n →∞ according to previous theorem, that the series is convergent.  Theorem 2.2.7 (Raabe’s test). Let  ∞ n = 1 u n be  a series with positive terms, and we un suppose that there exists lim n · u n + 1 − 1 = l. Then: n →∞

If l > 1 the series is convergent. If l < 1 the series is divergent.   Proof Since lim n · u nu+n 1 − 1 = l it results that ∀ ε > 0, ∃ n ε ∈ N∗ such that:

(1) (2)

n →∞

l −ε 1 we can choose ε > 0 such that α = l − ε > 1. Deleting (if necessary) a finite number of terms of the series we can suppose, according to Remark 2.1.2, that n ε = 1. Further, from (2.4) we deduce: n · u n − n · u n+1 ≥ α · u n+1 , ∀n ≥ 1. If we successively give to n the values 1, 2, . . . , n we get: u1 − u2 ≥ α · u1 2 · u2 − 2 · u3 ≥ α · u3 3 · u3 − 3 · u4 ≥ α · u4 ... n · u n − n · u n+1 ≥ α · u n+1 By suming the previous inequalities it results: Sn = u 1 + u 2 + · · · + u n ≥ α · (Sn − u 1 + u n+1 ) > α · (Sn − u 1 ) and further: Sn ≤

α · u1 , ∀n ∈ N∗ α−1

Therefore the sequence of the partial sums of the series is bounded above, hence the series is convergent. (2)

Since l < 1, we can choose ε > 0 such that l + ε < 1. From (2.4) it follows that:   n n · uun+1 − 1 < 1, hence n · u n ≤ (n + 1) · u n + 1 , ∀ n ≥ 1. Further we have: 1 n+1 1 n

≤

u n+1 , ∀n≥1 un

The assertion follows now from Theorem 2.2.3, since the harmonic series is divergent. Example 2.2.7 Find the nature of the series:

∞  n=1

3 · 6 · 9 · . . . · (3n) 1 · 2 · 5 · 8 · . . . · (3n − 1) n + 2

If we denote the general term of the series by:

∞

1 n=1 n

24

2 Real Number Series

un =

3 · 6 · 9 · . . . · (3n) 1 · 2 · 5 · 8 · . . . · (3n − 1) n + 2

then we have: lim n ·

n→∞

  un (3n + 2) · (n + 3) − 1 = lim n · −1 n→∞ u n+1 (3n + 3) · (n + 2) 2n 2 2 = < 1. = lim 2 n→∞ 3n + 9n + 6 3

From the previous theorem it resuts that the series is divergent.

2.3 Series with Arbitrary Terms In this section we investigate series whose terms are real numbers of any sign. The series that have only a finite number of positive (negative) terms can be assimilated with a series with positive terms. For the series with arbitrary terms we already have a convergence test, namely Cauchy’s test (Theorem 2.1.1). Further we present another important test. Theorem 2.3.1 (Dirichlet-Abel’s test). Let {an } be a decreasing sequence of positive numbers with lim an = 0, and let {vn } be a sequence such that there exists M > 0 n→∞

with the property:

|v1 + v2 + ... + vn | ≤ M, ∀n ∈ N∗ . Then the series

∞

n=1

an · vn is convergent.

Proof If we denote by Sn = v1 + v2 + ... + vn , ∀n ∈ N∗ , then by hypothesis there ∗ . Also, we shall denote by {Tn } the exists M > 0 such that |Sn | ≤ M, ∀n ∈ N ∞ sequence of the partial sums of the series n = 1 an · vn , i.e.: Tn = a1 · v1 + a2 · v2 + · · · + an · vn , ∀ ∈ N∗ First we remark that, since the sequence {an } is decreasing, it follows: |ak − ak+1 | = ak − ak+1 , for any k ∈ N∗ . Futher we have:

Tn+ p − Tn

2.3 Series with Arbitrary Terms

25

= an+1 · vn+1 + an+2 · vn+2 + · · · + an+ p · vn+ p

 = an+1 · (Sn+1 − Sn ) + an+2 · (Sn+2 − Sn+1 ) + · · · + an+ p · Sn+ p − Sn+ p −1 = |−an+1 · Sn + (an+1 − an+2 ) · Sn+1 + · · ·

 + an+ p−1 − an+ p · Sn+ p−1 + an+ p · Sn+ p ≤ an+1 · |Sn | + (an+1 − an+2 ) · |Sn+1 | + · · ·

 + an+ p−1 − an+ p · Sn+ p−1 + an+ p · Sn+ p  ≤ M · an+1 + an+ 1 − an+2 + · · · + an+ p−1 − an+ p + an+ p = 2 · M · an+1 .

Therefore for any n and p ∈ N∗ we have Tn + p − Tn ≤ 2 · M · an + 1 . As lim an = 0, it follows that for any ∀ ε > 0, ∃ n ε ∈ N∗ such that: n →∞

an
0 and vn = cos n(n − 1) − cos n(n + 1). Clearly,

the sequence {an } is decreasing and lim an = lim 2√1n+1 = 0. On the other hand n→∞ n→∞ ∞ {S } the sequence of the partial sums of the series n n = 1 vn has the property:

n |Sn | = k=1 vk = |1 − cos n(n + 1)| ≤ 2, ∀ n ∈ N∗ , hence it is bounded. From Theorem 2.3.1 it follows now that the series is convergent. Definition 2.3.1 A series is called alternating if its terms are alternately positive and negative. Therefore an alternating series has the form: ∞  n=1

(−1)n−1 · u n = u 1 − u 2 + u 3 − · · · + (−1)n−1 u n + · · · , u n > 0, n ∈ N∗

26

2 Real Number Series

 n −1 Theorem 2.3.2 (Leibniz’s test). Any alternatimg series ∞ · u n such n = 1 (−1) that the sequence {u n } is decreasing and convergent to 0, is convergent. Proof The assertion follows immediate from Theorem 2.3.1 for: an = u n and vn = (− 1)n − 1 . Indeed, by hypothesis, the sequence {an } is decreasing and lim an = 0. On the n→∞ other hand. Sn =

n 

 vk =

k=1

1 if n is odd 0 if n is even

hence |Sn | ≤ 1 , ∀ n ∈ N∗ . A series satisfying the conditions of Theorem 2.3.2 is often called Leibniz’s series. Example 2.3.2 Find the nature of the series:

1 1 1 1 + − + · · · + (−1)n−1 · + · · · 2 3 4 n 1 The series is convergent, since the sequence n is decreasing, and lim n1 = 0. n →∞ For calculate the sum of this series we establish in advance the following equality: 1−

1 1 1 1 1 + − + ··· + − 2 3 4 2n − 1 2 n

1 1 1 1 1 1 1 1 1 = 1 + + + + ··· + + −2· + + + ··· + 2 3 4 2n − 1 2n 2 4 6 2n

1 1 1 1 1 1 1 = 1 + + + + ··· + − 1 + + + ··· + 2 3 4 2n 2 3 n = ln(2 · n) + C + ε2·n − (ln n + C + εn ) = ln 2 + ε2·n − εn .

S2·n = 1 −

We mention that C is Euler’s constant and lim ε2 n = lim εn = 0 (Example n →∞ n →∞ 1.2.3). Therefore lim S2 n = ln 2. As the sequence {Sn } is convergent it follows n →∞ lim Sn = ln 2. n →∞ In conclusion we have: ln 2 = 1 −

1 1 1 1 + − + · · · + (−1)n−1 · + · · · 2 3 4 n

2.4 Approximating the Sum of a Leibniz’s Series

27

2.4 Approximating the Sum of a Leibniz’s Series Since the exact compute of the sum of a convergent series is generally not possible, it is customary to approximate the sum S by a partial sum Sn . The absolute error is: |Rn | = |S − Sn | In the particular series we have |Rn | = |S − Sn | < u n+1 .  case ofna−Leibniz’s 1 · u be a Leibniz’s series. Since the sequence {u n } is Indeed, let ∞ (−1) n n=1 decreasing it follows: S2n = S2n−2 + (u 2n−1 − u 2n ) ≥ S2n−2 S2n+1 = S2n−1 − (u 2n − u 2n+1 ) ≤ S2n−1 . If we denote by S the sum of the series then S2n S and S2n+1 S. Thus, we have: S2 < S4 < · · · < S2n < · · · < S < · · · < S2n+1 < · · · < S3 < S1 whence it results: 0 < S2 n + 1 − S < S2 n + 1 − S2 n + 2 = u 2 n + 2 0 < S − S2n < S2n+1 − S2n = u 2n+1 Therefore we have: |Rn | = |S − Sn | < u n+1 . The absolute error due to replacement of the sum of a Leibniz’s series by its n-th partial sum is not larger, in absolute value, than the first discarded terms. Example 2.4.1 Compute to exact four decimal the sum of the series:

∞ 

(−1)n−1 ·

n=1

1 1 1 1 = 1 − 2 + 3 − 4 + ··· n n 2 3 4

According to the above must be that u n+1 = n ≥ 5. Therefore we put approximately: S  S5 = 1 −

1 (n+1)n+1

< 10− , whence it results

1 1 1 1 + 3 − 4 + 5 = 0.78345 2 2 3 4 5

28

2 Real Number Series

2.5 Absolutely and Conditionally Convergent Series  Definition 2.5.1 A series arbitrary terms ∞ n = 1 u n is said to be absolutely with |u | |u | |u | = + + · · · + |u n | + · · · is convergent. convergent if the series ∞ n 1 2 n=1 Theorem 2.5.1 Any absolutely convergent series is convergent. ∞ Proof Indeed, ∞ if we suppose that the series n=1 u n is absolutely convergent; then the series n=1 |u n | is convergent and, by Cauchy’s test (Theorem 2.1.1) it results that ∀ε > 0,∃ n ε ∈ N∗ such that:

|u n+1 | + |u n+2 | + · · · + u n+ p < ε, ∀ n ≥ n ε , ∀ p ∈ N∗ On the other hand we have:

u n+1 + u n+2 + · · · + u n+ p ≤ |u n+1 | + |u n+2 | + · · ·

+ u n+ p < ε, ∀n ≥ n ε , ∀ p ∈ N∗ . From Cauchy’s criterion it follows now that the series

∞ n=1

u n is convergent.

Remark 2.5.1 The reverse statement is generally not true. There are series (with arbitrary terms) which are convergent but not convergent. This can be  absolutely n−1 1 · n which is convergent demonstrated by the example of the series ∞ (−1) n=1  ∞ 1 |u | (Example 2.3.2) but the series ∞ is divergent (Example 2.1.3). = n n=1 n=1 n ∞ Definition 2.5.2 The series ∞ n=1 u n is called conditionally convergent if it is convergent and the series n=1 |u n | is divergent. Remark 2.5.2 The series:

∞ 

(−1)n−1

n=1

1 1 1 1 1 = 1 − + − + ... + (−1)n · + · · · n 2 3 4 n

is conditionally convergent. One of the most important properties of the sum of a finite number of real summands is the commutativity property which asserts that a rearrangement of the summands does not affect the sum. A question naturally arises whether this property remains valid for the sum of a convergent series. The answer generally is negative. Example 2.5.2 Let us consider the following Leibniz’s series:

∞  n=1

(−1)n−1

1 1 1 1 1 1 = 1 − + − + ··· + − + ··· n 2 3 4 2n + 1 2n

2.5 Absolutely and Conditionally Convergent Series

29

This series is convergent and has the sum S = ln 2(Example 2.3.2). If {Sn } denoted the sequence of the partial sums of this series then lim Sn = ln 2. n→∞ Let us now rearrange the terms of series so:





1 1 1 1 1 1 1 1 1− − + − − + − − + ··· 2 4 3 6 8 5 10 12

1 1 1 − − + ··· + 2n − 1 4n − 2 4n

  If we denote by Sn the sequence of the partial sums of the modified series we have: n 

1 1 − − 2k − 1 4k −2 k=1

n 1 1 1 = − = 2 k=1 2k − 1 2k

 = S3·n

1 4k

=

n  k=1

1 1 − 4k − 2 4k



1 · S2n . 2

Obviously we have also the relations:  S3n−1 =

1 1 S2n + 2 4n

  = S3n−1 + S3n−2

1 . 4n − 2

Therefore there exists lim Sn = 21 ln 2 = S = ln 2. Thus we proved that, by n→∞ rearrangement of the terms of initial series the sum is modified. The concrete example we have considered shows that a conditionally convergent series does not posse the commutativity property. Further we present, without proof, the following theorem: Theorem 2.5.2 (Riemann’s theorem). If a series is conditionally convergent, then, for any specified number A, the terms of the series can be rearranged so that the resultant series will have the sum A. Moreover, the terms of a conditionally convergent series can be rearranged so that the resultant series will be divergent. From Riemann’s theorem we deduce that a conditionally convergent series does not posse the commutativity property. Further we shall prove that the commutativity property is valid for every absolutely convergent series. Theorem 2.5.2 Any series with positive terms convergent has the commutativity property.

30

2 Real Number Series

Let ∞

∞

and let S be its sum. Let n=1 u n be a series with positive terms convergent, ∞ v be the series obtaining from the series u by rearranging the terms. n=1 n n=1 n u kn . Obviously for any n ∈ N∗ there exists kn ∈ N∗ such that vn = ∞ series Since Tn = v1 + v2 + · · · + vn ≤ S it follows that the n=1 vn is convergent ∞ and that its sum T ≤ S. On the other hand, the series n=1 u n can be seen as a series obtaining from the series ∞ n=1 vn by rearranging the terms, where from it follows that S ≤ T , hence S = T . Theorem 2.5.4 (Cauchy’s theorem). If the terms of an absolutely convergent series are arbitrarily rearranged, then the series ramain absolutely convergent, and its sum remains the same.  Proof Let us consider the absolutely convergent series ∞ n=1 u n and let S be its sum. Let us put:  u+ n

=

u n if u n > 0 and u − n = 0 if u n < 0



−u n if u n < 0 0 if u n > 0

− + − We observe that we have u n = u + n − u n and |u n | = u n + un . ∞ −  ∞ + Let us consider the series with the positive terms n=1 u∞n and n=1 u n . We remark that this two series are convergent because the series n=1 |u n | is convergent − and u + n ≤ |u n |, u n ≤ |u n | (See Theorem 2.2.2).  If we denote by {Sn } the sequence of partial sums of the series ∞ n=1 u n then we have:

Sn =

n 

uk =

k=1

n  

u+ k

n=1

u n = S = lim Sn = lim n→∞



=

k=1

n→∞

n 

u+ k

k=1

Taking into account that the series ∞ 

−

u− k

∞ n=1

n 

u+ n and

u+ k − lim

k=1

n→∞

n=1

k=1

u− k

k=1

∞ n 

−

n 

u− n are convergent it rsults:

u− k =

∞  n=1

u+ n −

∞ 

u− n

n=1

Therefore we have: ∞  n=1

un =

∞  n=1

u+ n −

∞ 

u− n

(2.5)

n=1

 Let ∞ n=1 vn be the series obtained by rearranging the terms of the initial as above, the corresponding numbers series ∞ n=1 u n . Obviously we can construct,  v is also absolutely convergent because: vn+ and vn− . We remark that the series ∞ n=1 n

2.6 Operations on Convergent Series

31

∞ 



|v1 | + |v2 | + · · · + |vn | = u k1 + u k2 + · · · + u kn ≤ S = |u n | n=1

Then we have a similar formula with (1): ∞  n=1

vn =

∞ 

vn+ −

n=1

∞ 

vn−

(2.6)

n=1

 ∞ − + Of course,  the series ∞ n=1 vn and n=1 vn are series obtained from the series ∞ ∞ + − u by rearranging the to Theorem 2.5.3 it n n=1 u n and n=1 ∞ + ∞ terms. ∞According  + − − results that ∞ n=1 vn =  n=1 u n and n=1 vn = n=1 u n . Then from (1) and (2) we ∞ deduce that ∞ n=1 vn = n=1 u n and so the proof is finished.

2.6 Operations on Convergent Series We shall discuss here the problem of addition and multiplication of convergent series.   u n and ∞ Theorem 2.6.1 If the series ∞ n=1 n=1 vn are convergent, having the sums ∞ U respectively V , then the series n=1 (α · u n +β·vn ) is convergent for any α, β ∈ R, and has the sum α · U + β · V. Proof The assertion follows immediate from the equality:

n  k=1

(α · u k + β · vk ) = α ·

n  k=1

uk + β ·

n 

vk

k=1

∞ ∞  By the product of the series n=1 u n , n=1 vn is meant any series of the form k,l u k · vl . Therefore there are an infinity ways to multiply two series. Of these, the best known are the following two: u 1 · v1 + (u 1 · v2 + u 2 · v1 ) + · · · + (u 1 · vn + u 2 · vn−1 + · · · + u n · v1 ) + · · · (2.7) u 1 · v1 + (u 1 · v2 + u 2 · v2 + u 2 · v1 ) + · · · + (u 1 · vn + u 2 · vn + · · · + u n · vn + · · · + u n · v1 ) + · · · .

(2.8)

We mention that the product of two convergent series, generally, is not convergent. ∞ ∞ Theorem 2.6.2 If the series n=1 vn and n=1 u n are absolutely convergent, having the sums U respectively V , then any series product of these series is also absolutely convergent and has the sum U · V .

32

2 Real Number Series

Proof Let we have:

∞

k= 1

u ik · v jk be an arbitrary series product of these given series. Then





u i · v j + u i · v j + · · · + u i · v j 1 1 2 2 n n ≤ (|u 1 | + · · · + |u m |) · (|v1 | + · · · + |vm |)

 ∞ where m = max{i 1 , . . . , i n ; j1 , . .  . , jn }. Since the series ∞ n=1 |u n |, n=1 |vn | are convergent it rsults that the series ∞ u v is absolutely convergent, hence i j k k k=1 convergent. On the other hand, according to Theorem2.5.4, the series absolutely convergent are commutative, thus the sum of the series ∞ k=1 u i k · v jk is equal to the sum of the series product (4). But we observe immediate that the sequence {Pn } of partial sum of the series (4) is: Pn = (u 1 + u 2 + · · · + u n ) · (v1 + v2 + · · · + vn ) Therefore the sum of any series product of the series equal to lim Pn = U · V .

∞ n=1

u n and

∞ n=1

vn is

n→∞

2.7 Sequences and Series of Complex Numbers Further we shall denote by C the set of complex√ number  i.e.: C = z = a + i · b |a, b ∈ R, i = −1 . The modulus of the complex number z = a + i · b is denoted by |z| and is definite so:√ |z| = a 2 + b2 . Definition 2.7.1 A sequence of complex numbers {z n } is convergent in C and has C

the limit z ∈ C, if lim |z n − z| = 0. We shall denote this by z n −→ z. n→∞ C

Therefore z n −→ z if ∀ ε > 0, ∃ n ε ∈ N∗ such that |z n − z| < ε, ∀n ≥ n ε . Theorem 2.7.1 A sequence of complex numbers {z n }, z n = an + i · bn , ∀ n ∈ N∗ , is convergent in C, and has the limit z = a + i b, R

R

if and only if, an −→ a and bn −→ b. Proof The assertion follows immediate from the inequalities:

|an − a| ≤ |z n − z| =



(an − a)2 + (bn − b)2 ≤ |an − a| + |bn − b|

2.7 Sequences and Series of Complex Numbers

|bn − b| ≤ |z n − z| =



33

(an − a)2 + (bn − b)2 ≤ |an − a| + |bn − b|

Example 2.7.1 The sequence:

zn =

4· n + 5 1 n3 + 3 · n − 7 C 1 −→ z = + i · + i · 2· n − 3 7 · n3 + 1 2 7

Definition 2.7.2 A sequence of complex numbers {z n }, is fundamental in C if:

∀ ε > 0, ∃ n ε ∈ N∗ such that z n+ p − z n < ε , ∀ n ≥ n ε , ∀ p ∈ N∗ . Theorem 2.7.2 A sequence of complex numbers {z n }. z n = an + i bn , ∀ n ∈ N∗ is fundamental in C if the sequences of real numbers {an }, {bn } are fundamental in R. Proof The assertion follows immediate from the inequalities:



 2  2

an+ p − an ≤ z n+ p − z n = an+ p − an + bn+ p − bn



≤ an+ p − an + bn+ p − bn .





2  2

bn+ p − bn ≤ z n+ p − z n = an+ p − an + bn+ p − bn



≤ an+ p − an + bn+ p − bn . Theorem 2.7.3 A sequence of complex numbers {z n } is convergent in C if it is fundamental in C. The assertion follows from Theorems 2.7.1 and 2.7.2 and the general Cauchy’s test for convergence of real numbers sequences (Theorem 1.2.1).  Definition 2.7.3 A series of complex numbers ∞ n = 1 z n is called convergent in C if the sequence of its partial sums {Sn }, Sn = z 1 + z 2 + ... + z n , ∀n ∈ N∗ is convergent in C. If S = lim Sn , then S is the sum of the series and we shall n→∞  denote this by S = ∞ n=1 z n .  Theorem 2.7.4 For the series of complex numbers ∞ n=1 z n to be convergent it is necessary and sufficient that ∀ ε > 0, ∃n ε ∈ N∗ such that ∀n ≥ n ε and ∀ p ∈ N∗ we have:

z n+1 + z n+2 + · · · + z n+ p < ε The proof is similar to the proof of Theorem 2.1.1.

34

2 Real Number Series

∞ Definition 2.7.4 A series of complex numbers ∞ n=1 z n is said to be absolutely convergent if the series of positive numbers n=1 |z n | is convergent. Theorem 2.7.5 Any series of complex numbers absolutely convergent is convergent. The proof follows from Theorem 2.7.4 and the inequality:



z n+1 + z n+2 + · · · + z n+ p ≤ |z n+1 | + |z n+2 | + · · · + z n+ p As an application of the last theorem we present the exponential function e z e z .  z z2 zn zn Let us consider the series ∞ n=0 n ! = 1 + 1 ! + 2 ! + · · · + n! + · · · , z ∈ C. From Theorem 2.2.6 it results that this series is absolutely convergent. Indeed, we have: |z|n+1 (n+1)! lim n n→∞ |z| (n)!

= lim

n→∞

|z| =0 0, ∃n x,ε ∈ N∗ such that: D

| f n (x) − f (x)| < ε, ∀n ≥ n x,ε . Example 3.1.1 Let us consider the following sequence of functions: f n (x) = x n , x ∈ [0, 1], n ∈ N∗ . n ∗ We observe that for any x ∈ [0, 1), lim  n→∞ x = 0 and f n (1) = 1, ∀n ∈ N . s 0 for x ∈ [0, 1) Therefore f n −→ f , where f (x) = . [0,1] 1 for x = 1

Definition 3.1.2 We say that the sequence of functions { f n } is uniformly convergent to the function f on the set D, if ∀ε > 0, there exists n ε ∈ N∗ such that: | f n (x) − (x)| < ε, ∀n ≥ n ε and ∀x ∈ D. © The Author(s), under exclusive license to Springer Nature Singapore Pte Ltd. 2022 G. Paltineanu et al., Differential Calculus for Engineers, https://doi.org/10.1007/978-981-19-2553-5_3

37

38

3 Sequences of Functions (Functional Sequences) u

In this case we denote: f n −→ f . D

Remark 3.1.3 What is very essential to Definition 3.1.2, in contrast to Definition 3.1.1 is that n ε depend only on ε and is independent of x ∈ D. Example 3.1.2 Study the pointwise convergence and uniformly convergence of the following sequence of functions: f n : [−π, π ] → R, f n (x) =

cos n · x , n ∈ N∗ , x ∈ [−π, π ]. 2 · n2

For any x ∈ [−π, π ], we have limn→∞ f n (x) = limn→∞

cos n·x 2·n 2

= 0. s

If we denote by f the function f (x) = 0, ∀x ∈ [−π, π], then f n −→ f . [−π,π]

u

We shall prove that f n −→ f . [−π,π]

Indeed, for any x ∈ [−π, π] we have:  cos n x  |cos n x| 1   | f n (x) − f (x)| =  ≤ . = 2 2 2·n 2·n 2 · n2 = 0, it results that for any ε > 0 there exists n ε ∈ N∗ (we can   1 + 1), such that 2·n1 2 < ε, ∀n ≥ n ε . choose for example n ε = 2ε Therefore, for any ε > 0 there exists n ε ∈ N∗ such that: Since limn→∞

1 2·n 2

| f n (x) − f (x)| < ε, ∀n ≥ n ε , ∀x ∈ [−π, π], u

hence f n −→ f ≡ 0. [−π,π]

Remark 3.1.4 The geometrical interpretation of the uniform convergence is the following: for n ≥ n ε , the graph of the function f n is between the graphs of the functions f − ε and f + ε (Fig. 3.1). Further we shall denote by B(D) the vector space of all bounded real valued functions defined on D and for any f ∈ B(D) we note with: f = sup{| f (x)|; ∀x ∈ D}. Remark 3.1.5 We observe that the uniform convergence of the sequence { f n } on D to f is equivalent to the convergence of the number sequence { f n − f } to zero. That is: u

f n −→ f if lim f n − f = 0. D

n→∞

Indeed, the assertion follows immediate from the equivalence:

3.1 Simple and Uniformly Convergence

39

Fig. 3.1 Uniform convergence of a sequence of functions

| f n (x) − f (x)| < ε, ∀x ∈ D ⇔ f n − f < ε Remark 3.1.6 Obviously the uniform convergence involves the simple (pointwise) convergence. The reverse assertion, generally, is not true. Example 3.1.3 Let us consider the sequence of functions { f n }, f n : [0, 1] → R,  0 dac a x ∈ [0, 1] . f n (x) = x n , n ∈ N∗ and let f (x) =  1 dac a x = 1 S

In Example 3.1.1 we proved that f n −→ f . On the other hand, we have: [0,1]

f n − f = sup{| f n (x) − f (x)|; ∀x ∈ [0, 1]}  = sup x n ; ∀x ∈ [0, 1) ; 0 = 1, ∀n ∈ N∗ . It results that limn→∞ f n − f = 1 = 0, hence the sequence { f n } is not uniformly convergent to f on [0, 1]. Example 3.1.4 Study the uniformly convergence of the sequence of functions { f n }, x f n (x) = x+n , n ∈ N∗ , on the following sets: (a) (b)

the interval [a, b], 0 < a < b; the interval [0, ∞).

(a)

Since limn→∞ f n (x) = limn→∞

= 0, ∀x ∈ R, it results that

x x+n s

f n −→ f ≡ 0, R

For any 0 < a ≤ x ≤ b, and any n ∈ N∗ , we have:

40

3 Sequences of Functions (Functional Sequences)

a x x b b ≤ ≤ ≤ ≤ , b+n b+n x +n x +n a+n from which we deduce that: f n − f = sup{| f n (x) − f (x)|; x ∈ [a, b]} 

x b = sup ; x ∈ [a, b] ≤ . x +n a+n As limn→∞ (b)

b a+n

u

= 0, it follows that limn→∞ f n − f = 0, hence f n −→ 0. [a,b]

For any x ∈ [0, ∞) and any n ∈ N∗ , we have: 

x n 1 f n − f = sup ; ∀x ∈ [0, ∞) ≥ = , x +n n+n 2 from which results that limn→∞ f n − f = 0, hence the sequence { f n } is not uniformly convergent to f on [0, ∞). Theorem 3.1.1 (Cauchy’s uniformly convergence criterion). The necessary and sufficient condition for the sequence of functions { f n } to be uniformly convergent on the set D to the function f is that for any ε > 0 there exists n ε ∈ N∗ such that:    f n+ p (x) − f n (x) < ε, ∀x ∈ D, ∀n ≥ n ε , ∀ p ∈ N∗ . u

Proof Necessity. If f n −→ f , then ∀ε > 0, ∃n ε ∈ N∗ such that: D

| f n (x) − f (x)|
0, ∃n ε ∈ N∗ such that:    f n+ p (x) − f n (x) < ε, ∀n ≥ n ε , ∀ p ∈ N∗ , ∀x ∈ D.

(3.1)

3.1 Simple and Uniformly Convergence

41

On the other hand, it is obvious that for any fixed point x ∈ D, the real number sequence { f n (x)} is fundamental, hence convergent, according to Cauchy’s test for convergence of real number sequences (Theorem 1.2.1). Let us denote by f (x) = limn→∞ f n (x). If in the inequality (3.1) we pass to the limit on p (i.e., p → ∞), it results: | f (x) − f n (x)| < ε, ∀n ≥ n ε , ∀x ∈ D, u

hence f n −→ f . D

The following result establishes a sufficient condition for the uniformly convergence of a functional sequence. Proposition 3.1.1 If there exists a sequence of positive numbers {an } with the property limn→∞ an = 0, there exists also n 0 ∈ N∗ such that: | f n (x) − f (x)| ≤ an , ∀n ≥ n 0 , ∀x ∈ D, u

then f n −→ f . D

Proof Obviously we have: f n − f = sup{| f n (x) − f (x)|; x ∈ D} ≤ an , ∀n ≥ n 0 . u

As limn→∞ an = 0 it follows that limn→∞ f n − f = 0, hence f n −→ f . D

Example 3.1.5 Study the pointwise convergence and uniformly convergence of the functional sequence { f n }, where: f n (x) =

n 2 + cos n · x , n ∈ N∗ , x ∈ R. 3 · n2

For any x ∈ R, we have limn→∞ f n (x) = limn→∞

n 2 +cos n·x 3·n 2

s

= 13 , so f n −→ 31 . R

Also for any x ∈ R, we have:    f n (x) − 

Since an =

1 3·n 2

   1   cos n · x  1   n 2 + cos n · x − = =  3  3 · n2 3 3 · n2 |cos n · x| 1 = ≤ = an , ∀n ≥ 1 2 3·n 3 · n2

> 0, ∀n ∈ N∗ and limn→∞ an = limn→∞ u

Proposition 3.1.1 it results that f n −→ R

1 . 3

1 3·n 2

= 0, from

42

3 Sequences of Functions (Functional Sequences)

3.2 The Properties of the Uniformly Convergent Functional Sequences Further, we will examine the conditions under which a certain common property (continuity, derivability, integrability) of the terms of a sequence of functions is also transmitted to the limit of this sequence. We note that, as a rule, simple convergence is insufficient to achieve this transfer. Indeed, resuming Example 3.1.1, we find that although the functions f n are continuous on the interval [0, 1], its limit f is not a continuous function on [0, 1]. Theorem 3.2.1 If a functional sequence { f n } converges uniformly on the set D to the function f and if each term f n of the sequence is continuous on D, then f also is continuous on D. Proof Let x0 ∈ D be an arbitrary fixed point. For any x ∈ D and n ∈ N∗ , we have: | f (x) − f (x0 )| ≤ | f (x) − f n (x)| + | f n (x) − f n (x0 )| + | f n (x0 ) − f (x0 )|

(3.2)

u

Since f n −→ f , then ∀ε > 0, ∃n ε ∈ N∗ such that: D

| f n (x) − f (x)|
0, ∃δε > 0, such that, for any x ∈ D with the property |x − x0 | < δε we have: | f n (x) − f n (x0 )| < 3ε . If in the inequality (3.2) we suppose that n ≥ n ε and |x − x0 | < δε , then it results: | f (x) − f (x0 )| ≤

ε ε ε + + = ε, 3 3 3

hence f is continuous at the point x = x0 . Remark 3.2.1 If we suppose that x0 is an accumulation (limit) point of the set D (i.e. for any neighborhood V of x0 there exists x ∈ V ∩ D, x = x0 ), then from Theorem 3.2.1 it follows:    lim lim f n (x) = lim lim f n (x) . x→x0 n→∞

n→∞ x→x0

Indeed, taking into account of the continuity of f (respectively f n ) at the point x0 , it results: lim x→x0 f (x) = f (x0 ), respectively, lim x→x0 f n (x) = f n (x0 ).

3.2 The Properties of the Uniformly Convergent Functional Sequences

43

Therefore we have:  lim

x→x0

  lim f n (x) = lim f (x) = f (x0 ) = lim f n (x0 ) = lim lim f n (x) .

n→ ∞

x → x0

n→∞

n→ ∞

x → x0

Remark 3.2.2 Theorem 3.2.1 can be used in applications to show that a sequence of functions does is not uniformly convergent. Example 3.2.1 Show that the sequence: f n (x) =

1 , x ∈ R, ∀n ∈ N∗ , 1 + n2 · x 2

is not uniformly convergent on R.



s 0 if x = 0 , then we observe that f n −→ f . R 1 if x = 0 Therefore the functions, f n , n ∈ N∗ , are continuous on R, and the limit function f is not continuous on R. From Theorem 3.2.1 it results that { f n } is not uniformly convergent on R.

If we denote by f : R → R, f (x) =

Theorem 3.2.2 Let { f n } be a sequence of differentiable (derivable) functions on the u interval I ⊂ R with the properties that f n −→ f , and the sequence of derivatives I   f n is uniformly convergent on I to the function g. Then function f is differentiable on I and f  (x) = g(x), ∀x ∈ I . Proof For any h ∈ R such that x + h ∈ I , we have:



 f n+ p − f n (x + h) − f n+ p − f n (x) f n+ p (x + h) − f n+ p (x) − g(x) = h h  f n (x + h) − f n (x) − f n (x) + h    + f n (x) − g(x) , (∀)n, p ∈ N∗ . From Lagrange’s theorem it results that there exists a point c between x and x + h such that:





   f n+ p − f n (x + h) − f n+ p − f n (x) = f n+ p − f n (c) · h.

Therefore ∀x ∈ I and ∀n, p ∈ N∗ we have:       f n+ p (x + h) − f n+ p (x)   ≤  f (c) − f  (c)  − g(x) n+ p n   h     f n (x + h) − f n (x) − f n (x) +  h     (3.3) + f n (x) − g(x).

44

3 Sequences of Functions (Functional Sequences) u

Since f n −→ g, from Theorem 3.1.1, it results that ∀ε > 0, ∃n ε ∈ N∗ such that: I

  f

n+ p (c)

   ε − f n (c) +  f n (x) − g(x) < , 2

∀n ≥ n ε , ∀x ∈ I and ∀ p ∈ N∗ . f n (x) On the other hand, since lim h→0 fn (x+h)− = f n (x), it follows that there exists h δε > 0 such that:    f n (x + h) − f n (x)  ε   − f n (x) < ,  h 2 ∀x ∈ I, ∀h ∈ R, with the property x + h ∈ I and |h| < δε . If in the inequality (3.3) we suppose n ≥ n ε , p ∈ N∗ and |h| < δε , it results:    ε  f n+ p (x + h) − f n+ p (x)  ≤ + ε = ε.  − g(x)  2 2  h

(3.4)

If in the inequality (3.4) we pass to the limit on p (i.e. p → ∞), we obtain:    ε  f (x + h) − f (x) ≤ +ε =ε  − g(x)  2 2  h ∀x ∈ I, ∀h ∈ R, with x + h ∈ I and |h| < δε , where we deduce that there exists lim

h→0

f (x + h) − f (x) = g(x), ∀x ∈ I h

and thus, the proof is finished. 3.2.2 Establish the nature of the convergence of the sequences { f n } and Example f n , where f n (x) = arctann n·x , n ∈ N∗ , x ∈ R. Since limn→∞

arctan n·x n

s

= 0, ∀x ∈ R, it results that f n −→ f ≡ 0. R

We observe also that:    arctan n · x  |arctan n · x| π  | f n (x) − f (x)| =  − 0 = ≤ , ∀x ∈ R, ∀n ∈ N∗ n n 2·n As limn→∞

π 2·n

u

= 0, from Proposition 3.1.1 we deduce that f n −→ 0.

On the other hand we have f n (x) =  g(x) =

1 ,n 1+n 2 ·x 2

∈ N∗ , x ∈ R, and

0 if x = 0 , x ∈ R. 1 if x = 0

R f n

s

−→ g, where: R

3.2 The Properties of the Uniformly Convergent Functional Sequences

45

Since the functions f n are continuous on R and the limit function  g is not continuous on R, we deduce from Theorem 3.2.2 that the sequence f n is not uniformly convergent on R. Moreover, we have: (limn→∞ f n (x)) = f  (x) = 0, hence (limn→∞ f n (x)) = lim f n (x), since f  (0) = 0 and g(0) = 1. n→∞

Theorem 3.2.3 Let { f n } be a sequence of continuous functions on the interval [a, b] u such that f n −→ f . Then, there exist: [a,b]

⎡ lim ⎣

b

n→∞

⎤ f n (x)dx ⎦ =

a

b f (x)dx = a

b 

 lim f n (x) dx.

n→∞ a

Proof From Theorem 3.2.1 it follows that f is a continuous function on [a, b], hence f is an integrable function on [a, b]. Further we have:    b  b     b       f n (x)dx −  f (x)dx  =  [ f n (x) − f (x)]dx       a

a

a

b | f n (x) − f (x)|dx

≤ a

b ≤ fn − f ·

dx = (b − a) · f n − f . a

As limn→∞ f n − f = 0, it results that limn→∞

b a

f n (x)dx =

b a

f (x)dx.

Remark 3.2.3 The assertion of Theorem 3.2.3 remains true if it is replaced the continuity by the integrability of functions f n , n ∈ N∗ . Example 3.2.3 Establish the nature of the convergence of the sequences { f n }, where: f n (x) =

n·x , n ∈ N∗ , x ∈ [0, 1] en·x 2

Since for any x ∈ [0, 1], we have limn→∞ f n (x) = limn→∞ that: s

f n −→ f ≡ 0. [0,1]

n·x en·x 2

= 0, we deduce

46

3 Sequences of Functions (Functional Sequences)

Obviously, the functions, f and f n , n ∈ N∗ , are continuous functions on the interval [0, 1]. On the other hand we have: 1

1 f n (x)dx =

0

1

0





 n·x 1 1 1 1 and − 2 dx = − 2  = n ·x n·x 2 2 · en e 2·e 0 1 f (x)dx =

lim f n (x) dx =

n→∞ 0

1

0

0 dx = 0. 0

Therefore: 

1 f n (x)dx = lim

lim

n→∞

n→∞

0

=

1 1 − 2 2 · en

1 

 =

1 = 0 2

 lim f n (x) dx.

n→∞ 0

From Theorem 3.2.3 it follows that the sequence { f n } is not uniformly convergent to 0 on the interval [0, 1].

Chapter 4

Series of Functions (Functional Series)

4.1 Simple and Uniform Convergence Definition 4.1.1 Let {u n } be a  sequence of real valued functions defined on the set on D ⊂ R. The series of functions ∞ n=1 u n is called simply (uniformly) convergent  the set D if the functional sequence of partial sums {Sn }, defined by Sn = nk=1 u k = u 1 + u 2 + · · · + u n ; ∀n ∈ N∗ , is simply (uniformly) convergent on D. If S(x) = limn→∞ Sn (x), x ∈ D, then the function S : D → R is called the sum of the series and we will use notation: S=

∞ 

un = u1 + u2 + · · · + un + · · ·

n=1

 The series of functions ∞ called absolutely simply (uniformly) convern=1 u n is gent on D if the series of functions ∞ n=1 |u n | is simply (uniformly) convergent on the set D. Remark 4.1.1 Any series of functions, uniformly convergent on the set D is simply convergent on D. The reverse assertion, generally, is not true. Theorem 4.1.1 (Cauchy’s uniform convergence criterion). A necessary and suffi cient condition for the series of functions ∞ n=1 u n to be uniformly convergent on the set D, is that for any ε > 0, there exists n ε ∈ N∗ such that:   u n+1 (x) + · · · + u n+ p (x) < ε, ∀n ≥ n ε , ∀ p ∈ N∗ and ∀x ∈ D.  Proof The series of functions ∞ n=1 u n is uniformly convergent on D, if the sequence of partial sequence {Sn } is uniformly convergent on D. According to Cauchy’s uniform convergence criterion for sequences of functions (Theorem 3.1.1), the sequence {Sn } is uniformly convergent on D, if for any ε > 0, there exists n ε ∈ N∗ such that: © The Author(s), under exclusive license to Springer Nature Singapore Pte Ltd. 2022 G. Paltineanu et al., Differential Calculus for Engineers, https://doi.org/10.1007/978-981-19-2553-5_4

47

48

4 Series of Functions (Functional Series)

     Sn+ p (x) − Sn (x) = u n+1 (x) + · · · + u n+ p (x) < ε, ∀n ≥ n ε , ∀ p ∈ N∗ , ∀x ∈ D. Further we present a practical criterion of uniform convergence for series of functions.  Theorem 4.1.2 (Weierstrass’ uniform convergence criterion). Let ∞ n=1 u n be a ∗ ∈ N , and a convergent series of functions defined on the set D. If there exists n 0  a such that: number series ∞ n n=1 |u n (x)| ≤ an , ∀n ≥ n 0 , ∀x ∈ D, then the series of functions

∞

u n is uniformly and absolutely convergent on D. ∞ Proof Since the number series n=1 an is convergent, according to Theorem 2.1.1, it results that for any ε > 0, there exists n ε ∈ N∗ such that: n=1

an+1 + · · · + an+ p < ε, ∀n ≥ n ε and ∀ p ∈ N∗ .   Let n ε = max n ε , n 0 , n ≥ n ε and p ∈ N∗ . Then we have:     u n+1 (x) + · · · + u n+ p (x) ≤ |u n+1 (x)| + · · · + u n+ p (x) ≤ an+1 + · · · + an+ p < ε, ∀x ∈ D. The assertion results now from Theorem 4.1.1. Example 4.1.1 Study the uniform convergence of the series: ∞  sin(n + 1) · x √ , on R. x4 + n · n n=1

Since we have:    sin(n + 1) · x  |sin(n + 1) · x| 1 1 ∗   √  x 4 + n · n  = x 4 + n · √n ≤ x 4 + n · √n ≤ n · √n , ∀x ∈ R, ∀ n ∈ N and the series with positive terms

∞

1 √ n=1 n· n

is convergent (Example 2.2.2; α =

> 1), according to Theorem 4.1.2, the given series is uniformly and absolutely convergent on R. 3 2

4.2 Properties of the Uniformly Convergent Series of Functions

49

4.2 Properties of the Uniformly Convergent Series of Functions For the sum of a finite number of functions, the following properties are known: 1. 2. 3.

The sum of a finite number of continuous functions is also a continuous function. The sum of a finite number of differentiable functions is also a differentiable function and the derivative of the sum is the sum of the derivatives. The sum of a finite number of integrable functions is also an integrable function and the integral of the sum is the sum of the integrals.

Further we will show under what conditions these properties are kept for the sum of a series of functions (the “sum” of an infinity number of functions).  Theorem 4.2.1 If the series of functions ∞ n=1 u n converges uniformly on the set D ⊂ R and all its terms are continuous functions on D, then its sum S is also a continuous function on D. Proof Since any sum of a finite number of continuous functions is also a continuous function, it results that the sequence {Sn } of partial sums of the given series is a u sequence of continuous functions on D. As Sn −→ S, from Theorem 3.2.1 we deduce D

that S is a continuous function on D.  Remark 4.2.1 If the series ∞ n=1 u n , whose terms are continuous on D, but which do not uniformly converge on D, may have as its sum a discontinuous function.  n−1 , x ∈ [0, 1]. Example 4.2.1 Consider the series ∞ n=1 (1 − x) · x We have: Sn (x) =

n 

(1 − x) · x k−1 = (1 − x) · (1 + x + x 2 + · · · + x n−1 )

k=1

= (1 − x) ·

1 − xn = 1 − x n , ∀x ∈ [0, 1) (1 − x) 

1 if x ∈ [0, 1) . 0 if x = 1 Therefore the sum S is a discontinuous function on [0, 1].  Theorem 4.2.2 (Term-by-term differentiation theorem). Let ∞ n = 1 u n be a series of functions uniformly convergent on the interval I ⊂ R, and all its terms are 

u of these derivatives is also differentiable on I. Suppose that the series ∞ n=1 n  u , and T is the sum of uniformly  convergent on I. If S is the sum of the series ∞ n n=1

u , then S is differentiable on I and: the series ∞ n=1 n s

and Sn (1) = 0. Obviously, Sn −→ S, where S(x) = [0,1]



S (x) =

∞  n=1



u n (x)

=

∞  n=1

u n (x) = T (x), ∀ x ∈ I.

50

4 Series of Functions (Functional Series)

Proof If Sn (x) =

n k=1

u k (x), ∀x ∈ I , then:

Sn (x) =

n 

u k (x) = Tn (x), ∀x ∈ I

k=1 u

u

I

I

On the other hand, by hypothesis Sn −→ S and Sn = Tn −→ T . From Theorem 3.2.2 we deduce that S is differentiable on I and S = T on I. Example 4.2.2 Study the differentiability of the function: f (x) =

∞  cos n · x , x ∈ R. 2 · (n + 1) n n=1

  cos n·x According to Theorem 4.1.2, the functional series ∞ u n (x) = ∞ n=1 n=1 n 2 ·(n+1)    n·x  1 1 converges uniformly on R because  ncos 2 ·(n+1)  ≤ n 2 ·(n+1) ≤ n 3 , ∀x ∈ R, ∀n ≥ 1 and ∞ 1 the number series n=1 n 3 is convergent (Example 2.2.2; α = 3 > 1). ∞

∞ − sin n·x The series of derivatives n=1 u n (x) = n=1 n·(n+1) is also uniformly convergent on R, because:    − sin n · x  1 1 ∗    n · (n + 1)  ≤ n · (n + 1) ≤ n 2 , ∀x ∈ R, ∀n ∈ N  1 and the series with positive terms ∞ n=1 n 2 is convergent (Example 2.2.2; α = 2 > 1). From Theorem 4.2.2 it results that f is differentiable on R and: f (x) =

∞  − sin n · x , x ∈ R. n · (n + 1) n=1

 Theorem 4.2.3 (Term-by-term integration theorem). Let ∞ n=1 u n be a series of continuous functions, uniformly convergent on the interval [a, b]. Then its sum S is integrable on [a, b] and we have:

b S(x)dx = a

b  ∞ a

n=1

⎞ ⎛ b

∞  ⎝ u n (x)dx ⎠ u n (x) dx =

n=1

(4.1)

a

Proof From Theorem 4.2.1, it results that S is continuous on [a, b], hence it is integrable on [a, b].  If we denote by {Sn } the sequence of the partial sums of the series ∞ n=1 u n , then u Sn −→ S, and according to the Theorem 3.2.3 we deduce that: [a,b]

4.3 Power Series

51

b

b S(x)dx = lim

Sn (x)dx.

n→∞

a

a

On the other hand, we have:

b Sn (x)dx = a

b  n

u k (x) dx =

k=1

a

n



b

u k (x)dx.

k=1 a

Therefore:

b  ∞

u n (x) dx =

n=1

a

b S(x)dx a

= lim

n→∞

n



b

u k (x)dx =

∞



b

u n (x)dx.

n=1 a

k=1 a

Remark 4.2.2 The assertion of Theorem 4.2.3 rests true if the continuity of the functions u n , n ∈ N∗ , is replaced with the integrability of this functions.  sin n·x Example 4.2.3 Show that the function f (x) = ∞ n=1 n·(n+1) is continuous on R and  π2 ∞ 1 2 compute 0 f (x)dx, using Euler’s result n=1 n 2 = π6 .  sin n·x According to the Example 4.2.1, the series of continuous functions ∞ n=1 n·(n+1) is uniformly convergent on R and its sum f is continuous on R (Theorem 4.2.1). Further, taking into account from the Euler’s result and from the Example 2.1.1 we obtain: π

2 0

π π π 

2  ∞ ∞ 2 2   sin n · x sin n · x cos n · x  π 2 f (x)dx = − 2 dx = dx = n · (n + 1) n · (n + 1) n · (n + 1)  0 n=1 n=1 n=1

0

=

∞  n=1

0

∞ ∞   1 1 1 π2 − = = − 1. n 2 · (n + 1) n 2 n=1 n · (n + 1) 6 n=1

4.3 Power Series A power series is a particular case of a functional series, so all results obtained for functional series rest valid also for power series. Definition 4.3.1 A power series is a series of functions of the type:

52

4 Series of Functions (Functional Series) ∞ 

an · x n = a0 + a1 · x + a2 · x 2 + · · · + an · x n + · · · , x ∈ R

(4.2)

n=0

where {an } is a sequence of real numbers.

 n The set Mc of all points x0 ∈ R such that the number series ∞ n=0 an · x 0 is convergent is called the set of convergence of power series (4.2); Therefore:  Mc = x0 ∈ R;

∞ 

 an · x0n is convergent .

n=0

We observe that 0 ∈ Mc , hence Mc = ∅. Lemma 4.3.1 If the series (4.2) is convergent at the point x0 ∈ R, then this series is absolutely convergent at every point x ∈ R with the property |x| < |x0 |. ∞ n Proof Since the number series n=0an · x 0 is convergent, from Proposition 2.1.1  n it follows that the sequence an · x0 is convergent and has the limit 0. Because any sequence is bounded, it results that there exists M > 0 such that   convergent an · x n  ≤ M, ∀n ∈ N. 0 On the other hand we have:    n      n   an · x n  = an · x n  ·  x  ≤ M ·  x  = vn , ∀n ∈ N. 0 x  x  0 0  n ∞ x v = M ·  x0  is convergent because n=0 n   n=0   is a geometric series with the ratio q =  xx0  < 1 (Example 2.1.2). According to ∞ Theorem 2.2.2 it results that the series n=0 an · x n is absolutely convergent.  n Theorem 4.3.1 (Abel’s first theorem). For any power series ∞ n=0 an · x there exists 0 ≤ R ≤ ∞ with the properties:  n 1. The series ∞ n=0 an · x is absolutely convergent for any x ∈ R, |x| < R. ∞ 2. The series n=0 an · x n is divergent if x ∈ R, |x| > R. n 3. The series ∞ n=0 an · x is uniformly and absolutely convergent on the interval [−r, r ], ∀0 < r < R. If |x| < |x0 |, then the series

∞

R is called the radius of convergence of the power series and the interval (−R, R) is called the convergence interval of the series. Proof If the convergence set Mc = {0}, then R = 0 and the proof is finished. Suppose that Mc = {0}. Case 1. If Mc is unbounded, then Mc = R and R = +∞. there Indeed, let x1 ∈ R, be an arbitrary point. Since the set Mc is unbounded,  n exists x0 ∈ Mc such that |x1 | < x0 = |x0 |, hence the series ∞ n=0 an · x 1 is absolutely convergent according to Lemma 4.3.1.

4.3 Power Series

53

Case 2. If the set Mc is bounded above, then R = sup Mc > 0. such that |x| < x0 = |x0 | < R. Indeed, if |x| < R then there exists x0 ∈ Mc n According to Lemma 4.3.1, the power series ∞ n=0 an · x is convergent. Let x ∈ R such that |x| > R. Obviously, there exists y > 0 such that |x| > y > R. If  n a · we suppose that the series ∞ n=0 n x is convergent, then from Lemma 4.3.1 it results thaty ∈ Mc and this contradicts the definition of R = sup Mc . Therefore n the series ∞ n=0 an · x is divergent for |x| > R.  n Finally, the fact that the series ∞ n=0 an · x is uniformly and absolutely convergent on every interval [−r, r ], 0 < r < R, follows from Weierstrass’s test (Theorem 4.1.2), because |an · x n | < |an · r n |, ∀x ∈ [−r, r ], and the number series ∞ n n=0 |an · r | is convergent. The next theorem gives a practical computational formula for the convergence radius.  n Theorem 4.3.2 (Cauchy–Hadamard). Let ∞ n=0 an · x be a power series and let √ n |a |. ω = limn→∞ n Then: 1. 2. 3.

R = ω1 , if 0 < ω < ∞, R = 0, if ω = ∞, R = ∞, if ω = 0.

Proof  Applying Cauchy’s test (Theorem 2.2.5) to the number series with positive n terms ∞ n=0 |an · x | we get: L = lim

n→∞

1.

2.

3.

 n

|an · x n | = ω · |x|.

If 0 ω1 then there exists |x| > y > ω1 . If we suppose that ∞ n=0 an n· x |a | is convergent, then from Lemma 4.3.1 it results that n=0 n · y is √ the series n = ω · y > 1, where | · y convergent. On the other hand, we have limn→∞ n |a n  n results, according to Cauchy’s test, that the series ∞ n=0 |an | · y is divergent. We come thus to a contradiction.  n Let ω = ∞ and let x0 = 0 be arbitrary. We will show that the series ∞ n=0 an · x 0 n is divergent. Indeed, let 0 < y < |x0 |. If we suppose that the series ∞ a · x 0 n=0 n n |a | is convergent, then from Lemma 4.3.1 it results that√the series ∞ · y n=0 n n n = ω · y > 1, |a | is convergent. On the other hand we have lim · y n→∞ n  n whence it results that the series ∞ n=0 |an | · y is divergent. Thus we come to a contradiction. Therefore, Mc =√{0}, hence R = 0. n | = ω · |x| = 0 < 1, ∀x ∈ R, and If ω = 0, then L = limn→∞ n |an · x n according to Cauchy’s test, the series ∞ n=0 |an · x | is convergent. Therefore Mc = R, and R = ∞.

54

4 Series of Functions (Functional Series)

Remark 4.3.1 The convergence radius R of the power series compute by the formula:   an R = lim  n→∞ a

n+1

∞ n=0

an · x n can be

  , 

if this limit exists.

     n  a  Proof Indeed, if there exists limn→∞  aan+1 , then there exists limn→∞  an+1  (with the n 1 conventions 10 = ∞ and ∞ = 0). But it is known that if there exists limn→∞ √ then there exists limn→∞ n |an | = l.

|an+1 | |an |

= l,

Example 4.3.1 Find the radius of convergence R and the set of convergence Mc for the following power series: ∞ (−1)n−1 n n−1 2 3 · x = x − x2 + x3 − · · · + (−1)n · x n + · · · . 1. n=1 n Since an =

(−1)n−1 ,n n

≥ 1, by Remark 4.3.1 we deduce:

  an R = lim  n→∞ a

n+1

   = lim n = 1.  n→∞ n + 1

From Abel’s theorem (Theorem 4.3.1), it results that the series is absolutely convergent on the interval (−1, 1) and divergent on the set (−∞, −1) ∪ (1, ∞).  (−1)n−1 , which is convergent according to For x = R = 1, we get the series ∞ n=1 n Leibniz’s test for alternating series (Theorem 2.3.2). ∞ 1  −1 For x = −R = −1, we get the series ∞ n=1 n = − n=1 n which is divergent. Therefore the set of convergence of the series is Mc = (−1, 1]. ∞ n 2 n 2. n=0 n! · x = 1 + 1! · x + 2! · x + · · · + n! · x + · · · . Since an = n!, n ≥ 0 we have:    an  n! 1  = lim = 0. R = lim  = lim n→∞ an+1  n→∞ (n + 1)! n→∞ n + 1 Therefore Mc = {0}. ∞ x n x 3. n=0 n! = 1 + 1! + Since an =

1 ,n n!

x2 2!

+ ··· +

xn n!

+ ··· .

≥ 0, we have:

   an   = lim (n + 1)! = lim (n + 1) = +∞. R = lim  n→∞ an+1  n→∞ n→∞ n! Therefore Mc = R.

4.3 Power Series

55

∞

x 3n n=0 2n

Series

5.

∞ n=1

(−1)n−1 n

=1+

x3 2

+

x6 22

3n

+ · · · + x2n + · · · .   0 dac a n = 3k We notice that in this case an = , n ≥ 0.  2−n dac a n = 3k √  √  3n 2−n hence: The sequence n an is composed from the subsequences {0} and   √ √ ω = limn→∞ n |an | = max 0, √312 = √312 , whence it follows that R = ω1 = 3 2. From Abel’s theorem (Theorem it results that the series is absolutely  √ √4.3.1),  √ 3 3 convergent on the interval − 2, 2 , and divergent on the set (−∞, − 3 2) ∪ √ ( 3 2, ∞). √  ∞ 2n For x = R = 3 2, we obtain the series ∞ n=0 2n = n=0 1, which is divergent. √  3 ∞ n For x = −R  = − 2, we get the divergent series n=0 (−1) , hence Mc =  √ √ 3 3 − 2, 2 . 4.

· (x − 3)n = (x − 3) − 21 · (x − 3)2 + · · · + (−1)n

n−1

· (x − 3)n + · · ·

 (−1)n−1 · yn . If we use the notation x − 3 = y, then we obtain the series ∞ n=1 n ∞ (−1)n−1 n According to Example 4.3.1 (1), the series n=1 n · y is convergent on the interval (−1, 1].  (−1)n−1 ·(x − 3)n Since y ∈ (−1, 1] ⇔ x ∈ (2, 4], we deduce that the series ∞ n=1 n is convergent for x ∈ (2, 4], hence Mc = (2, 4]. Theorem 4.3.3 The sum of a power series is a continuous function on the convergence interval of the series. Proof Let x ∈ (−R, R) be an arbitrary fixed point. Obviously, there exists r ∈ R, such that |x| < r < R. According to Theorem 4.3.1, the power series converges uniformly on the interval [−r, r ]. Taking into account to Theorem 4.2.1, we deduce that the sum S of the series is a continuous function on the interval [−r, r ]. Particularly, the sum S is continuous in the point x. Theorem 4.3.4 Any power series can be differentiated term-by-term at any point x of its convergence interval (−R, R) and the series of derivatives has the same radius of convergence as the initial series. Proof The series of derivatives is: a1 + 2 · a2 · x + 3 · a3 · x 2 + · · · + n · an · x n−1 + · · · =

∞  n=1

n · an · x n−1 =

∞ 

(n + 1) · an+1 · x n

n=0

If we denote by R the radius of convergence of the series of derivatives, then we have:

56

4 Series of Functions (Functional Series)

R = lim

 n

n→∞

= lim

 n

n→∞

(n + 1) · |an+1 |

 (n + 1) · lim sup n |an+1 | = 1 · R = R. n→∞

Therefore, from Theorem 4.2.2 it results that: 

∞ 



an · x

=

n

n=0

∞ 

n · an · x n−1 , ∀x ∈ (−R, R).

n=1

Theorem 4.3.5 Any power series can be integrated term-by-term in its convergence interval (−R, R) and the radius of convergence of the series obtained by term wise integration is also R. Specifically, there holds:

x  ∞ 0

an · x

dx =

n

n=1

∞  an · x n+1 , x ∈ (−R, R) n + 1 n=0

∞ ∞ n n Proof Let S(x) = n=0 an · x , ∀x ∈ (−R, R). Since the series n=0 an · x converges uniformly on the interval [−r, r ], for any 0 < r < R, it results, from Theorem 4.2.3, that:

x S(t)dt = 0

∞  n=0

x an ·

∞  an t dt = · x n+1 , ∀x ∈ (−R, R). n + 1 n=0 n

0

On the other hand we have: √   lim n |an |    a n n→∞ = lim n  = lim n |an | = R, √ n n→∞  n + 1  n→∞ lim n + 1 n→∞

whence it results that the radius of convergence of the series obtained by integration is also R. Remark 4.3.2 A power series can be differentiated and integrated term-by-term on its convergence interval, whenever we want, and the radius of convergence of the new obtained series is equal with the radius of convergence of the initial series. Can prove the following theorem ([10], Theorem 2.4.6):  n Theorem 4.3.6 (Abel’s second theorem). Let ∞ n=0 an x be a power series with the radius of convergence R < ∞ and with the sum S. If this power series converges at the point x = R (respectively x = −R), then S(R) = lim x→R x−R S(x)).

4.3 Power Series

57

Example 4.3.2 (Logarithmic series). Find the set of convergence and the sum of the logarithmic series i.e.: ∞  (−1)n−1 n=1

· xn = x −

n

x2 x3 (−1)n−1 n + − ··· + · x + ··· . 2 3 n

According to Example 4.3.1 (1), the set of convergence is Mc = (−1, 1]. Let S(x) =

∞  (−1)n−1 n=1

+

(−1) n

n

· xn = x −

n−1

x2 x3 + − ··· 2 3 .

· x n + · · · , x ∈ (−1, 1]

From Theorem 4.3.4 it results: S (x) = 1 − x + x 2 − x 3 + · · · =

1 , ∀x ∈ (−1, 1) 1+x

and further:

S(x) =

1 dx = ln(1 + x) + C, ∀x ∈ (−1, 1). 1+x

Since S(0) = 0, it follows that C = 0, hence S(x) = ln(1 + x), ∀x ∈ (−1, 1). Because x = 1 ∈ Mc , from Theorem 4.3.6, we deduce that: S(1) = lim ln(x + 1) = ln 2. x→1 x a be an arbitrary fixed point.. We remark that g is continuous on the interval [a, x], differentiable on the open interval (a, x) and g(a) = g(x) = f (x). From Rolle’s Theorem it results that there exists ξ ∈ (a, x) such that: g (ξ ) = 0.

(4.7)

Further we have: g (t) = f (t) +

f (t) f (n+1) (t) f

(t) · (x − t) − + ··· + · (x − t)n 1! 1! n!

f (n) (t) · n · (x − t)n−1 − (n + 1) · (x − t)n · K (x) n! f (n+1) (t) = · (x − t)n − (n + 1) · (x−)n · K (x). n! −

According to (4.7) it results: K (x) =

f (n+1) (ξ ) f (n+1) (ξ ) = . (n + 1) · n ! (n + 1)!

(4.8)

From (4.5) and (4.8) we deduce the Lagrange’s remainder of Taylor’s formula: Rn (x) =

f (n+1) (ξ ) (x − a)n+1 , (n + 1)!

and so the proof of the theorem is finished.

(4.9)

4.4 Taylor’s Formula

61

Remark 4.4.1 If f is a polynomial function of order n, then the nth remainder Rn (x) = 0, ∀x ∈ I and Taylor’s formula becomes: f (a) f

(a) (x − a) + (x − a)2 1! 2! f (n) (a) + ··· + (x − a)n n! = c0 + c1 (x − a) + c2 (x − a)2 + · · · + cn (x − a)n .

f (x) = f (a) +

Thus, in the case of a polynomial function, the Taylor’s formula returns to its representation as a polynomial in its powers (x − a). Theorem 4.4.2 (Taylor’s formula with Cauchy’s remainder). Let f ∈ C n+1 (I ) and a ∈ I . Then, for any x ∈ I , there exists ξ between a and x such that: f (a) f

(a) · (x − a) + · (x − a)2 + · · · 1! 2! f (n) (a) f (n+1) (ξ ) + · (x − a)n + · (x − a) · (x − ξ )n . n! n!

f (x) = f (a) +

(4.10)

Proof The proof is similar to the proof of Theorem 4.4.1 except that we will search the remainder of Taylor’s formula (4.3) of the form: Rn (x) = (x − a) · K (x). Proceeding as in the proof of Theorem 4.4.1 we obtain the following expresion for Cauchy’s remainder: Rn (x) =

f (n+1) (ξ ) · (x − a) · (x − ξ )n . n!

(4.11)

Remark 4.4.2 Since ξ is between a and x, there exists 0 < θ < 1 such that ξ − a = θ · (x − a). If we denote by h = x − a, it results that x = h + a, ξ = a + θ · h and x − ξ = (1 − θ) · h. Thus the Taylor’s formula (4.3) is written: f (a) f

(a) 2 ·h+ ·h 1! 2! f (n) (a) n + ··· + · h + Rn (x) n!

f (a + h) = f (a) +

where the remainder Rn has one of the forms: Rn (x) = Rn (x) =

f (n+1) (a+θ·h) (n+1)! f (n+1) (a+θ·h) n!

· h n+1 (Lagrange’s form) · h n+1 · (1 − θ)n (Cauchy’s form).

62

4 Series of Functions (Functional Series)

If a = 0 ∈ I , then the Taylor’s formula is called Maclaurin’s formula. Therefore, the Maclaurin’s formula with the Lagrange’s remainder is: f (0) f (n) (0) n · x + ··· + ·x 1! n f (n+1) (θ · x) n+1 + ·x . (n + 1)!

f (x) = f (0) +

and the Maclaurin’s formula with the Cauchy’s remainder is: f (0) f (n) (0) n · x + ··· + ·x 1! n! f (n+1) (θ · x) n+1 + · (1 − θ )n . ·x n!

f (x) = f (0) +

Further we will write the Maclaurin’s formula, with the Lagrange’s remainder, for some basic functions: f (x) = ex , x ∈ R.

1.

Since f (n) (x) = ex and f (n) (0) = 1, ∀n ∈ N, we obtain: ex = 1 +

x x2 xn x n+1 + + ··· + + · eθ·x , 0 < θ < 1. 1! 2! n! (n + 1)!

f (x) = sin x, x ∈ R.

2.

We have:  n·π f (n) (x) = sin x + , ∀n ∈ N, 2 and f

(n)

n·π = (0) = sin 2



0 if n = 2 · k . n−1 2 if n = 4 · k + 1(or 4 · k + 3) (−1)

Therefore we obtain: x3 x 2n+1 x5 + + · · · + (−1)n · 3! 5! (2n + 1)! x 2n+2 · sin(θ · x + (n + 1) · π) + (2n + 2)!

sin x = x −

3.

f (x) = cos x, x ∈ R.

4.4 Taylor’s Formula

63

We have:  n·π f (n) (x) = cos x + and 2  n·π 0 if n = 2 · k + 1 = . f (n) (0) = cos k (−1) if n = 4 · k 2 Therefore we obtain: x2 x 2n x4 + + · · · + (−1)n · 2! 4! (2n)!  2n+1 x π + . · sin θ · x + (2n + 1) · 2 (2n + 1)!

cos x = 1 −

4.

f (x) = ln(1 + x), x ∈ (−1, ∞)

Since f (n) (x) = (−1)n−1 · we obtain:

(n−1)! , (1+x)n

f (0) = 0, and f (n) (0) = (−1)n−1 · (n − 1)!,

x2 x3 x4 + − 2 3 4 xn x n+1 1 . + (−1)n · · + · · · + (−1)n−1 · n n + 1 (1 + θ · x)n

ln(1 + x) = x −

Further we will present two important applications of Taylor’s formula for study of real functions. Let I ⊂ R be an open interval, a ∈ I and let f : I → R, be a function differentiable at x = a. It is well known that a necessary condition for the point x = a to be a local extremum point for f is that f (a) = 0. The next theorem states a sufficient condition for the existence of the local extremum points of a function. Theorem 4.4.3 Let I ⊂ R be an open interval, a ∈ I and f ∈ C n (I ), such that: f (a) = f

(a) = f

(a) = · · · = f (n−1) (a) = 0 and f (n) (a) = 0. If n is an even number, then x = a is a point of local extremum point for f , namely, x = a is a point of local maximum if f (n) (a) < 0 and a point of local minimum if f (n) (a) > 0. If n is an odd number, then x = a isn’t a point of local extremum for f. Proof From Taylor’s formula with Lagrange’s remainder it results: f (x) = f (a) +

f (n) (ξ ) (x − a)n , (∀)x ∈ I n!

64

4 Series of Functions (Functional Series)

where ξ is between a and x. Suppose that n is an even number and f (n) (a) < 0. Since f (n) is a continuous function at the point in x = a, it results that there exists an open interval J such that a ∈ J ⊂ I and f (n) (t) < 0, ∀t ∈ J . Therefore, if x ∈ J we have: f (x) − f (a) =

f (n) (ξ ) (x − a)n ≤ 0, n!

whence it results that f (x) − f (a) ≤ 0, ∀x ∈ J , hence x = a is a point of local maximum for f . Similarly, we prove that if f (n) (a) > 0, then x = a is a point of local minimum for f . If n is an odd number, then the expression f (x) − f (a) has no constant sign on any neighborhood of the point x = a, hence x = a does not be a point of local extremum for f . Corollary 4.4.1 If f (a) = 0 and f

(a) = 0, then x = a is a point of local extremum for f, namely, a point of local minimum if f

(a) > 0, respectively a point of local maximum if f

(a) < 0. If f (a) = f

(a) = 0 and f

(a) = 0, then x = a isn’t point of local extremum for f (it is a point of inflexion for f ). Definition 4.4.3 A function f ∈ C 2 (I ) is said to be convex (concave) on the interval I if for any a, x ∈ I we have: f (x) ≥ f (a) + f (a)(x − a), respectively: f (x) ≤ f (a) + f (a)(x − a). From a geometric point of view, the function is convex (respectively, concave) if its graph is located above (respectively, below) to the tangent of the graph at any its point.   Proposition 4.4.1 If f ∈ C 2 (I ) and f

(x) ≥ 0 f

(x) ≤ 0 for any x ∈ I , then f is convex (concave) on the interval I. Proof Let a, x ∈ I . From Taylor’s formula with Lagrange’s remainder and n = 1, it results: f (x) = f (a) +

f (a) f

(ξ ) (x − a) + (x − a)2 , 1! 2!

with ξ between a and x. If f

≥ 0 on I, it results:   f

(ξ ) f (x) − f (a) + f (a)(x − a) = (x − a)2 ≥ 0, ∀x ∈ I, 2!

4.5 Taylor’s and Maclaurin’s Series

65

therefore f is convex function on the interval I. Similarly, if f

≤ 0 on I it results: f (x) ≤ f (a) + f (a)(x − a), ∀x ∈ I, hence, f is concave function on the interval I.

4.5 Taylor’s and Maclaurin’s Series Further we will denote by C ∞ (I )—the class of all real valued functions indefinitely differentiable on the open interval I (i.e. f has derivatives of all orders on I). Definition 4.5.1 Let I be an open interval, a ∈ I and f ∈ C ∞ (I ). The power series: f (a) f

(a) · (x − a) + · (x − a)2 + · · · 1! 2! f (n) (a) + · (x − a)n + · · · n!

f (a) +

(4.12)

is called the Taylor’s series attached to the function f at the point x = a. In the particular case a = 0 ∈ I , the series (4.12) becomes: f (0) +

f (n) (0) n f

(0) 2 f (0) ·x+ · x + ··· + · x + ··· 1! 2! n!

(4.13)

and is called the Maclaurin’s series attached to the function f . If we denote by Mc the set of convergence of series (4.12), then we remark that a ∈ Mc , hence Mc ∩ I = φ (because a ∈ Mc ∩ I ). Definition 4.5.2 A function f ∈ C ∞ (I ) is said to be expanded into the Taylor’s series around point x = a on the set A ⊂ Mc ∩ I , if ∀x ∈ A we have: f (x) = f (a) +

f (a) f (n) (a) · (x − a) + · · · + · (x − a)n + · · · 1! n!

(4.14)

Remark 4.5.1 Let f ∈ C ∞ (I ) and a ∈ I . Generally is not true that the function f can be expanded into the Taylor’s series around point x = a on a set A ⊂ Mc ∩ I , as it results from the following Cauchy’s example.  1 e− x 2 if x = 0 Example 4.5.1 Let f : R → R be defined by f (x) = . 0 if x = 0 We will show that f is indefinitely differentiable on R and that for any n ∈ N∗ we have:

66

4 Series of Functions (Functional Series)

f (n) (0) = 0. Indeed, if x = 0, then f (x) = we have:

2 x3

f (0) = lim f (x) = lim x→0

x→0

= 3 · lim

x→0

− x12 − x23

·e

1 x2

· e− x 2 . Since f is continuous at the point x = 0, 1

2 x3

= lim

1

x→0

e x2 =

− x64 1

− x23 · e x 2

= 3 · lim

x→0

1 x 1

e x2

x 3 · lim 1 = 0. x→0 2 e x2

Generally we have, f (n) (x) = P(x) e− x 2 , x = 0, where P is a polynomial function. xm Applying L’Hospital rule a sufficient number of times, it results: 1

1 e− x 2 xm = P(0) · lim n ≥ 1. 1 = 0, x→0 x m x→0 x 2 e 1

f (n) (0) = lim f (n) (x) = P(0) · lim x→0

The Maclaurin’s series (4.13) attached to f , is convergent on R, and has the sum S(x) = f (0) = 0, ∀x ∈ R. It results that f (x) = S(x), ∀x ∈ R\{0}, hence, f is not expanded into Maclaurin’s series on any set A ⊂ R, A = {0}. Theorem 4.5.1 The necessary and sufficient condition for a function f ∈ C ∞ (I ) to be expanded into the Taylor’s series around the point x = a ∈ I on the set A ∈ Mc ∩ I is limn→∞ Rn (x) = 0, ∀x ∈ A. Proof We remark that the Taylor’s polynomial Tn , of order n, attached to function f at the point x = a, coincides with the partial sum Sn of the Taylor’s series (4.12). If we denote by S the sum of the series (4.12), then: S(x) = lim Tn (x), ∀x ∈ Mc . n→∞

On the other hand, from Taylor’s formula (4.3) we have: f (x) = Tn (x) + Rn (x), ∀x ∈ I. The fact that f can be expanded into Taylor’s series on the set A ⊂ Mc ∩ I , arround the point x = a, means that f (x) = S(x), ∀x ∈ A. Obviously this happens if and only if limn→∞ Rn (x) = 0, ∀ ∈ A. Corollary 4.5.1 A sufficient condition for a function f ∈ C ∞ (I ) to be expanded into the Taylor’s series around the point x = a ∈ I , on the set A ⊂ Mc ∩ I is there exists M > 0 such that:  (n)   f (x) ≤ M, ∀n ∈ N∗ , ∀x ∈ A ⊂ Mc ∩ I.

4.5 Taylor’s and Maclaurin’s Series

67

Proof For any fixed point x ∈ A, from the formula (4.9) we have: |Rn (x)| =

|x − a|n+1  (n+1)  |x − a|n+1 · f = un , (ξ ) ≤ M · (n + 1)! (n + 1)!

where ξ is between a and x. We observe that limn→∞ u n = 0, since the number series with positive terms ∞ |x−a|n+1 n=1 M · (n+1)! is convergent, as it results immediate from Theorem 2.2.6 because: |x − a| u n+1 = lim = 0 < 1. n→∞ u n n→∞ n + 2 lim

Therefore, limn→∞ Rn (x) = 0, ∀x ∈ A and the assertion follows now from Theorem 4.5.1. Example 4.5.2 Maclaurin’s expansion for some elementary functions. 1.

f (x) = ex , x ∈ R. For any r > 0, we have:  (n)   f (x) = ex ≤ er = M, ∀x ∈ [−r, r ], ∀n ∈ N.

From Corollary 4.5.1 it results that the function f (x) = ex can be expanded into Taylor’s series on the interval [−r, r ] around any point a ∈ [−r, r ]. As r > 0 was arbitrary, we deduce that the expansion is valid on R. Therefore we have: ∞

 xn x x2 xn e =1+ + + ··· + + ··· = , ∀x ∈ R. 1! 2! n! n! n=0 x

2.

f (x) = sin x, x ∈ R. We have:   (n)     f (x) = sin x + n · π  ≤ 1, ∀x ∈ R, ∀n ∈ N∗ . 2

From Corollary 4.5.1 it results that the function f (x) = sin x can be expanded into Taylor’s series on R. The Maclaurin’s series for the function f (x) = sin x, x ∈ R is: x3 x 2n+1 x5 + + · · · + (−1)n · + ··· 3! 5! (2n + 1)! ∞  x 2n+1 = , ∀x ∈ R. (−1)n · (2n + 1)! n=0

sin x = x −

68

4 Series of Functions (Functional Series)

3.

f (x) = cos x, x ∈ R, can be expanded into Taylor’s series on R around any point a ∈ R, because:   (n)     f (x) = cos x + nπ  ≤ 1, ∀x ∈ R, ∀n ∈ N∗ . 2 The Maclaurin’s series for the function f (x) = cos x, x ∈ R is: cos x = 1 − =

∞ 

x 2n x4 x2 + + · · · + (−1)n · + ··· 2! 4! (2n)! (−1)n ·

n=0

4.

x 2n , ∀x ∈ R. (2n)!

The function f (x) = (1 + x)α , α ∈ R\N, x ∈ (−1, ∞), can be expanded into Maclaurin’s series on the interval (−1, 1). According to Theorem 4.5.1, it is sufficient to prove that: lim Rn (x) = 0, ∀x ∈ (−1, 1).

n→∞

Since f (n) (x) = α · (α − 1) · . . . · (α − n + 1) · (1 + x)α−n , n ∈ N∗ , it results from (4.11) that the Cauchy’s remainder is: Rn (x) =

α · (α − 1) · . . . · (α − n) · x n+1 · n!



1−θ 1 + θx

n

· (1 + θ x)α−1 ,

depends of n.  where θ ∈ (0, 1) and, generally,    If we denote by u n =  α·(α−1)·...·(α−n) · x n+1  and suppose that |x| < 1, then the n! ∞ series n=1 u n is convergent as it results immediate from Theorem 2.2.6. Indeed we have: lim

n→∞

|α − n − 1| u n+1 |x| = |x| < 1. = lim n→∞ un n+1

From Proposition 2.1.1, it results that limn→∞ u n = 0. On the other hand, if |x| < 1 we have 0 < 1 − θ < 1 + θ · x, whence we deduce that: 0
1 . |1 + θ x|α−1 ≤ (1 − |x|)α−1 if α < 1

(4.15)

From (4.15) we obtain:  |Rn (x)| = u n ·

1−θ 1 + θx

n

· (1 + θ x)α−1 < u n · (1 + θ x)α−1

(4.16)

Taking into account to (4.16) and to the fact that limn→∞ u n = 0, we deduce now that limn→∞ Rn (x) = 0, ∀x ∈ (−1, 1), and so the proof is finished. Therefore we have the expansion: α α · (α − 1) · . . . · (α − n + 1) n α · (α − 1) 2 ·x+ · x + ··· + · x + ··· 1! 2! n! ∞  α · (α − 1) · . . . · (α − n + 1) n =1+ · x , x ∈ (−1, 1) n! n=1

(1 + x)α = 1 +

also known as the binomial series. In the particular case α = − 21 we have: 1 · 3 · 5 · . . . · (2n − 1) n 1 1·3 2 1 · x + · · · + (−1)n · x + ..., =1− ·x + √ 2 2 · 4 2 · 4 · 6 · . . . · (2n) 1+x ∀x ∈ (−1, 1). At the end of this section we recapitulate the Maclaurin’s expansions of the basic functions:  2 n xn 1. ex = 1 + 1x! + x2 ! + · · · + xn! + · · · = ∞ n=0 n! , x ∈ R. x3 x 2n+1 x5 sin x = x − + + · · · + (−1)n · + ··· 3! 5! (2n + 1)! 2. ∞ x 2n+1 = , x ∈ R. (−1)n · n=0 (2n + 1)! x2 x 2n x4 cos x = 1 − + + · · · + (−1)n · + ··· 2! 4! (2n)! 3. ∞ x 2n , x ∈ R. = (−1)n · n=0 (2n)! α α · (α − 1) 2 ·x (1 + x)α = 1 + · x + 1! 2! α · (α − 1) · . . . · (α − n + 1) n 4. + ··· + · x + ··· n! ∞ α · (α − 1) · . . . · (α − n + 1) · x n , x ∈ (−1, 1). =1+ n=1 n!

70

5.

6.

4 Series of Functions (Functional Series)

(binomial series) x2 xn x3 ln(1 + x) = x − + + · · · + (−1)n−1 · + ··· 2 3 n n ∞ x = (−1)n−1 · , x ∈ (−1, 1]. n=1 n (logarithmic series) (i.e. Example 4.3.2).  1 n = 1 + x + x2 + x3 + · · · + xn + · · · = ∞ n=0 x , x ∈ (−1, 1). 1−x (geometric series)

4.6 Elementary Functions. Euler’s Formulas. Hyperbolic Trigonometric Functions Euler’s formulas play an important role in mathematics, especially in the theory of complex functions and the differential equations. We recall (see Chap. 2, Sect. 2.7) that the exponential function z → ez : C → C is definite by: z z2 zn + + ··· + + · · · , z ∈ C. 1! 2! n! √ For z = i · x, x ∈ R, i = −1 we get: ez = 1 +

ei·x = 1 + i

x x2 x3 x4 − −i + + ··· . 1! 2! 3! 4!

Similarly, for z = −i · x, x ∈ R we get: e−i·x = 1 − i

x x2 x3 x4 − +i + − ··· . 1! 2! 3! 4!

Summing and subtracting this above relations it results: i·x

e

−i·x

+e

ei·x − e−i·x

  x2 x4 =2· 1− + − · · · = 2 · cos x 2! 4!   x3 x5 =2·i · x − + − · · · = 2 · i · sin x. 3! 5!

From the last relations we deduce the Euler’s formulas: ei·x + e−i·x 2 ei·x − e−i·x sin x = . 2·i

cos x =

(4.17)

4.6 Elementary Functions. Euler’s Formulas. Hyperbolic …

71

An equivalent form of formulas (4.17), are the following formulas: ei·x = cos x + i · sin x e−i·x = cos x − i · sin x In particular, from Euler’s formulas follow this two somewhat surprising results: ei·k·π = e−i·k·π = −1 and ei·2·k·π = e−i·2·k·π = 1, ∀k ∈ Z In the following we will define two other important elementary functions, namely: z 2·n z2 z4 + − · · · + (−1)n + · · · , z ∈ C, 2! 4! (2 · n)!

(4.18)

z 2·n−1 z3 z5 + − · · · + (−1)n−1 + · · · , z ∈ C. 3! 5! (2 · n − 1)!

(4.19)

cos z = 1 − sin z = z −

The definitions make sense because the series on the right side is absolutely convergent as it immediately follows from Cauchy’s root test. Taking into account the Maclaurin’s expansions of the basic real functions presented in the previous paragraph we notice that the complex variable functions (4.18), (4.19) are generalizations of the real variable functions: x → cos x : R → R, x → sin x : R → R. From Euler’s formulas and the property of the exponential function ez+u = ez · eu it results: sin(z ± u) = sin z · cos u ± sin u · cos z. cos(z ± u) = cos z · cos u ∓ sin z · sin u. Indeed, for example: ei·z − e−i·z ei·u + e−i·u ei·u − e−i·u ei·z + e−i·z · + · ·i 2  2·i 2 2 i·(z+u) −i(z+u) 2· e −e = = sin(z + u). 4·i

sin z · cos u + sin u · cos z =

By analogy with formulas (4.17), are definite the hyperbolic trigonometric functions: cosh : R → (0, ∞), coshx =

ex + e−x 2

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4 Series of Functions (Functional Series)

sinh : R → R, sinhx =

ex − e−x . 2

(4.20)

The main property of this functions is: ch2 x − sh2 x = 1, ∀x ∈ R

(4.21)

The hyperbolic trigonometric functions are used for parametric equations of the hyperbola. Indeed, the canonical equation of the hyperbola is: x2 y2 − = 1, a, b > 0. a2 b2 We observe that the parametric equations of the branch of the hyperbola situated into the first quadrant are: 

x = a · cht , t ≥ 0, y = b · sht

because:  x 2 a

−

 y 2 b

= ch2 t − sh2 t = 1.

The following properties of the hyperbolic trigonometric functions are easily verified: cosh(0) = 1, cosh(−x) = cosh(x), (cosh x) = sinh x, ∀x ∈ R sinh(0) = 0, sin h(−x) = − sin h(x), (sinh x) = cosh x, ∀x ∈ R. The Maclaurin’s expansions of this functions are: ∞

cosh x = 1 +

 x 2·n x2 x4 x 2·n + + ··· + + ··· = , x ∈R 2! 4! (2 · n)! (2 · n)! n=0

sinh x = x +

 x 2·n+1 x5 x 2·n+1 x3 + + ··· + + ··· = , x ∈ R. 3! 5! (2 · n + 1)! (2 · n + 1)! n=0

∞

Chapter 5

Functions of Several Variables

5.1 Vector Space Rn . Basic Notions and Notations The n-dimensional space Rn isn−Rn defined as follows: Rn = R × R × · · · × R = {x = (x1 , . . . , xn ); x1 , . . . , xn ∈ R} . It is easy to verify that Rn is a real vector space with respect to the operations: x + y = (x1 + y1 , . . . , xn + yn ), ∀ x = (x1 , . . . , xn ), ∀y = (y1 , . . . , yn ) ∈ Rn λ · x = (λ · x1 , . . . , λ · xn ), ∀ x = (x1 , . . . , xn ) ∈ Rn , ∀λ ∈ R. Any ordered system x = (x1 , x2 , . . . , xn ) is called vector (point) of Rn , and x1 , . . . , xn are called the coordinates (components) of the vector x. The vector (point) 0Rn = (0, 0, . . . , 0) is called the zero vector of Rn and x + x + 0Rn = x, ∀x ∈ Rn . Definition 5.1.1 The scalar (inner) product of two vectors x, y ∈ Rn is defined as: x, y =

n 

xi · yi = x1 · y1 + · · · + xn · yn ,

i=1

∀ x = (x1 , . . . , xn ) ∈ Rn , ∀y = (y1 , ..., yn ) ∈ Rn . The scalar product obviously possesses the following properties: (i)

x, y = y, x, ∀ x , y ∈ Rn ;

© The Author(s), under exclusive license to Springer Nature Singapore Pte Ltd. 2022 G. Paltineanu et al., Differential Calculus for Engineers, https://doi.org/10.1007/978-981-19-2553-5_5

73

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5 Functions of Several Variables

(ii) (iii)

λ · x + μ · y, z = λ · x, z + μ · y, z, ∀ x, y, z ∈ Rn , ∀λ, μ ∈ R; x, x ≥ 0, ∀ x ∈ Rn and x, x = 0 iff x = 0Rn .

Theorem 5.1.1 (Inequality Cauchy–Buniakovski–Schwarz). For any x, y ∈ Rn we have:  n 2  n   n     2 2 2 x, y = xi · yi ≤ xi · yi . i=1 i=1 i=1 = x, x · y, y Proof From the property (iii) of the scalar product it results: λ · x + y, λ · x + y ≥ 0 , ∀ x , y ∈ Rn and ∀ λ ∈ R. Using also the other properties of the scalar product we obtain: λ2 · x, x + 2 · λ · x , y + y, y ≥ 0 , ∀ λ ∈ R. For this inequality to be true for any λ ∈ R, it must that: x, y2 − x , x · y, y ≤ 0, hence x, y2 ≤ x , x · y, y i.e.:  n 

2 xi · yi

≤

 n 

i =1

i =1

xi2

  n   2 yi . · i =1

Definition 5.1.2 Any map  ·  : Rn → R, with the properties: (i) (ii) (iii)

 λ · x  = | λ | ·  x  , ∀ x ∈ Rn , ∀ λ ∈ R; x + y ≤  x  +  y , ∀ x , y ∈ Rn ;  x  ≥ 0 , ∀ x ∈ Rn and  x  = 0 if x = 0Rn ,

is called a norm on Rn . Several norms can be introduced on the space Rn . Two of these are most commonly used, namely  x ∞ = max{|xi |; 1 ≤ i ≤ n} , ∀ x = (x1 , ..., xn ) ∈ Rn

(5.1)

and  x 2 =



  n  x , x =  xi2 = x12 + ... + xn2 , i =1

(5.2)

∀ x = (x1 , ..., xn ) ∈ Rn (The norm defined by formula (5.2) is also called the Euclidean norm). The fact that formula (5.1) defines a norm results immediately from the properties of the module. Further we will prove that formula (5.2) defines a norm on Rn . Since

5.2 Convergent Sequences of Vectors in …

75

the properties (i) and (ii) are obvious, it remains to be proved the property (iii). For this we use the Cauchy–Buniakovski–Schwarz’s inequality: x + y22 = x, x + 2 · x, y + y, y ≤ x, x + 2 · |x, y| + y, y 2

≤ x22 + 2 · x2 · y2 + y22 = x2 + y2 . Therefore: x + y2 ≤ x2 + y2 , ∀ x , y ∈ Rn . Proposition 5.1.1 The two norms are equivalent because:  x ∞ ≤  x  2 ≤

√

n ·  x  ∞ , ∀ x ∈ Rn .

Proof Let x = (x1 , . . . , xn ) ∈ Rn be an arbitrary fixed vector. Obviously there is i 0 ∈ {1 , 2 , ... , n} such that  x ∞ = max{|xi |; 1 ≤ i ≤ n} = xi0 . Further we have: √ 2 x∞ = xi0 ≤ x2 = x12 + · · · + xi20 + · · · + xn2 ≤ n · xi0 = n · x∞

5.2 Convergent Sequences of Vectors in Rn Definition 5.2.1 The Rn sequence of vectors {xk } from Rn is said to be convergent in Rn if there exists a vector x ∈ Rn such that lim xk − x∞ = 0. We will denote k →∞

this by: Rn

xk −→ x. k →∞

Rn

Therefore xk −→ x, if ∀ ε > 0, ∃ kε ∈ N∗ such that: k →∞

xk − x∞ < ε, ∀k ≥ kε . From Proposition 5.1.1 it results that: Remark 5.2.1 The sequence of vectors {xk } converges to the vector x in Rn if and only if lim xk − x2 = 0. k →∞

Theorem 5.2.1 The sequence of vectors {xk } from Rn ,where for any k ∈ N∗ , xk = (xk 1 , xk 2 , ... , xk n ), xk i ∈ R , ∀ i = 1 , n, converges in Rn to the vector

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5 Functions of Several Variables

x = (x1 , ... , xn ) ∈ Rn , if and only if, ∀ i = 1 , n the sequence of real numbers {xk i } converges in R to xi . Proof The assertion follows immediately from (5.1) and Definition 5.2.1. From Theorem 5.2.1 we deduce that the convergence of a sequence of vectors from Rn is reduced to the study of the convergence of the real number sequences formed by their components. Example 5.2.1 (1) (2)

 2  R2 The sequence of vectors n n+2 1 , √1n −→ (1, 0).

  R3



 n n n2 The sequence 3n−1 , 5n 2 −4n+1 , 1 + n1 −→ 13 , 15 , e

Remark 5.2.2 From Theorem 5.2.1 it results also: (1) (2)

Any convergent sequence of vectors from Rn has a unique limit. Any subsequence of a convergent sequence of vectors from Rn is also convergent and has the same limit with the initial sequence.

5.3 Topology Elements on Rn Definition 5.3.1 Let a = (a1 , ... , an ) ∈ Rn be o fixed point and r > 0.The set of all x ∈ Rn such that x − a < r (respectivelyx − a ≤ r ) is calledar the n-dimensional open (respectively, closed) ball with center at a and radius r . Further we will denote the n-dimensional open ball by:   B(a, r ) = x ∈ Rn ; x − a < r , respectively, the n-dimensional closed ball by:   B (a, r ) = x ∈ Rn ; x − a ≤ r



As we introduced on the space Rn two norms ( · 2 and  · ∞ ), it results that we have two types of open (respectively, closed) balls, namely   B2 (a, r ) = x = (x1 , . . . , xn ) ∈ Rn ; x − a2 < r   = x = (x1 , . . . , xn ) ∈ Rn ; (x1 − a1 )2 + · · · + (xn − an )2 < r 2 respectively:   B∞ (a, r ) = x = (x 1 , . . . , xn ) ∈ Rn ; x − a∞ < r   = x = (x1 , . . . , xn ) ∈ Rn ; |x1 − a1 | < r, . . . , |xn − an | < r .

5.3 Topology Elements on ...

77

In a similar way we have:   Bˆ 2 [a, r ] = x = (x1 , . . . , xn ) ∈ Rn ; (x1 − a1 )2 + · · · + (xn − an )2 ≤ r 2   Bˆ ∞ [a, r ] = x = (x1 , . . . , xn ) ∈ Rn ; |x1 − a1 | ≤ r, . . . , |xn − an | ≤ r . Example 5.3.1 (1)

Since on R we have x2 = x∞ = |x| , ∀ x ∈ R, it results that for any a ∈ R and r > 0, we have: B2 (a, r ) = B∞ (a, r ) = {x ∈ R; |x − a| < r } = (a − r, a + r ) 



B 2 (a, r ) = B ∞ (a, r ) = {x ∈ R; |x − a| ≤ r } = [a − r, a + r ]

From a geometric point of view, the open (respectively, closed) ball with center at a and radius r represents the symmetric open (closed) interval (a − r , a + r ). (respectively, [a − r, a + r ]). (2)

Let a = (a1 , a2 ) ∈ R2 be a fixed point and r > 0. Then we have:   B2 (a, r ) = x = (x1 , x2 ) ∈ R2 ; (x1 − a1 )2 + (x2 − a2 )2 < r 2 and   B∞ (a, r ) = x = (x1 , x2 ) ∈ R2 ; |x1 − a1 | < r , |x2 − a2 | < r .

From geometric point of view, B2 (a, r ) represents the interior of the circle with center at a and radius r (without circumference) (Fig. 5.1), and B∞ (a, r ) is the interior of the square with center at a and side 2 · r (without sides) (Fig. 5.2). (3)

On the space R3 let a = (a1 , a2 , a3 ) ∈ R3 and r > 0.Then we have: B2 (a, r ) = {x = (x1 , x2 , x3 ) ∈ R3 ; (x1 − a1 )2 + (x2 − a2 )2 + (x3 − a3 )2 < r 2 } and B∞ (a, r ) = {x = (x 1 , x2 , x3 ) ∈ R3 ; |x1 − a1 | < r, |x2 − a2 | < r, . |x3 − a3 | < r }

Geometrically speaking, B2 (a, r ) represents the interior (inside) of the sphere with the center at a = (a1 , a2 , a3 ) and the radius r and B∞ (a, r ) represents the interior (inside) of the cube with the center at a = (a1 , a2 , a3 ) and the length of the edge 2 · r .

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5 Functions of Several Variables

Fig. 5.1 B2 (a, r ) origin and radious R = B2 (a, r )

Fig. 5.2 B∞ (a, r )

Generally, on Rn we will call B2 (a, r ) the n-dimensional open sphere and B∞ (a, r ) the n-dimensional open cube. Remark 5.3.1 The two types of balls of Rn satisfy the following inclusions:   r ⊂ B2 (a, r ) ⊂ B∞ (a, r ) B∞ a, √ n

5.3 Topology Elements on ...

79

 Indeed, if x = (x1 , ... , xn ) ∈ B∞ a, results (x1 − a1 ) + ... + (xn − an ) On the other hand, if x = (x1 , ..., 2

2

√r n



, then |xi − ai | < 2

< n · rn = r 2 , hence xn ) ∈ B2 (a, r ), then:

√r , ∀i, n

whence it

x ∈ B2 (a, r ).

(x1 − a1 )2 + ... + (xn − an )2 < r 2 .  Taking into account that |xi − ai | ≤ (x1 − a1 )2 + ... + (xn − an )2 < r ,∀ i = 1 , n, we deduce that x ∈ B∞ (a, r ). In the particular case n = 2 we obtain the inclusions:   r ⊂ B2 (a, r ) ⊂ B∞ (a, r ), B∞ a, √ 2 which are represented geometrically in Fig. 5.3. Definition 5.3.2 Any subset V ⊂ Rn with the property that there exists r > 0 such that V ⊃ B(a, r ) is called a neighborhood of the point a ∈ Rn . The family of all neighborhoods of the point a ∈ Rn will be denoted by V(a). According to Definition 5.3.2, on R, any subset V ⊂ R which contains an open interval (a − r , a + r ), r > 0, is a neighborhood of the point a ∈ R. In particular the interval (a − r , a + r ) itself is a neighborhood of the point a ∈ R. It would seem that on Rn (n ≥ 2) there are two types of neighborhoods for a point a ∈ Rn , namely neighborhoods containing open balls of the type B2 (a, r ), respectively, neighborhoods containing open balls of type B∞ (a, r ). From Remark 5.3.1, it results that the two types of neighborhoods coincide. Fig. 5.3 B∞ (a, √r ) ⊂ 2 B2 (a, r ) ⊂ B∞ (a, r )

80

5 Functions of Several Variables

Therefore, further, by the neighborhood of the point a ∈ Rn we mean any subset of Rn which contains either an open n-dimensional sphere or an open n-dimensional cube. Proposition 5.3.1 The family V(a) of all neighborhoods of the point a ∈ Rn has the following properties: (1) (2) (3) (4) (5)

a ∈ V for any V ∈ V(a). If V ∈ V(a) and U ⊃ V , then U ∈ V(a). If a = b ∈ Rn , then there are V ∈ V(a) and W ∈ V(b) such that V ∩ W = ∅. If Vi ∈ V(a) , i ∈ {1, 2, ..., m}, then im= 1 Vi ∈ V(a). For any V ∈ V(a), there exists W ∈ V(a), such that V ∈ V(b), for any b ∈ W .

Proof The properties If a = b, then a − b = r > 0. It  (2) are obvious. 

(1) and is immediate that B a , r3 ∩ B b , r2 = ∅, hence (3) is true. Let Vi ∈ V(a) and 0 such that Vi ⊃ B(a, ri ), i = 1, m. If we denote ri >   by: r = min{ri : 1 ≤ i ≤ m}, then im= 1 Vi ⊃ B(a, r ), whence it results that im= 1 Vi ∈ V(a). Finally,

let V ∈ V(a) and r > 0 such that V ⊃ B(a, r ). If we denote by W = B a, r2 , then for any b ∈ W and any x ∈ B b, r2 we have: x − a ≤ x − b + b − a
0 is an open set.

5.3 Topology Elements on ...

81

Let (α , β) ⊂ R be an arbitrary open interval. If we denote by a = α +2 β and by r = β −2 α , then (α , β) = (a − r , a + r ). It results that any open interval from R is an open set. (2) (3)

On R2 , the interior of any circle (square) is an open set. On R3 , the interior of any sphere (cube) is an open set.

Proposition 5.3.3 The open sets have the following properties: (1) (2) (3)

Any union of open sets is also an open set. Any finite intersection of open sets is also an open set. The sets Rn and φ(empty set) are open sets.

Proof (1)

Let {Di }i

∈I

be an arbitrary family of open sets and let D =

 i ∈I

Di

If a ∈ D, then there exists i 0 ∈ I such that a ∈ Di0 . As Di 0 is an open set, it results that there exists a neighborhood V ∈ V(a) such that V ⊂ Di0 . Obviously V ⊂ D, whence it results that a is an interior point of D, hence D is an open set.  (2) Let D1 , ..., Dm be a family of open sets and let D = im= 1 Di . If a ∈ D, then a ∈ Di , ∀ i ∈ {1, 2, ..., m}. Since Di is an open set, it follows that there  exists a neighborhood Vi ∈ V(a) such that Vi ⊂ Di . If we denote by V = im= 1 Vi , then V ∈ V(a) and V ⊂ D. It results that a is an interior point of D, hence D is an open set. The property (3) is obvious. Proposition 5.3.2 allows us to give more varied examples of open sets. For example, on R, any union and any finite intersection of open intervals there is an open set. On R2 , various unions and finite intersections of the interiors of circles or squares are examples of open sets and so on. Definition 5.3.4 A point a ∈ Rn is called an adherent point of the set A ⊂ Rn if for any neighborhood V ∈ V(a) we have V ∩ A = ∅. The set of all adherent points of the set A is denoted by A and is called the closure of the set A. Obviously A ⊂ A. The reverse inclusion, generally, is not true. The set A ⊂ Rn is called closed if A ⊂ A , i.e. A = A. Remark 5.3.2 Any point a ∈ A is an adherent point of the set A; there may be adherent points of A which belong not to A. Theorem 5.3.1 The necessary and sufficient condition for a set A ⊂ Rn to be a closed set is that its complement set  A = Rn \ A to be an open set. Proof. Necessity: / A = A, thus Suppose that A ⊂ Rn is a closed set, i.e. A = A. V ⊂ A then b ∈ b is not an adherent point of A. It results that there exists a neighborhood V ∈ V(b) such that V ∩ A = ∅, hence V ⊂ A. Therefore b is an interior point of  A so  A is an open set.

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5 Functions of Several Variables

Sufficiency: Suppose that  A is an open set. I f b ∈  A, , then there exists a neighborhood V ∈ V(b) such that V ⊂ A . Then we have V ∩ A = ∅, whence it results that b is not an adherent point of A, hence b ∈ A . Therefore we proved that  A ⊂ A whence we deduce A ⊂ A, hence A = A. ˆ Proposition 5.3.4 The closed ball B[a, r ] is a closed set. Proof From Theorem 5.3.1 it follows that it is sufficient to show that the set ˆ  B[a, r ] = {x ∈ Rn ; x − a > r } is an open set. ˆ Let b ∈  B[a, r ] and let 0 < ε < b − a − r . If x ∈ B(b, ε), then x − b < ε. Further we have b − a ≤ b − x + x − a < b − a − r + x − a, ˆ r ].Therefore, B(b, ε) ⊂ whence it results that x − a > r , hence x ∈  B[a, ˆ ˆ n B[a, r ], so b is an interior point of  B[a, r ]. As b was arbitrary, it results that ˆ  B[a, r ]. is an open set. Example 5.3.3 (1)

On R, for any a ∈ R and r > 0, the closed ball. ˆ B[a, r ] = [a − r , a + r ].

From Proposition 5.3.4 we deduce that any symmetric closed interval is a closed set. Since any closed interval [α , β] ⊂ R can be represented as a symmetric closed interval, it follows that any closed interval of R is a closed set. (2)

On R2 , for any a = (a1 , a2 ) ∈ R2 and r > 0 we have:   Bˆ 2 [a, r ] = x = (x1 , x2 ) ∈ R2 ; (x1 − a1 )2 + (x2 − a2 )2 ≤ r 2 , which geometrically represents the closed disk with center at a and radius r (consisting of the interior of the circle and its circumference, Fig. 5.4).and   Bˆ ∞ [a, r ] = x = (x1 , x2 ) ∈ R2 ; |x1 − a1 | ≤ r , |x2 − a2 | ≤ r , which geometrically represents the closed square with center at a and side 2 · r. (containing besides the interior of the square and its sides, Fig. 5.5). Both the sets B 2 (a, r ) and B ∞ (a, r ) are closed sets. 

(3)



On R3 , examples of closed sets are Bˆ 2 [a, r ] which represents the solid sphere with center at a and radius r (including the all interior points of the sphere and all points that constitute the sphere) and Bˆ ∞ [a, r ] represents the solid cube with center at a and side 2 · r (including the interior of the cube and its faces).

Proposition 5.3.3 The family of closed sets has the following properties: (1)

Any finite union of closed sets is also a closed set.

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83

Fig. 5.4 Bˆ 2 (a, r )−the closed ball centered in origin and radius R

Fig. 5.5 Bˆ ∞ (a, r )−the closed square with center at a and sides 2r.2r = Bˆ ∞ (a, r )

(2) (3)

Any intersection of closed sets is also a closed set. The set Rn and the empty set ∅ are closed sets.

Proof The assertions follow from Theorem 5.3.1, Proposition 5.3.2, and De Morgan relations. For example: m (1) Let F1 , . . . , Fm be a finite number of closed sets and let F = i = 1 Fi . From Theorem 5.3.1, it results that it is sufficient to prove that the complement set  F isan open set. On the other hand, from De Morgan relation we have m F = i = 1  Fi . As  Fi is an open set for any i ∈ {1, 2, . . . , m}, from Proposition 5.3.2 it results that  F = im= 1  Fi is an open set.

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5 Functions of Several Variables

Definition 5.3.5 is called the boundary of the set A ⊂ Rn and is denoted by Fr A the following set: Fr A = A ∩  A Example 5.3.4 (1) On R, for any closed interval [α , β], we have Fr([α, β]) = {α; β}. (2) On R2 , for any closed ball Bˆ 2 [a, r ] we have:     Fr Bˆ 2 [a, r ] = (x1 , x2 ) ∈ R2 ; x12 + x22 = r 2 , which represents the circumference of the circle with center at a and radius r , and   Fr Bˆ ∞ [a, r ]   (x1 , a2 − r ) ∪ (x1 , a2 + r ) ∪ (a1 − r, x2 ) ∪ (a1 + r, x2 ); = |x1 − a1 | ≤ r, |x2 − a2 | ≤ r which represents the union of the sides of the square: 

 x = (x1 , x2 ) ∈ R2 ; |x1 − a1 | ≤ r , |x2 − a2 | ≤ r .

3 ˆ of the  solid sphere B2 [a, r ] is the sphere  (3) On R , 3the 2 boundary 2 2 2 (x1 , x2 , x3 ) ∈ R ; x1 + x2 + x3 = r and the boundary of the solid cube Bˆ 2 [a, r ] consists in the union of its faces.

Lemma 5.3.1 For any subset A ⊂ Rn we have: ◦ 

◦

 A =  A and A = A . ◦

◦

Proof If b ∈  A , then b ∈ / A and so for any neighborhood V ∈ V(b) we have V ∩  A = ∅ , whence it results that b ∈  A . Reciprocal, if b ∈  A , then for any neighborhood V ∈ V(b) we have avem V ∩  A = ∅ , whence it results that ◦

◦

b∈ / A, hence  A ⊂  A . The other equality is similarly proved. Proposition 5.3.4 Let A be an arbitrary subset of Rn . Then: ◦

(i) (ii)

A, the interior of A, is an open set. A, the closure of A, is a closed set.

(iii)

Fr A, the boundary of A, is a closed set and Fr A = A \ A.

◦

Proof (i)

◦

If a ∈ A, then there exists r > 0 such that V = B(a, r ) ⊂ A. According to ◦ ◦ Remark 5.3.2, V is an open set. Therefore we have V = V ⊂ A, whence it ◦ ◦ results that a is an interior point of A, hence A is an open set.

5.3 Topology Elements on ...

85 ◦ 

(ii)

From Lemma 5.3.1 it results that  A = A . On the other hand, according

(iii)

to (i) A is an open set, hence  A is an open set. From Theorem 5.3.1, it follows that A is a closed set. From Lemma 5.3.1 it results that:

◦ 

◦

◦

Fr A = A ∩  A = A ∩  A = A \ A As, according to (ii), A and  A are closed sets, we deduce from Proposition 5.3.3 that Fr A is a closed set. Theorem 5.3.2 Let A ⊂ Rn be an arbitrary subset. Then: (i)

A point b ∈ Rn is an adherent point for A iff there exists a convergent sequence

(ii)

{ak }, ak ∈ A, ∀k ∈ N∗ , such that ak −→ b. The subset A is closed iff the limit of any convergent sequence of vectors of A belongs to A.

Rn

Proof If b ∈ A, then A ∩ B(b, r ) = ∅, ∀ r > 0. In particular, for any k ∈ N∗ there exists ak ∈ A ∩ B a, k1 . Therefore ak ∈ A and ak − a < k1 , ∀ k ∈ N∗ ,

(i)

Rn

whence it results that ak −→ b. Rn

Reciprocal, if ak ∈ A, ∀ k ∈ N∗ and s¸i ak −→ b, then ∀ r > 0, ∃ kr ∈ N∗ such that ak − b < r , ∀ k ≥ kr , whence it results that ak ∈ A ∩ B(b, r ) , ∀k ≥ kr , hence b ∈ A. (ii)

Let A be a closed set and let {ak } be a convergent sequence of vectors of A, Rn

such that ak −→ b. From (i) it results that b ∈ A = A. Reciprocal, if b ∈ A, then from (i) it results that there exists a convergent sequence Rn

of vectors {ak }, ak ∈ A, ∀k ∈ N∗ , such that cu ak −→ b. By hypothesis it follows that b ∈ A, whence we deduce that A is closed. Remark 5.3.3 From Theorem 5.3.2, (ii), it results that a set is closed if it contains all of its limit points, in other words if it is “closed” with respect to the limit crossing operation. Definition 5.3.6 A point b ∈ Rn is called an accumulation point of the set A ⊂ Rn if for any neighborhood V ∈ V(b) there exists a ∈ V ∩ A such that b = a. The set of all accumulations points of A is called the derived set of A and is denoted by A . Obviously A ⊂ A. Remark 5.3.4 An accumulation point of A may or may not belong to A.

86

5 Functions of Several Variables

Theorem 5.3.3 Let A ⊂ Rn be an arbitrary subset. Then: A point b ∈ Rn is an accumulation point of A iff there exists a convergent sequence of vectors {ak }, ak ∈ A, ∀k ∈ N∗ ,ak = am if k = m, such that

(i)

Rn

(ii)

ak −→ b. A is closed iff A ⊂ A.

Proof If b ∈ A , then there exists a1 ∈ A ∩ B(b, 1), a1 = b. 

Let r1 = = a1 − b < 1 and a2 ∈ A ∩ B b, r21 . Obviously a2 = a1 and

 a2 − b < 21 . Let r2 = = a2 − b and a3 ∈ A ∩ B b, r22 , a3 = b. Obviously a3 = a2 , a3 = a1 s¸i a3 − b < r22 < 212 . By complete mathematical induction it can be shown that there exists a sequence {ak }, ak ∈ A, ∀k ∈ N∗ ,ak = am if k = m, (i)

Rn

such that ak − b < 2k1− 1 , whence it results that ak −→ b. The reverse assertion is obvious. (ii)

If A is closed, then A ⊂ A. As A ⊂ A it follows that A ⊂ A.

Reciprocal, suppose that A ⊂ A and let b ∈ A. From Theorem 5.3.2, it results Rn

that there exists a sequence {ak }, ak ∈ A, ∀k ∈ N∗ such that ak −→ b. If the sequence {ak } has an infinity number of terms different two by two, it results according to (i) that b ∈ A . Since A ⊂ A, it follows that b ∈ A. If {ak } does not have an infinity number of terms different two by two, then ak = b for an infinity of indices k, hence b ∈ A. Therefore A ⊂ A, and so A is closed. Remark 5.3.5 From Theorem 5.3.3 it follows that, if b is an accumulation point of A, then every neighborhood V ∈ V(b) contains an infinity number of different elements of A. It results that the finite sets do not have accumulation points, so, according to Theorem 5.3.3, any finite set is a closed set. The sets which do not have accumulation points are called discrete sets. There are infinite discrete sets. For example, the set Z of all integers is a discrete subset of R. Definition 5.3.7 A subset A ⊂ Rn is called bounded if there exists M > 0 such that x ≤ M, ∀x ∈ A. Lemma 5.3.2 (Cesàro). Any bounded sequence of vectors from Rn contains a convergent subsequence. Proof We present the proof for the particular case n = 2. Let {z k } = {(xk , yk )}, k ∈ N∗ , be a bounded sequence of vectorsR2 of R2 . Then there exists M > 0 such that z k ∞ ≤ M , ∀ k ∈ N∗ . Since |xk |, |yk | ≤ z k ∞ ≤ M, ∀k ∈ N∗ , it results that the real number sequences {xk } and {yk } is bounded. From Cesàro’s lemma for real number   sequences (Lemma 1.2.1), it follows that there exists a convergent subsequence xk p . Let x = lim xk p . Applying again Lemma 1.2.1 for p→∞

   sequence yk p it results that there exists a subsequence yk pl which converges in R

5.4 Limits of Functions of Several Variables

87

   

to y. From Theorem 5.2.1 we deduce that the sum sequence z k pl = xk pl , yk pl is convergent in R2 and has the limit z = (x, y). For the theory of limits of functions of several variables, it is important to know when a subset A ⊂ Rn has accumulation points. The following theorem gives us a sufficient condition for a subset of Rn to have accumulation points. Theorem 5.3.4 (Weierstrass–Bolzano). Every bounded and infinite set of Rn has at least one accumulation point. Proof Let A ⊂ Rn be a bounded and infinite subset. Since A is infinite, it results that / A, xk = xl if k = l. Taking into account that A is there exists a sequence {xk },xk ∈ From Lemma 5.3.2, it follows bounded, it results that the sequence {xk } is bounded.  that there exists a convergent subsequence xk p . If we denote by b = lim xk p , then, p→∞

according to Theorem 5.3.2, b is an accumulation point of A. Definition 5.3.8 A subset K ⊂ Rn is called a compact setRn if it is bounded and closed. From Proposition 5.3.3, it follows that a finite union of compact sets is also a compact set and any intersection of compact set is also a compact set. It is also obvious that any finite set is a compact set (Remark 5.3.7).

5.4 Limits of Functions of Several Variables Definition 5.4.1 By vector function we mean any function F defined on a set A ⊂ Rn with values in Rm . So, for any x = (x1 , ..., xn ) ∈ A ⊂ Rn , the image y = F(x) ∈ Rm it is a vector of form F(x) = (y1 , ..., ym ) ∈ Rm , If we denote by f i (x) = yi , ∀ x = (x1 , ..., xn ) ∈ A, ∀ i = 1 , m , then we obtain m scalar functions f i : A ⊂ Rn → R, i = 1 , m, which we call the scalar components of the vector function F. Therefore, we have: F = ( f 1 , ..., f m ) : A ⊂ Rn → Rm , F(x) = ( f 1 (x), ..., f m (x)) , ∀ x ∈ A. Example 5.4.1 (1)

The function r : [0, 2 π] ⊂ R → R2 defined by:

r (t) = (a · cos t, a · sin t) , t ∈ [0, 2 π ], a > 0, is a vector function. The image of this function is the circle with center at origin and radius a. (2)

The function r : [0, 2 π ] × [0, π ] ⊂ R2 → R3 , r (u, v) = (a · sin u · cos v, a · sin u · sin v, a · cos u), u ∈ [0, π ], v ∈ [0, 2π ]a > 0

88

5 Functions of Several Variables

is a vector function, whose image represents the sphere with center at origin and radius a. Definition 5.4.2 Let F : A ⊂ Rn → Rm be a vector function, a = (a1 , ... , an ) ∈ Rn an accumulation point of A and L ∈ Rm . It is said that L is the limit of function F at the point a, and we will note this by lim F(x) = L, if for any neighborhood x →a

U ⊂ Rm of L, there exists a neighborhood V ⊂ Rn of a, such that F(x) ∈ U ,∀x ∈ V ∩ A , x = a. Theorem 5.4.1 Let F, a and L be as in Definition 5.4.2. Then the following statements are equivalent: lim F(x) = L;

(1)

x →a

(2)

For any ε > 0, ther e is δε > 0 such that for any x ∈ A , x = a with the property x − a < δε , it results that F(x) − L < ε (here the norm  ·  is any of the norms  · ∞ sau  · 2 ).

(3)

For any sequence of vectors {xk }, xk ∈ A xk = a , ∀ k ∈ N∗ , xk −→ a, it

Rn

Rm

follows that F(xk ) −→ L. Proof (1)⇒(2). If ε > 0 and U = B(L , ε), then obviously U ∈ V(L) and according to (1), there exists a neighborhood V ∈ V(a)(which generally depends on ε) such that F(x) ∈ U, ∀ x ∈ V ∩ A, x = a. Since V ∈ V(a), there exists δε > 0 with the property V ⊃ B(a, δε ). If x ∈ A, x = a, and x − a < δε , it results that x ∈ B(a, δε ), hence x ∈ V ∩ A, and x = a. According to (1) we have F(x) ∈ U , so F(x) − L < ε (2)⇒(3) Let ε > 0 and δε > 0 with the properties from (2). If {xk } is a convergent Rn

sequence of vectors from A, xk = a , ∀ k ∈ N∗ and xk −→ a, then there exists kε ∈ N∗ such that xk − a < δε , for any k ≥ kε . From (2) we deduce that Rm

F(xk ) − L < ε, ∀ k ≥ kε , hence F(xk ) −→ L. (3)⇒(1) If we suppose that (1) is not true, then there is a neighborhood U0 ∈ V(L) such that for any neighborhood V ∈ V(a), there exists. x_V ∈ V ∩ A, x V = a, with the property F(x V ) ∈ / U0 .

1 In particular, for Vk = B a, k , ∀ k ∈ N∗ , it results that there exists xk ∈ A ∩

1 B a, k , xk = a, such that F(xk ) ∈ / U0 , ∀ k ∈ N∗ .

1 Rn As xk ∈ B a, k , ∀ k ∈ N∗ , it follows that xk − a < k1 , hence that xk −→ a. Rm

From (3) it results that F(xk ) −→ L and this contradicts the fact that F(xk ) ∈ / U0 , ∀ k ∈ N∗ . Remark 5.4.1 (1)

lim F(x) = L can also be written in the form:

x →a

5.4 Limits of Functions of Several Variables

89

lim F(x1 , . . . , xn ) = L .

x1 →a1 ... xn →an

In the second notation it is highlighted the fact that the components xi of the vector x = (x1 , x2 , . . . , xn ) converge simultaneously and independently to the components ai of the vector a = (a1 , . . . , an ). (2)

For to prove that the function F : A ⊂ Rn → Rm does not have limit at the point a ∈ A , it is sufficient to find two sequences {xk } and {yk } of vectors from Rn

Rn

A, such that xk −→ a, yk −→ a, and the sequences {F(xk )} and {F(yk )} have different limits. Remark 5.4.2 If f : A ⊂ R → R, a ∈ R is an accumulation point of A and l ∈ R, then, from Theorem 5.4.1, it results that lim f (x) = l if ∀ ε > 0, ∃ δε > 0 x →a such that for any x ∈ A , x = a, with the property |x − a| < δε it follows that | f (x) − l| < ε (here we use the fact that on R, x2 = x∞ = |x|). We thus found the definition of the limit of a function of one variable. Theorem 5.4.2 Let F = ( f 1 , ..., f m ) : A ⊂ Rn → Rm be a vector function, a ∈ Rn be an accumulation point of A and L = (l1 , ..., lm ) ∈ Rm . Then, the necessary and sufficient condition for lim F(x) = L is that lim f i (x) = li , for x →a

any i = 1 , m.

x →a

Rn

Proof Let {xk } be a sequence of vectors from A, xk = a , ∀ k ∈ N∗ , xk −→ a. From Theorem 5.4.1, it follows that lim F(x) = L if and only if: x →a

Rm

F(xk ) = ( f 1 (xk ), . . . , f m (xk )) −→ L = (l1 , . . . , lm ) On the other hand, according to Theorem 5.2.1, this is equivalent with the fact R

that ∀ i = 1 , m f i (xk ) −→ li . Applying again Theorem 5.4.1, it results that lim f i (x) = li . x →a

Remark 5.4.3 From Theorem 5.4.2 it follows that the study of the limit of a vector function returns to the study of the limits of its scalar components. For this reason, it is sufficient to further study only limits of scalar functions. Remark 5.4.4 Let f : A ⊂ Rn → R be a scalar function, a = (a1 , . . . , an ) an accumulation point of A and l ∈ R. If we use the norm  · ∞ , then from Theorem 5.4.1, it results that the following statements are equivalent with lim f (x) = l: x →a

(1) (2)

For any neighborhood U ∈ V(l), U ⊂ R, there is a neighborhood V ∈ V(a), V ⊂ Rn such that f (x) ∈ U , ∀ x ∈ V ∩ A , x = a; ∀ ε > 0, ∃ δε > 0 such that ∀ x = (x1 , ... , xn ) ∈ A , x = a, with the property |x1 − a1 | < δε , ... , |xn − an | < δε , it results | f (x1 , ..., xn ) − l | < ε;

90

(3)

5 Functions of Several Variables Rn

For any sequence {xk } of vectors from A, xk = a , ∀ k ∈ N∗ , xk −→ a, we R

have f (xk ) −→ l. Let us now consider the case even simpler when n = 2. Let f : A ⊂ R2 → R be a scalar function of two variables and l ∈ R. We will use the notations (x , y) instead (x1 , x2 ) and (a , b) instead (a1 , a2 ), with (a , b) ∈ A . It follows from the above that the following statements are equivalent: f (x, y) = lim f (x, y) = l;

(2)

→ (1)

(3)

→ (2) ∀ ε > 0, ∃ δε > 0 such that for any (x , y) ∈ A with the properties |x − a| < δε , |y − b| < δε , it follows | f (x , y) − l | < ε. For any sequence {(xn , yn )} of elements of A,

(4)

lim

(x,y)→(a,b)

x→a y→b

R2

(xn , yn ) = ( a , b) ∀ n ∈ N∗ , (xn , yn ) −→(a, b), it results that: R

f (xn , yn ) −→ l. Example 5.4.2 Compute

lim

(x,y)→(0,0)

f (x, y) = lim f (x, y), where: x→0 y→0

 sin x 3 + y 3 f (x, y) = , (x, y) ∈ R2 \{(0, 0)} x 2 + y2 Since |sin x| ≤ |x| , ∀ x ∈ R, it results:

sin x 3 + y 3  | f (x, y)| = ≤ x 2 + y2

3 x + y 3 x 2 + y2

≤

3 3 x + y x 2 + y2

x 2 |x| + y 2 |y| x 2 + y2 x2 y2 |y| = |x| 2 + ≤ |x| + |y|. x + y2 x 2 + y2 =

R2

For any sequence {(xn , yn )}, xn = 0 , yn = 0, (xn , yn ) −→(0, 0), we have R

R

|xn | + |yn | −→ 0 and so f (xn , yn ) −→ 0. Therefore lim f (x, y) = 0. x→0 y→0

Example 5.4.3 Let the scalar function: f (x, y) = To show that does not exist

x 2 − y2 , (x, y) ∈ R2 \{(0, 0)} x 2 + y2 lim

(x,y)→(0,0)

f (x, y) = lim f (x, y) x→0 y→0

5.4 Limits of Functions of Several Variables

91

 

 

Consider the sequences n1 , − n1 , respectively, n2 , − n1 from R2 .  R2  R2



Obviously n1 , − n1 −→(0, 0) and n2 , − n1 −→(0, 0).On the other hand we  R  R



have f n1 , − n1 −→ 0 and f n2 , − n1 −→ 35 , whence it results that does not exists 2 2 lim x 2 −y 2 . x→0 x +y y→0

Theorem 5.4.3 (Cauchy–Bolzano). Let f : A ⊂ Rn → R be a scalar function and let a = (a1 , ... , an ) ∈ Rn be an accumulation point of A. The necessary and sufficient condition to exist lim f (x) = l ∈ R is: x →a

  for any ε > 0 there exists a neighborhood 

  Vε ∈ V(a) such that ∀x , x ∈ A ∩ Vε ,    < ε. x = a , x = a we have f x − f x

Proof. Necessity. If l = lim f (x) ∈ R, then ∀ ε > 0 there exists a neighborhood Vε ∈ V(a) such x →a

that | f (x) − l | < 2ε for any x ∈ Vε ∩ A , x = a. For any x  , x  ∈ Vε ∩ A, x  = a, x  = a, we have: 

 

 f x − f x  ≤ f x  − l + l − f x  < ε + ε < ε 2 2 Sufficiency: Let ε > 0 and Vε ∈ V(a) be as in the statement of the Theorem, and Rn

let {xk } be a sequence of vectors from A, xk = a, ∀k ∈ N∗ ,xk −→ a. Then there exists a rank kε ∈ N∗ (this rank depends turn depends on ε) such on Vε , which

 in that xk ∈ Vε , ∀ k ≥ kε . It follows that f (xk ) − f x p < ε, for any k, p ≥ kε , whence we deduce that the sequence { f (xk )} is a fundamental sequence of real numbers. From Cauchy’s criterion for sequences of real numbers (Theorem  1.2.1), it results that the sequence { f (xk )} converges in R to a number l ∈ R. If x k is an Rn

  ∗ sequence of vectors from A, xk = a, ∀k ∈ N , xk −→ a, then the sequence another  f (xk ) is also convergent. Let us note with l its limit. On the other hand it is obvious that the sequence:



xk , xk

 k

  Rn = x1 , x1 , . . . , xk , xk , . . . −→ a and xk = a, xk = a.

the above we deduce that the sequence  From   { f (xk )} ), . . . , f (x ), f (x ), . . . is convergent. As the sequences f (x1 ), f (x k k 1 and f xk are subsequences of this convergent sequence, it follows that they have the same limit so that l = l  . According to Theorem 5.4.1, (3), it follows that there exists lim f (x) = l. x →a

For a function f : A ⊂ Rn → R we can consider, in addition to the previously defined limit, in which the variables x1 , x2 , ... , xn tend independently, but simultaneously to their limiting values a1 , a2 , ... , an , and iterated limits, in which the variables x1 , x2 , ... , xn tend to their limiting values a1 , a2 , ... , an in succession. By way of illustration we consider a scalar function of two variables.

92

5 Functions of Several Variables

  Let f : D ⊂ R2 → R, where D = (x, y) ∈ R2 ; |x − a| < h, |y − b| < k is a rectangle. Suppose that for any x ∈ (a − h, a + h) there exists lim f (x, y). Obviy →b

ously this limit depends on x. We will denote by ϕ(x) = lim f (x, y), ∀x ∈ y →b

(a − h, a + h). If we assume in addition that there exists lim ϕ(x), then this limit is denoted by x →a

lim lim f (x, y) and is called the (x, y)− iterated (repeated) limit of the func-

x →a y →b

tion f at point (a, b). Similarly is defined (y, x)− iterated (repeated) limit of the function f at point (a, b) which is denoted by lim lim f (x, y). y →b x →a

The iterated limits are generally not equal, according to the following examples: Example 5.4.4 For the function: f (x, y) =

x 2 − y2 , (x, y) ∈ R2 \{(0, 0)} x 2 + y2

we have lim lim f (x, y) = 1= lim lim f (x, y) = −1 x →0 y →0

y →0 x →0

We remark that in this case the lim f (x, y) does not exists (see Example 5.4.3). x→0 y→0

Example 5.4.5 Consider the function: f (x, y) = x sin

1 , (x, y) ∈ R × R∗ y

Since x sin 1y ≤ |x|, it follows that lim f (x, y) = 0. x→0 y→0

We notice that lim lim f (x, y) = 0, while the limit y →0 x →0

lim lim f (x, y)

x →0 y →0

does not exist. The connection between the iterated limits and the limit with respect to both variables, i.e. lim f (x, y) is highlighted by the following proposition. x→∞ y→b

Proposition 5.4.1 Let f : A ⊂ R2 → R be a scalar function and let (a , b) be an accumulation point of A. If we suppose that there exists lim f (x, y) = l and for any x→a y→b

x ∈ (a − h, a + h) there exists also ϕ(x) = lim f (x, y), then there exists the y →b

(x, y)− iterated limit lim lim f (x, y) and lim lim f (x, y) = lim ϕ(x) = x →a y →b

l.

x →a y →b

x →a

5.5 Continuous Functions of Several Variables

93

Proof Since lim f (x, y) = l, it results that for any ε > 0 there is δε > 0 such that x→a y→b

∀ (x , y) ∈ A, with |x − a| < δε , |y − b| < δε we have: | f (x , y) − l | < ε. When y → b we obtain |ϕ(x) − l | ≤ ε, for any x ∈ (a − h, a + h), with the property |x − a| < δε . Therefore there exists lim ϕ(x) = l, so there exists x →a

lim lim f (x, y) = l.

x →a y →b

Corollary 5.4.1 If the iterated limits exist and are different then lim f (x, y) does x→a y→b

not exists. Corollary 5.4.2 If there exists only one of the three limits lim f (x, y), x→a y→b

lim lim f (x, y) or lim lim f (x, y), then it does not results that there are the

x →a y →b

y →b x →a

other two.

5.5 Continuous Functions of Several Variables For any vector function F : Rn → Rm and any subset A ⊂ Rn , we will denote by F(A) = {F(x) ; x ∈ A}. Obviously F(A) ⊂ Rm and is called the direct image of A under F. For any subset B ⊂ Rm we denote by F −1 (B) = { x ∈ Rn ; F(x) ∈ B}. The subset F −1 (B) is called the inverse image of B under F. Definition 5.5.1 Let F : A ⊂ Rn → Rm be a vector function and a ∈ A. On said that the function F is continuous at the point a if for any neighborhood U of F(a), there exists a neighborhood V of a, such that F(V ∩ A) ⊂ U . The function F is continuous on A if it is continuous at every point of A. Remark 5.5.1 If a ∈ A is an accumulation point of A, then F is continuous at the point a iff there exists lim F(x) and lim F(x) = F(a). If a ∈ A does not an x →a

x →a

accumulation point of A (i.e.a is an isolated point), then there is a neighborhood V ∈ V(a) such that V ∩ A = {a} and is obvious that F(V ∩ A) ⊂ U , ∀ U ∈ V(F(a)), whence it results that F is continuous at a. Therefore, any function is continuous at every isolated point of its domain. Theorem 5.5.1 Let F : A ⊂ Rn → Rm be a vector function and a ∈ A.The following statements are equivalent: (1)

F is continuous at a;

94

(2) (3)

5 Functions of Several Variables

For any ε > 0, there is δε > 0 such that ∀ x ∈ A with x − a < δε ,it results F(x) − F(a) < ε; Rn For any convergent sequence {x k } of vectors from A, xk −→ a, it results Rm

F(xk ) −→ F(a). Proof The proof results from Theorem 5.4.1 and Observation 5.5.1, nothing that if a is an isolated point, then any of the statements (1)–(3) is obvious. Theorem 5.5.2 The vector function F = ( f 1 , ..., f m ) : A ⊂ Rn → Rm is continuous at the point a ∈ A iff each of its scalar component f i : A ⊂ Rn → R is continuous at a, for any i ∈ {1, 2, ..., m}. Proof The proof follows from Theorem 5.4.2 and Remark 5.5.1. From above theorem we deduce that it is sufficient to study the continuity of scalar functions. Let f : A ⊂ Rn → R be a scalar function and a = (a1 , ... , an ) ∈ A. If we use the norm  · ∞ , then the function f is continuous at the point a if ∀ ε > 0, ∃ δε > 0 such that for any x = (x1 , ..., xn ) ∈ A with property |x1 − a1 | < δε , ... , |xn − an | < δε it results: | f (x1 , ..., xn ) − f (a1 , ..., an ) | < ε Further we consider the case of functions of two variables and we use the notations (x , y) instead of (x 1 , x2 ) and (a , b) instead of (a1 , a2 ). Remark 5.5.2 If f : A ⊂ R2 → R and (a , b) ∈ A, then the following statements are equivalent: (1) (2) (3)

f is continuous at (a , b); ∀ ε > 0, ∃ δε > 0 such that for any (x , y) ∈ A with the property |x − a| < δε , |y − b| < δε , it results | f (x , y) − f (a , b) | < ε; For any sequence {(xn , yn )}, (xn , yn ) ∈ A, ∀n ∈ N∗ , R2

(xn , yn ) −→(a, b) it results: R

f (xn , yn ) −→ f (a, b) Example 5.5.1 (1)

The function f : R2 → R, f (x, y) = x 2 − 3 x · y + y + 1 is continuous R2

on R2 , because ∀ (a, b) ∈ R2 and ∀ (xn , yn ) −→(a, b), we have:

5.5 Continuous Functions of Several Variables

95

R

f (xn , yn ) −→ f (a, b) (2)



The function f : R → R , f (x, y) = 2

sin(x 3 + y 3 ) x 2 + y2

0

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

is continuous on R2 . Indeed, if (a, b) = (0, 0), then the statement follows from Remark 5.5.2, (3). The continuity at the origin follows since, as we proved in Example 5.4.2, 

sin x 3 + y 3 lim f (x, y) = lim = 0 = f (0, 0). x→0 x→0 x 2 + y2 y→0 y→0  (3)

The function f : R → R, f (x, y) = 2

x 2 −y 2 x 2 +y 2

0



dac a (x, y) = (0, 0) 

dac a (x, y) = (0, 0)

is not continuous at the origin, since, as we proved in Example 5.4.3, this function does not have limit at origin. Definition 5.5.2 Let f : A ⊂ Rn → R be a scalar function,a ∈ A where a = (a1 , ... , an ) and Ai = {t ∈ R ; (a1 , ..., ai − 1 , t, ai + 1 , ..., an ) ∈ A} , i = 1, n. The function f is said to be continuous at the point x = a with respect to the variable xi , if the function of one variable ϕi : Ai → R defined by: ϕi (t) = f (a1 , ..., ai − 1 , t, ai + 1 , ..., an ) is continuous at the point t = ai . Let f : A ⊂ R2 → R, be a function of two variables and (a, b) ∈ A. Then f is continuous at the point (a, b) with respect to the variable x (respectively, y) if the function of one variable ϕ(x) = f (x, b), ∀ x ∈ R, is continuous at the point x = a (respectively, if the function ψ(y) = f (a, y), ∀ y ∈ R, is continuous at the point y = b). Remark 5.5.3 If f : A ⊂ Rn → R is continuous at the point a ∈ A with respect to all variables (Definitia 5.5.1), then f is continuous at the point x = a with respect to each variable xi , i = 1 , n. The reverse statement, generally, is not true. Example 5.5.2 Let:



x2 y x 4 + y2

i f (x, y) = (0, 0) . 0 i f (x, y) = (0, 0) We remark that this function is not continuous at the origin (0, 0)(with respect to the both variables), because lim f (x, y) does not exist. f : R → R , f (x, y) = 2

x→0 y→0

Indeed, the following sequences 

origin (0, 0), but lim f n1 , n12 = n→∞

 1 1 2

 

 , n12 and n2 , n12 converge in R2 to the

2 1  4 = lim f n , n 2 = 17 .

n

n→∞

96

5 Functions of Several Variables

On the other hand, we have ϕ(x) = f (x, 0) = 0 , ∀ x ∈ R, whence it results that the function ϕ : R → R is continuous at x = 0, hence f is continuous at the point (0, 0) with respect to the variable x. Similarly it turns out that f is continuous at the point (0, 0) with respect to the variable y. Theorem 5.5.3 Assume that the function F : A ⊂ Rn → B ⊂ Rm is continuous at the point a ∈ A and the function G : B ⊂ Rm → R p is continuous at the point b = F(a) ∈ B. Then the composite function H = G ◦ F : A ⊂ Rn → R p is continuous at the point a. Proof The proof results from Theorem 5.5.1, (3). Indeed, let {xk } be a sequence Rn

of elements of A, with xk −→ a.Then, taking into account of the continuity of the Rm

Rp

functions F and G, it results that F(xk ) −→ F(a) = b and G(F(xk )) −→ G(F(a)). Rp Therefore H (xk ) −→ H (a), hence H is continuous at the a. Theorem 5.5.4 If f : A ⊂ Rn → R is continuous at the point a ∈ A and f (a) > 0 (respectively, f (a) < 0), then there exists a neighborhood V ∈ V(a) such that f (x) > 0 (respectively, f (x) < 0), for any x ∈ V ∩ A. Proof Suppose f (a) > 0 and denote by ε = 21 · f (a) > 0. Since f is continuous at a, it results that there is δε > 0 such that for any x ∈ A, with x − a < δε we have | f (x) − f (a) | < ε. If we will denote by V = B(a, δε ), then V ∈ V(a) and for any x ∈ V ∩ A we have: 1 · f (a) = f (a) − ε < f (x) < f (a) + ε = 23 · f (a), 2 hence f (x) > 0. Theorem 5.5.5 Assume that the functions, f, g : A ⊂ Rn → R, are continuous at the point a ∈ A.Then, for any α , β ∈ R the functions α · f + β · g and f · g are continuous at a. If g(a) = 0, then the function gf is also continuous at a. Rn

Proof Let {xk } be a sequence of elements of A, xk −→ a. By hypothesis it results R

R

that f (xk ) −→ f (a) and g(xk ) −→ g(a). Taking into account by the properties of the real number sequences, it results: (α · f + β · g)(xk ) = α · f (xk ) + β · g(xk ) R

−→ α · f (a) + β · g(a) = (α · f + β · g)(a).

R

( f · g)(x k ) = f (xk ) · g(xk ) −→ f (a) · g(a) = ( f · g)(a) whence it follows that c˘a α · f + β · g and f · g are continuous at a.

5.6 Properties of Continuous Functions Defined …

97

If g(a) = 0, then, according to Theorem 5.5.4, it results that there exists a neighborhood V ∈ V(a) such that g(x) = 0, ∀ x ∈ V ∩ A. Possibly removing a finite number of terms of the sequence {xk }, and denoting its terms, we can assume ∗ that xk∈ V ∩ A , ∀ k ∈ N∗ , hencethat  g(xk ) = 0 , ∀ k ∈ N . Then we have: f g

(xk ) =

f (xk ) g(xk )

R

−→

f (a) g(a)

=

f g

whence it results that the function

(a),

f g

is continuous at the point a.

Theorem 5.5.6 The following statements are equivalent: (i) (ii) (iii)

F : Rn → Rm is continuous on Rn ; F −1 (D) is an open set for any open set D ⊂ Rm ; F −1 (B) is a closed set for any closed set B ⊂ Rm .

Proof (i)⇒(ii) Let D ⊂ Rm be an open set and let a ∈ F −1 (D) be an arbitrary point. Then b = F(a) ∈ D and because D is open there exists a neighborhood U ∈ V(b) such that U ⊂ D. As F is continuous at a, it results that there exists a neighborhood V ∈ V(a) with the property F(V ) ⊂ U ⊂ D, so V ⊂ F −1 (U ) ⊂ F −1 (D). Therefore a is an interior point of F −1 (D), hence F −1 (D) is an open set. (ii)⇒(i) Let a ∈ Rn be arbitrary and let b = F(a). If U ∈ V(b), then there exists r > 0 such that U ⊃ B(b, r ). As B(b, r ) is open (Proposition 5.3.2), from (ii) it results that V = F −1 (B(b, r )) is also open. Obviously V ∈ V(a) and F(V ) ⊂ U , hence F is continuous at the point a. from Theorem 5.3.1 and from the immediate (ii)⇔(iii) This equivalence

follows  remark  F −1 (B) = F −1  B , for any B ⊂ Rm .

5.6 Properties of Continuous Functions Defined on Compact or Connected Sets We recall that a subset of Rn is compact if it is closed and bounded. Theorem 5.6.1 Let F : Rn → Rm be a continuous vector function. If K ⊂ Rn is a compact set, then its direct image F(K ) is also a compact set. Proof We will prove that the set F(K ) = {F(x) ; x ∈ K } is closed and bounded. First we shall prove that the set F(K ) is bounded. If us suppose that F(K ) is unbounded, then for any M > 0 there exists x M ∈ K such that F(x M ) > M. In particular, for M = p , p ∈ N∗ there is x p ∈ K such that:   F x p  > p

(5.1)

98

5 Functions of Several Variables

  Since the subset K is bounded, it follows  the sequence x p is bounded;  that hence it contains a convergent subsequence x pl (Lemma 5.3.2). Let a = lim x pl . l →∞

As K is closed, it results that a ∈ K , according to Theorem 5.3.2. Finally, taking into account

 from the continuity of F we deduce, from Theorem 5.5.1, that lim F x pl = F(a).On the other hand, from (5.1) we have l →∞      F x p  > pl , whence it results that lim  F x p  = ∞. We have thus come to l

l →∞

l

a contradiction, because F(a) it is a finite number. We will prove now that F(K ) is a closed set. Let b ∈ F(K  ) be an arbitrary point. From Theorem 5.3.2, it results that there exists a sequence y p of elements of F(K ) Rm

such that y p −→ b.

 Let x p ∈ K be with the property that y p = F x p . As we proved in the first part

 Rn of the proof, there exists a convergent subsequence x pi , x pl −→ a and a ∈ K .

 Rm As F is a continuous function, it follows that y pl = F x pl −→ F(a) On the other Rm

hand, y pl −→ b, whence we deduce that b = F(a) ∈ F(K ). Therefore F(K ) ⊂ F(K ), hence F(K ) is a closed set. Definition 5.6.1 Let f : A ⊂ Rn → R be a bounded scalar function. We will say that f attains its supremum and its infimum on A, if there are x M ∈ A and xm ∈ A such that: f (x M ) = M = sup f (A) and f (xm ) = m = inf f (A). Theorem 5.6.2 Let f : K ⊂ Rn → R be a continuous function. If K is a compact subset then f is bounded on K and attains its supremum and infimum on K . Proof The fact that f is bounded on K follows from Theorem 5.6.1. If we denote by M = sup f (K ) and by m = inf f (K ), then M , m ∈ f (K ). Indeed, if V is a neighborhood of M, then there exists ε > 0 such that: V ⊃ (M − ε , M + ε). From the definition of the supremum it results that there exists xε ∈ K such that M − ε < f (xε ). (Remark 1.1.2). Therefore, V ∩ f (K )  = ∅. As V was an arbitrary neighborhood of M, it results that M is an adherent point for f (K ), i.e.M ∈ f (K ). Analogously it is shown that m ∈ f (K ). On the other hand, according to Theorem 5.6.1, f (K ) is closed, hence M , m ∈ f (K ). Therefore there are x M ∈ K and xm ∈ K such that M = f (x M ) and m = f (xm ). Definition 5.6.2 A function F : A ⊂ Rn → Rm is called uniformly continuous on   the set A if  for any ε > 0 thereis δε >   0 such

that  ∀ x , x ∈ A with the property x  − x   < δε it results that  F x  − F x   < ε. We recall that the function F is continuous on the set A if it is continuous at each point x of A, hence for any x ∈ A  and any  ε > 0 there exists δx , ε > 0 such that for any x  ∈ A with the property x  − x  < δx , ε it results that:

5.6 Properties of Continuous Functions Defined …

99

    F x − F(x) < ε. Therefore, the difference between uniformly continuity and continuity of the function F on the set A is that in the case of uniform continuity, for there is  ε > 0, 

any  F x  − F x   < ε that a number δε > 0, the same for all points x ∈ A, such  for any x  , x  ∈ A satisfies the condition x  − x   < δε . Obviously any uniformly continuous function on the set A is a continuous function on A. The reverse statement, generally, is not true. Example 5.6.1 The function f : R → R , given by f (x) = x 2 is continuous R, but is not uniformly continuous on R. For to prove that f is not uniformly continuous on R, we will have to show that there exists ε0 > 0 such that ∀ δ > 0, ∃xδ , xδ ∈ R with the properties:  

 x − x  < δ and f x  − f x  ≥ ε0 . δ δ δ δ Let ε0 = 21 and δ > 0 arbitrary. We observe that if x  =



 √ and x  = n, then f x  − f x  = 1, ∀ n ∈ N∗ . On the other hand, we have: x  − x  =

√

n+1−

√ n=√

1 n+1+

√

n + 1.

√ →0 n

whence it follows that there exists n δ ∈ N∗ such that: 1 √ √ < δ , ∀ n ≥ nδ n +1+ n Now if we choose xδ =

√ √ n δ + 1 and xδ = n δ , then we observe that:

 

 x − x  < δ and f x  − f x  = 1 > ε0 = 1 . δ δ δ δ 2 Therefore, the function f is not uniformly continuous on R. Proposition 5.6.1 If f : I ⊂ R → R is differentiable on the interval I , and has bounded derivative on this interval I , then f is uniformly continuous on I . Proof Let M > 0 be such that f  (x) ≤ M , ∀ x ∈ I . From Lagrange’s theorem it results that for any two points x , y ∈ I , there exists a point ξ between x and y such that: f (x) − f (y) = f  (ξ )(x − y). Further we have: | f (x) − f (y)| = f  (ξ ) |x − y| ≤ M|x − y|, ∀x, y ∈ I

100

5 Functions of Several Variables ε M

Let us denote by δε = δε then:

> 0 for any ε > 0.We observe now that if |x − y|
0 such that for any δ > 0 , ∃ xδ , xδ ∈ K with the properties:     

 x − x   < δ and  F x  − F x   ≥ ε0 . δ δ δ δ In particular, for δ = with the properties:

1 p

, p ∈ N∗ , it results that there are two points x p , x p ∈ K    x − x   < 1 p p p

(5.2)

 

  F x − F x   ≥ ε0 p p

(5.3)

and

  Since K is bounded, it results that the sequence x p of elements of K is also bounded so, according to Lemma 5.3.2, there exists a convergent subseRn

quence x pl −→ a. Taking into account that K is closed, it follows that a ∈ K . Rn

(Theorem 5.3.2). On the other hand, from (5.2) we deduce that x pl −→ a. As F is continuous, from Theorem 5.5.1 we have:

 Rm

 Rm F x pl −→ F(a) and F x pl −→ F(a). Therefore:



 Rm F x pl − F x pl −→ F(a) − F(a) = 0.

5.6 Properties of Continuous Functions Defined …

101

 We

 have thus come to a contradiction, because according to (5.3),  F x  − F x   ≥ ε0 , ∀ l ∈ N∗ . pl pl Definition 5.6.3 A subset A ⊂ Rn is called connected if does not exist two nonempty open sets D1 and D2 with properties: A ⊂ D1 ∪ D2 , D1 ∩ D2 ∩ A = ∅ , D1 ∩ A  = ∅ , D2 ∩ A  = ∅ Example 5.6.3 The set A = (0, 1)∪(2, 3) is disconnected, while the set A = (−1, 1) is connected. The following characterization theorem for connected subsets of R can be proved. Theorem 5.6.4 A subset non-void A ⊂ R is connected iff it is an interval, namely if it has the property: ∀ x , y ∈ A and any x < z < y, it results that z ∈ A. Theorem 5.6.5 Let F : Rn → Rm be a continuous function. If the subset A ⊂ Rn is connected, then F(A) ⊂ Rm is connected. Proof If we suppose that F(A) ⊂ Rm is disconnected, then there exist two non-void open sets Di ⊂ Rm , i = 1, 2 with properties: F(A) ⊂ D1 ∪ D2 , D1 ∩ D2 ∩ F(A) = ∅, D1 ∩ F(A) = ∅, D2 ∩ F(A) = ∅ From Theorem 5.5.6 it results that the sets F −1 (Di ), i = 1, 2 are open sets. It easy to see that F −1 (Di ) = φ, i = 1, 2 and A ⊂ F −1 (D1 ) ∪ F −1 (D2 ), F −1 (D1 ) ∩ F −1 (D2 ) ∩ A = ∅, F −1 (Di ) ∩ A = ∅, i = 1, 2 As F −1 (Di ), i = 1, 2 are open sets, it results that A is disconnected which is a contradiction. Definition 5.6.4 Let I ⊂ R be an interval. We say that the function f : I → R has the Darboux’s property on I if for any interval J ⊂ I its image f (J ) is also an interval. Corollary 5.6.1 If f : I ⊂ R → R is a continuous function on the interval I , then f has the Darboux’s property on I . Proof If J ⊂ I is an interval, then from Theorem 5.6.4 it follows that J is a connected subset of R. Taking into account to Theorem 5.6.5, it results that f (J ) is also a connected subsets of R. Applying again Theorem 5.6.4, we deduce that f (J ) is an interval, so f has Darboux’s property on I . Corollary 5.6.2 (Bolzano’s theorem). If f : [a , b] → R is continuous and f (a) · f (b) < 0, then there exists a point a < c < b such that f (c) = 0.

102

5 Functions of Several Variables

Proof From Corollary 5.6.1 it results that the set J = f ([a , b]) ⊂ R is an interval. Since f (a) · f (b) < 0 we can suppose, for example, that f (a) < 0 < f (b). As J is an interval and 0 ∈ J , it follows that there exists c ∈ (a , b) that that f (c) = 0. We remark that Corollary 5.6.2 is useful in solving equations. Suppose that the equation f (x) = 0 has an unique root α in the interval (a , b). For to make a choice we shall assume that f (a) > 0 and f (b) < 0 (Fig. 5.1). The simplest method of approximating this root is the bisection method which consists of the following: We divide the segment [a , b] into two equal parts by the point c = a +2 b . If f (c) = 0, then α = c is the sought-after root. If f (c) < 0 than the root α ∈ [a, c] and we shall denote by a1 = a and b1 = c. 1 be the middle point of the interval [a1 , b1 ]. If f (c1 ) = 0, then α = c1 Let c1 = a1 +b 2 and the algorithm terminates. Let us suppose that f (c1 ) < 0. Then the required root α ∈ [a1 , c1 ] and we shall denote by a2 = a1 and b2 = c1 . Now we shall deal with the interval [a2 , b2 ] in the same way as with the interval [a1 , b1 ]. Let c2 be the middle point of the interval [a2 , b2 ]. If f (c2 ) = 0, then α = c2 and the algorithm terminates. If f (c2 ) > 0, then the required root α ∈ [c2 , b2 ], and we shall denote by a3 = c2 and b3 = b2 .Proceeding in the same way, we shall have two possibilities: 1 2

Either the algorithm described above will terminate because in the middle point cn of the interval [an , bn ] we have f (cn ) = 0 (in this case α = cn ). Or the process described can be continued indefinitely and we shall receive a nested system of closed intervals [a1 , b1 ] ⊃ [a2 , b2 ] ⊃ .... ⊃ [an , bn ] ⊃ .... with f (an ) · f (bn ) < 0.This nested system has a common point α to which each of the sequences {an } and {bn } converges.

Since f is continuous, it follows that f (an ) → f (α) and f (bn ) → f (α). The n , the number cn = an +b differs from length of the interval [an , bn ] being equal to b−a 2n 2 b−a the root α by not more than 2n+1 . Therefore it possible to compute the root α with any pre-assigned accuracy. Example 5.6.4 Find the root of the equation x 3 − x − 2 = 0, lies in the interval [1, 2]. We have f (x) = x 3 − x − 2, f (1) = −2, f (2) = 4, so the equation f (x) = 0 has a root α in the closed interval closed [1, 2]. We observe that α is the unique root of the equation in the interval [1, 2], because the function f is strictly increasing on this interval. Indeed, f  (x) = 3x 2 − 1 > 0, ∀x ∈ [1, 2]. According to the algorithm of bisection method we have: a = 1 b = 2 c = 1.5

a1 = 1.5 b1 = 2 c1 = 1.75

f (c) = −0.125

f (c1 ) = 1.6093750

5.7 Linear Continuous Maps from …

103

a2 = 1.5 b2 = 1.75 c2 = 1.625

f (c2 ) = 0.6660156

a3 = 1.5 b3 = 1.625 c3 = 1.5625

a4 = 1.5 b4 = 1.5625 c4 = 1.53125

a5 = 1.5 b5 = 1.53125 c5 = 1.515625

f (c3 ) = 0.2521973

f (c4 ) = 0.0591125

f (c5 ) = −0.0340538

a6 = 1.515625 b6 = 1.53125 c6 = 1.5234375

f (c6 ) = 0.0122504

...

a15 = 1.5213623 b15 = 1.5214233 c15 = 1.5213928

f (c15 ) = 0.0000780

5.7 Linear Continuous Maps from Rn to Rm Definition 5.7.1 A function T : Rn → Rm is called linear map (transformation) if it satisfies the following condition: T (λ · x + μ · y) = λ · T (x) + μ · T (y) , ∀ x , y ∈ Rn , ∀ λ , μ ∈ R. Further we will denote by L(Rn , Rm ) the set of all linear maps from Rn to Rm . Proposition 5.7.1 The set L(Rn , Rm ) has the following properties: (i) (ii)

T (0Rn ) = 0Rm , ∀ T ∈ L(Rn , Rm ). If T , U ∈ L(Rn , Rm ), then: 

α · T + β · U ∈ L Rn , Rm , ∀α , β ∈ R

104

5 Functions of Several Variables

(iii)

If T ∈ L(Rn , Rm ) and U ∈ L(Rm , R p ), then: 

U ◦ T ∈ L Rn , R p

The verification is immediate. The following theorem for characterizing linear maps from Rn to Rm is also well known. Theorem 5.7.1 The map T : Rn → Rm is linear iff there exists a matrix A with m rows and n columns A ∈ Mm , n (R) such that: T (x) = x At , ∀ x ∈ Rn where At denoted the transpose of the matrix A. (A is called the matrix associated with the linear map T , with respect to the canonical bases from Rn and Rm ). If: ⎛

λ11 ⎜ λ21

 A = λi j 1 ≤ i ≤ m = ⎜ ⎝ ... 1≤ j≤n λm1

λ12 λ22 ... λm 2

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

⎞ λ1n λ2n ⎟ ⎟ ∈ Mm,n (R) ...⎠ λm n

and x = (x1 , x2 , . . . , xn ) ∈ Rn , then: T (x) = (λ11 x1 + λ12 x2 + . . . + λ1 n xn , . . . , λm1 x1 + λm2 x2 + · · · + λmn xn ). Corollary 5.7.1 A map T : R → R is linear iff there exists λ ∈ R such that: T (x) = λ · x , ∀ x ∈ R. Corollary 5.7.2 A map T : Rn → R is linear iff there are λ1 , λ2 , ..., λn ∈ R such that: T (x) = λ1 · x1 + λ2 · x2 + ... + λ

n

· xn , ∀ x = (x1 , x2 , ..., xn ) ∈ Rn

Let T : Rn → Rm be a linear map and let A be its matrix with respect to the canonical bases from Rn and Rm . According to (5.4), for any x = (x1 , ..., xn ) ∈ Rn , we have: T (x) = (λ11 · x1 + ... + λ1

n

· xn , ... , λm

whence it results that: T (x)2 = (λ11 · x1 + · · · + λ1

n

· xn )2 + · · · + (λm

1

· x1 + ... + λm

1

n

· xn )

· x1 + · · · + λm n · xn )2 (5.4)

5.7 Linear Continuous Maps from …

105

Further, taking into account to inequality Cauchy–Buniakovski, we have:   n n n    n 2  T (x)2 ≤  λ1 j · x 2j + · · · + λ2m j · x 2j j=1

j=1

j=1

  n  m  = x2 ·  λi2 j .

j=1

(5.5)

i=1 j=1

Also we have: T (x)∞ = max(|λ11 · x1 + · · · + λ1 n · xn |, · · · , |λm 1 · x1 + · · · + λm n · xn |) ≤ max(|λ11 | · |x1 | + · · · + |λ1 n | · |xn |, . . . , |λm 1 | · |x1 | + · · · + |λm n | · |xn |) ≤ x∞ · max(|λ11 | + · · · + |λ1n |, . . . , |λm1 | + · · · + |λm n |).

(5.6)

Several norms can be entered on the set Mm , n (R) of all matrices with m rows and n columns. Further, we consider the following two norms for any ⎞ ⎛ λ11 · · · λ1 n A = ⎝ · · · · · · · · · ⎠ ∈ Mm,n (R) : λm 1 · · · λm n   n  m  A2 =  λi2 j

(5.7)

i =1 j =1

A∞ = max(|λ11 | + · · · + |λ1 n |, . . . , |λm 1 | + · · · + |λmn |).

(5.8)

From the relations (5.5)–(5.8), we deduce the following result: Theorem 5.7.2 Let T : Rn → Rm be a linear map and let A be its matrix with respect to the canonical bases from Rn and Rm . Then we have: T (x)2 ≤ A2 · x2 and T (x)∞ ≤ A∞ · x∞ , ∀ x ∈ Rn . (Therefore, T (x) ≤ A · x , ∀ x ∈ Rn , for any of the norms  · 2 or  · ∞ ). Corollary 5.7.3 Any linear map T : Rn → Rm is continuous on Rn . Proof Let a ∈ Rn be arbitrary and let {xk } be a sequence of vectors of Rn such that Rn

xk −→ a. From Theorem 5.7.2 and from the linearity of T it results: T (xk ) − T (a) = T (x k − a) ≤ A · xk − a. Rm

As lim xk − a = 0, it follows that T (xk ) −→ T (a), hence T is continuous at k →∞

the point x = a.

Chapter 6

Differential Calculus of Functions of Several Variables

6.1 Partial Derivatives. Differentiability of a Function of Several Variables Definition 6.1.1 LetA ⊂ R2 be an open subset and let f : A ⊂ R2 → R be a function of two variables. The partial derivative of f , with respect to x, at the point (a , b) ∈ A is defined as the limit: lim

x →a

f (x, b) − f (a, b) f (a + h, b) − f (a, b) = lim , h →0 x − a h

provided that it exists and is finite. We shall denote the partial derivative of the function f at the point (a, b) by ∂ f b) or by f x (a, b). ∂ x (a, If there exists ∂∂ xf (a, b) at any point (a , b) ∈ A, then the function of two variables (a , b) → ∂∂ xf (a, b) : A → R is denoted by ∂∂ xf and named the partial derivative of the function f with respect to x on A. Similarly defined the partial derivative of the function f with respect to y at the point (a , b), namely: f (a, y) − f (a, b) ∂ f (a, b) = f y (a, b) = lim y →b ∂ y y − b f (a , b + k) − f (a, b) . = lim k →0 k Remark 6.1.1 The partial derivative ∂∂ xf (a, b) is nothing but the ordinary derivative of function of on variable t → f (t , b) at the point t = a.Therefore the partial derivative ∂∂ xf is an ordinary derivative with respect to x by regarding y as a constant and similarly, the partial derivative ∂∂ yf is an ordinary derivative with respect to y computed by regarding x as a constant.Therefore, partial derivatives can be © The Author(s), under exclusive license to Springer Nature Singapore Pte Ltd. 2022 G. Paltineanu et al., Differential Calculus for Engineers, https://doi.org/10.1007/978-981-19-2553-5_6

107

108

6 Differential Calculus of Functions of Several Variables

compute in concordance with the rules governing the camputation of the derivatives of functions of one variable.   Example 6.1.1 Let A = (x , y) ∈ R2 ; x > 0 , y > 0 and f : A → R, defined by: f (x, y) = x 2 · y 3 − 2 · y · ln x + y x . Compute the partial derivatives ∂∂ xf and ∂∂ yf on A. Taking into account of Remark 6.1.1 we have: y ∂ f + y x ln y. (x, y) = 2 · x · y 3 − 2 · ∂ x x And ∂ f (x, y) = 3 · x 2 · y 2 − 2 · ln x + x · y x − 1 . ∂ y For an arbitrary function of several variable we have the following definition: Definition 6.1.2 Let A ⊂ Rn be an open subset, f : A ⊂ Rn → R be a real function of n variables and a = (a1 , . . . , an ) ∈ A. The partial derivative of function f with respect to the variable xi at the point a is defined by: ∂f ∂f (a) = (a1 , . . . , ai , . . . , an ) ∂ xi ∂ xi f (a1 , . . . , ai − 1 , xi , ai = lim

+ 1,

. . . , an ) − f (a1 , . . . , ai , . . . , an ) xi − ai f (a1 , . . . , ai − 1 , ai + h, ai + 1 , . . . , an ) − f (a1 , . . . , ai , . . . , an ) = lim h →0 h = f xi (a). xi → ai

provided this limit exists and is finite. Remark 6.1.2 The partial derivative ∂∂ xfi is the ordinary derivative of function f with respect to xi computed by regarding the other variables x j , j = 1, n, j = i as constants. Proposition 6.1.1 Let A ⊂ Rn be an open subset, f : A ⊂ Rn → R be a real function of n variables and a = (a1 , . . . , an ) ∈ A. If ∂∂ xf (a) exists, then f is continuous with respect to xi at the point a. Proof Since the function of one variable: xi → f (a1 , . . . , ai

− 1,

xi , ai

+ 1,

. . . , an )

6.1 Partial Derivatives. Differentiability of a Function …

109

is derivable at the point ai ∈ R, it results that this function is continuous at the point ai , hence the function of n variables f is continuous with respect to xi at the point a. Remark 6.1.3 The existence of all partial derivatives of a function at a given point does not imply, in general, the continuity of the function at that point. Indeed, we proved that the function:  f (x, y) =

x2· y x 4 + y2

0

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

does not continuous at the origin (0, 0),because lim f (x, y) does not exists(Example x→0 y→0

5.5.2). On the other hand it is clearly that: ∂f ∂f (0, 0) = (0, 0) = 0. ∂x ∂y Further, we will present the notion of derivative of a function of one variable in an equivalent form, that will allow the generalization of this notion for vector functions. Let I ⊂ R be on open iterval, a ∈ I and f : I → R.We recall that the function f is derivable at the point a if the following limit: lim

x →a

f (x) − f (a) f (a + h) − f (a) = lim = f  (a) , h →0 x − a h

exists and is finite. Proposition 6.1.2 Let I ⊂ R be an interval, f : I → R and a ∈ I an interior point. The following statements are equivalent: (i) (ii)

f is derivable (differentiable) at the point a. There exists a linear map T = Ta : R → R such that: lim

h →0

f (a + h) − f (a) − T (h) = 0. h

Proof (i) ⇒(ii) Since f is derivable at the point a, it results that the limit: f  (a) = lim

h →0

f (a + h) − f (a) . h

exists and it is finite. If we denote by T (h) = f  (a) · h, for any h ∈ I , then T : R → R is a linear map (obviously continuous) on R and we have: lim

h →0

f (a + h) − f (a) − T (h) f (a + h) − f (a) = lim − f  (a) = 0. h h h →0

110

6 Differential Calculus of Functions of Several Variables

(ii) ⇒(i) According to the hypothesis there exists a linear map T : R → R such that: lim

h →0

f (a + h) − f (a) − T (h) = 0 h

(6.1)

On the other hand, from Corollary 5.7.1 it results that there exists λ ∈ R such that T (h) = λ · h , ∀ h ∈ R. Taking into account of (6.1) we deduce that: lim

h →0

f (a + h) − f (a) = λ ∈ R, h

hence f is derivable at a and f  (a) = λ. Remark 6.1.4 From the proof of Proposition 6.1.2, it follows that if f : I → R is derivable at the point a, then there exists an unique linear map T : R → R for which it takes place (6.1), namely T (h) = f  (a) · h , h ∈ R. This linear map is called the differential of f at the point a and is denoted by d f (a). Definition 6.1.3 Let I ⊂ R be an interval, f : I → R and a ∈ I an interior point. It’s called the differential of f at the point a, the following linear map on R : d f (a) : R → R , d f (a)(h) = f  (a) · h , ∀ h ∈ R. Further we present the geometric interpretation of the differential. Let (C) be the graph of the function f and M0 (a, f (a)) ∈ C(Fig. 6.1). It is known that f  (a) = tg α(the slope of the tangent line to the graph at the point M0 ). From the triangle M0 P T we have:

Fig. 6.1 The geometrical interpretation of differential

6.1 Partial Derivatives. Differentiability of a Function …

111

PT PT = , M0 P h

f  (a) = tg α =

whence it results that P T = f  (a) · h = d f (a)(h). Therefore, d f (a)(h) is equal with the segment length P T , where T is the point belongs to the the tangent line to the graph at the point M0 , corresponding to the abscissa a + h. On the other hand, the increment of the function f i.e., f (a + h) − f (a) is equal to the the segment lenght P M, where M(a + h, f (a + h))) ∈ C. Clearly, P M = f (a + h) − f (a) = P T = d f (a)(h), but we observe that for a small increment h  of the argument x, the increment:   f a + h  − f (a) = P  M  ≈ P  T  = d f (a)(h  ) . Therefore, for small values (| h | 0 such that B(a , r ) ⊂ A. For any h = (h 1 , h 2 ) ∈ R2 such that a + h ∈ A we have: f (a1 + h 1 , a2 + h 2 ) − f (a1 , a2 ) = f (a1 + h 1 , a2 + h 2 ) − f (a1 , a2 + h 2 ) + f (a1 , a2 + h 2 ) − f (a1 , a2 ). From Lagrange’s Theorem it results that there are 0 < θi < 1 , i = 1 , 2, such that:

116

6 Differential Calculus of Functions of Several Variables

f (a1 + h 1 , a2 + h 2 ) − f (a1 , a2 ) ∂ f = (a1 + θ1 · h 1 , a2 + h 2 ) · h 1 + ∂ x

∂ f (a1 , a2 + θ2 · h 2 ) · h 2 . ∂ y (6.4)

If we denote by: ω1 (h 1 , h 2 ) =

∂ f ∂ f (a1 + θ1 · h 1 , a2 + h 2 ) − (a1 , a2 ), ∂ x ∂ x

and ω2 (h 1 , h 2 ) =

∂ f ∂ f (a1 , a2 + θ2 · h 2 ) − (a1 , a2 ), ∂ y ∂ y

then, from (6.4) it follows: f (a + h) − f (a) ∂ f ∂ f = (a) · h 1 + (a) · h 2 + ω1 (h) · h 1 + ω2 (h) · h 2 ∂ x ∂ y Finally, if we denote by ϕ(h) = ω1 (h) · h 1 + ω2 (h) · h 2 , then: f (a + h) − f (a) = d f (a)(h) + ϕ(h).

(6.5)

If we will prove that ϕ is an infinitely small (ϕ = o(h)) then will results that f is differentiable at the point a. Since ∂∂ xfi are continuous we deduce that lim ωi (h) = 0, i = 1 , 2. On the other hand, we have:

h →0

|h 1 | |h 2 | |ϕ(h)| ≤ |ω1 (h)| · + |ω2 (h)| · ≤ |ω1 (h)| + |ω2 (h)|, h h h |ϕ(h)| h →0 h 

whence it results that lim

= 0, and so the proof is finished.

Remark 6.1.7 From Theorem 6.1.3 it results that the elementary functions of several variables are differentiable on any open subset in their domain of definition.

6.2 Differentiability of Vector Functions. Jacobian Matrix

117

6.2 Differentiability of Vector Functions. Jacobian Matrix Definition 6.2.1 Let F = ( f 1 , . . . , f m ) : A ⊂ Rn → Rm be a vector function of n variables and let a ∈ A be an interior point. We say that F is differentiable at the point a if there exists a linear map T = Ta : Rn → Rm such that: F(a + h) − F(a) − T (h) = 0. h

lim

h →0

where  ·  is any of the norms  · ∞ or  · 2 . Similar to the statement in the Theorem 6.1.2. We have: Remark 6.2.1 A vector function F : A ⊂ Rn → Rm is differentiable at the interior point a ∈ A iff there exist a linear map T = Ta : Rn → Rm , a neighborhood V of the origine and an infinitely small function  : V → Rm (i.e. = 0(h)) such that: F(a + h) − F(a) = T (h) + (h) , ∀ h ∈ V. We specify that  = 0(h) means: lim (h) = (0Rn ) = 0Rm and lim

h→0Rn

h→0Rn

(h) = 0Rm . h

Theorem 6.2.1 The vector function F = ( f 1 , . . . , f m ) : A ⊂ Rn → Rm is differentiable at the interior point a ∈ A, if and only if, any its scalar component f i : A ⊂ Rn → R, i = 1 , m, is differentiable at the point a. Proof Since F is differentiable at the point a, then there exists a linear map T = Ta : Rn → Rm such that: lim

h →0

F(a + h) − F(a) − T (h)∞ = 0.  h ∞

(6.6)

 the other hand, from Theorem 5.7.1, it results that there exists a matrix A =  On λi j 1 ≤ i ≤ m ∈ Mm , n (R) such that: 1 ≤j ≤n

T (h) = h · At = (λ11 · h 1 + · · · + λ1 n · h n , . . . , λm 1 · h 1 + · · · + λm n · h n )

for any h = (h 1 , . . . , h n ) ∈ Rn . If we denote by ti (h) = λi 1 · h 1 + · · · + λi n · h n , ∀ h = (h 1 , . . . , h n ) ∈ Rn , then for any i = 1 , m, ti : Rn → R is a linear map and T = (t1 , . . . , tm ). Because we have: | f i (a + h) − f i (a) − ti (h)| ≤ F(a + h) − F(a) − T (h)∞ , it result from (6.6) that:

118

6 Differential Calculus of Functions of Several Variables

lim

h →0

| f i (a + h) − f i (a) − ti (h)| = 0 , (∀) i = 1 , m  h ∞

(6.7)

Therefore, the scalar component f i is differentiable at the point a, for any i = 1 , m. Reciprocal, if any scalar component f i of the vector function F is differentiable at the point a, then the relation (6.7) is true for any i = 1 , m. As: F(a + h) − F(a) − T (h)∞ =

max | f i (a + h) − f i (a) − ti (h)|

1≤i ≤ m

it follows that: lim

h →0

F(a + h) − F(a) − T (h)∞ = 0,  h ∞

hence the vector function F is differentiable at the point a. Remark 6.2.2 From Theorem 6.1.1, we deduce that ti is the differential of first order of the scalar function f i at the point a, so for any i = 1 , m we have: ∂ fi ∂ fi (a) · h 1 + · · · + (a) · h n , ∂ x1 ∂ xn ∀ h = (h 1 , . . . , h n ) ∈ Rn .

ti (h) = d f i (a)(h) =

Therefore, if the vector function F = ( f 1 , . . . , f m ) is differentiable at the point a, then linear map T : Rn → Rm is uniquely determined, namely: T (h) = (d f 1 (a)(h) , . . . , d f m (a)(h)) , (∀) h ∈ Rn . We will call this linear map the differential of first order of the vector function F at the point a and we will denote them by d F(a). Therefore d F(a) : Rn → Rm , and: d F(a)(h) = (d f 1 (a)(h) , . . . , ∂ f1 ∂ = (a) · h 1 + · · · + ∂ x1 ∂

d f m (a)(h))

f1 ∂ fm ∂ fm (a) · h n , . . . , (a) · h 1 + . . . + (a) · h n , xn ∂ x1 ∂ xn

∀ h = (h 1 , . . . , h n ) ∈ Rn .

Definition 6.2.2 The matrix associated with the linear application d F(a) : Rn → Rm , with respect to the canonical bases in Rn and Rm is denoted by: ⎛

∂ f1 ∂ x1 (a)

... ⎜ JF (a) = ⎝ . . . . . . ∂ fm ... ∂ x1 (a)

∂ f1 ∂ xn (a) ∂ ∂

⎞

⎟ ... ⎠ fm xn (a)

6.3 Differentiability of Composite Functions

119

and is called the jacobian matrix associated with the vector function F at the point a. We remark that we have: d F(a)(h) = h · (JF (a))t where, for any B we denoted by B t the transpose of the matrix B. Remark 6.2.3. From Theorem 6.2.1 it results that the study of the differentiability of a vector function returns to the study of the differentiability of its scalar components. Therefore it is sufficient to consider functions of the form f : A ⊂ Rn → R. If m = n, then the jacobian matrix JF (a) is a square matrix and its determinant is called jacobian determinant ( or simply jacobian) and is denoted by: D( f 1 , . . . , f n ) (a) = D(x1 , . . . , xn )

∂( f 1 , . . . , f n ) (a). ∂(x1 , . . . , xn )

Therefore we have:

∂ f1 (a) . . .

∂ x1 D( f 1 , . . . , f n )

(a) = . . . . . .

∂ fn D(x1 , . . . , xn )

∂ x (a) . . . 1



∂ f1

∂ xn (a)

...



.

∂ fn

∂ xn (a)

6.3 Differentiability of Composite Functions Theorem 6.3.1 Let A ⊂ Rn and B ⊂ Rm be two open subsets. If F : A → B is differentiable at the point a ∈ A, and G : B → R p is differentiable at the point b = F(a) ∈ B, then the composite function H = G ◦ F : A ⊂ Rn → R p is differentiable at the point a ∈ A and: d H (a) = d G(b) ◦ d F(a). J H (a) = JG (b) · JF (a). Proof By Remark 6.2.1 we have: F(a + h) − F(a) = d F(a)(h) + (h), . (h) h→0 h

where lim (h) = (0Rn ) = 0Rm ,lim h→0

and:

= 0Rm .

(6.8)

120

6 Differential Calculus of Functions of Several Variables

G(b + k) − G(b) = d G(b)(k) + (k) (k) k→0 k

where lim (k) = (0Rn ) = 0Rm ,lim k→0

(6.9)

= 0Rm .

Since H = G ◦ F, further it results: H (a + h) − H (a) = G(F(a + h)) − G(F(a)) = = G(F(a) + d F(a)(h) + (h)) − G(b) . If we denote by k(h) = d F(a)(h) + (h) then, from (6.9) we deduce that: H (a + h) − H (a) = G(b + k(h)) − G(b) = d G(b)(k(h)) + (k(h)). Taking into account that d G(b) is a linear map it results: H (a + h) − H (a) = d G(b)(d F(a)(h)) + d G(b)((h)) + (k(h)) = (d G(b) ◦ d F(a))(h) + χ(h). where we denoted by: χ (h) = d G(b)((h)) + (k(h)). As d G(b) ◦ d F(a) : Rn → R p is also a linear map it results that it is sufficient to prove that χ = 0(h), i.e.χ is an infinitely small function. m Obviously lim χ (h) = χ (0Rn ) = 0Rm . It remains to show that lim χ(h)  h  = 0R . h →0

h→0

Using Theorem 5.7.2, and taking into account that JG (b) is the matrix associated with the linear map d G(b) : Rm → R p , and JF (a) is the matrix associated with the linear map d F(a) : Rn → Rm , it results: χ (h)  (k(h)) k(h) (h) ≤ JG (b) · + · h h  k(h)  h

 (k(h)) (h) (h) + · JF (a) + . ≤ JG (b) · h k(h) h Since lim

h →0 lim χ (h) h h →0

(h) h

= 0, lim k(h) = 0 and lim

= 0, hence

h →0

lim χ(h) h →0 h 

= 0Rm .

k →0

 (k) k 

= 0, we deduce that:

Therefore we proved that:

H (a + h) − H (a) = (d G(b) ◦ d F(a))(h) + χ(h), where χ is an infinitely small, whence it results that H is differentiable at the point a and:

6.3 Differentiability of Composite Functions

121

d H (a) = d G(b) ◦ d F(a). As to the operation of composing linear maps corresponds the operation of multiplying their associated matrices, it follows that: J H (a) = JG (b) · JF (a). Exemple 6.3.1 Let’s consider the subset:   A = (x , y) ∈ R2 ; x > 0 , y > − 2 ⊂ R2 and the following functions: F : A ⊂ R2 → R3 , F(x, y) =

√

G : R3 → R2 , G(u, v , w) =

x,



x 2 + 3 · y2 ,



 y + 2 ,



u 2 + v 2 + 2 · w2 , u 2 − v 2



H = G ◦ F : A ⊂ R2 → R2 . Let also be the points: a = (1 , − 1) ∈ A and b = F(a) = (1 , 2 , 1) ∈ R3 . Prove that: J H (a) = JG (b) · JF (a). We have successive: ⎛

1√

0

⎞

⎟ ⎜ 2· x x √ √ 3 ·y ⎟ JF (x , y) = ⎜ ⎝ x 2 + 3 ·y 2 x 2 + 3· y 2 ⎠, √1 0 2· y + 2 ⎛1

JF (1 , − 1) = ⎝

2 1 2

0 −

0 JG (u, v , w) =

3 2 1 2

⎞

⎠,

2 u 2 v 4w ; JG (1 , 2 , 1) = 2u −2v 0



2 44 2 −4 0

and: JG (1, 2 , 1) · JF (1 , − 1) =

3 −4 . −1 6

122

6 Differential Calculus of Functions of Several Variables

On the other hand, for any point (x, y) ∈ A, we have:  √   x , x 2 + 3 · y2 , y + 2 H (x, y) = G(F(x, y)) = G   = x 2 + 3 · y2 + x + 2 · y + 4 , − x 2 − 3 y2 + x . The jacobian matrix associated with the function H (x, y) ∈ A is: J H (x , y) =

2 ·x + 16 ·y + 2 −2 ·x + 1 −6 ·y

= G ◦ F at the point

and so J H (1 , − 1) =

3 −4 −1 6

Theorem 6.3.2 Let A , B ⊂ R2 be two open subsets, F = (u , v) : A ⊂ R2 → B ⊂ R2 be a vector function of C 1 class on A and let f : B ⊂ R2 → R be a scalar function of C 1 class on B.Then the composite function h = f ◦ F : A ⊂ R2 → R, given by: h(x, y) = ( f ◦ F)(x, y) = f (F(x, y)) = f (u(x, y) , v(x, y)) , ∀ (x, y) ∈ A. is of C 1 class on A and we have the following formulas: ∂h ∂ f ∂u ∂ f ∂v = · + · ∂ x ∂u ∂ x ∂v ∂ x ∂h ∂ f ∂u ∂ f ∂v = · + · . ∂ y ∂u ∂ y ∂v ∂ y

(6.10)

Proof The scalar functions u , v and f are differentiable, because, by hypothesis they are of C 1 class (Theorem 6.1.3). From Theorem 6.2.1, it results that the vector function F is differentiable on A, and from Theorem 6.3.1 we deduce that the composite function h = f ◦ F is differentiable on A. Let (a, b) ∈ A, be an arbitrary point. If we denote by c = u (a, b) and by d = v (a, b), then (c, d) ∈ B. From Theorem 6.3.1, it results: Jh (a, b) = J f (c, d) · JF (a, b), that is: 

∂h ∂ x (a,

b)

∂h ∂ y (a,

 b)

=

∂

f ∂ u (c,

Following the computes we obtain:

d)

∂ f ∂ v (c,

d)



 ·

∂ ∂ ∂ ∂

u x (a, v x (a,

b) b)

∂ ∂ ∂ ∂

u y (a, v y (a,

 b) . b)

6.3 Differentiability of Composite Functions

∂h ∂ (a, b) = ∂ x ∂ ∂h ∂ (a, b) = ∂ y ∂

f ∂u ∂ (c, d) · (a, b) + u ∂ x ∂ f ∂u ∂ (c, d) · (a, b) + u ∂ y ∂

123

f ∂v (c, d) · (a, b) v ∂ x f ∂v (c, d) · (a, b) . v ∂ y

Example 6.3.2 Let f : R2 → R be a function of C 1 class on R2 and   h(x, y) = f 2 · x + y 2 , 1 − y · cos x , (x, y) ∈ R2 . Compute ∂∂ hx (x, y) and ∂∂ hy (x, y). If we denote by u(x, y) = 2 · x + y 2 and by v(x, y) = 1 − y · cos x, then h(x, y) = f (u(x, y), v(x, y)) and: ∂ f ∂u ∂ f ∂v ∂ f ∂ f ∂h = · + · = · 2 + · y · sin x ∂ x ∂u ∂ x ∂v ∂ x ∂u ∂v ∂h ∂ f ∂u ∂ f ∂v ∂ f ∂ f = · + · = · 2 ·y + · (− cos x). ∂ y ∂u ∂ y ∂v ∂ y ∂u ∂v Remark 6.3.1 The formulas (6.10) generalizes the known derivation formula of composite functions of ane variable, i.e. if h(x) = f (u(x)) then: h  (x) = f  (u(x)) · u  (x). Remark 6.3.2 The formulas (6.10) admit the following generalization: If h(x1 , . . . , xn ) = f (u 1 (x1 , . . . , xn ) , . . . , u m (x1 , . . . , xn )),then: ∂h ∂ f ∂ u1 ∂ f ∂ um = · + ··· + · ∂ x1 ∂ u1 ∂ x1 ∂ um ∂ x1 .................................................................. ∂h ∂ f ∂ u1 ∂ f ∂ um = · + ... + · . ∂ xn ∂ u1 ∂ xn ∂ um ∂ xn

(6.11)

Definition 6.3.1 Let K ⊂ Rn be an open cone (i.e. a subset with the property that for any x ∈ K and any t ∈ R , t = 0, it results that t · x ∈ K ). A function f : K → R is called homogeneous p of order p,where p ∈ R, if: f (t · x1 , . . . , t · xn ) = t p · f (x1 , . . . , xn ) , ∀ x = (x1 , . . . , xn ) ∈ K . Exemple 6.3.3 1.

The function f : R2 → R , given by: f (x, y) = x 2 + 4 · y 2 , ∀(x, y) ∈ R2

124

6 Differential Calculus of Functions of Several Variables

is homogeneous of order 2 este R2 , because:   f (t · x, t · y) = (t · x)2 + 4 · (t · y)2 = t 2 · x 2 + 4 · y 2 = t 2 · f (x , y) , ∀ (x, y) ∈ R2 2.

  The function f : (x , y) ∈ R2 ; x = 0 → R , given by: f (x, y) = (x + y) · arctg

3.

y , x

is homogeneous of order 1, because: f (t · x, t · y) = (t · x + t · y) ·   arctg tt ·· xy = t · (x + y) · arctg xy = t · f (x√, y). √ The function f : R2 → R , given by: f (x, y) = 3 x + 3 y 1 is omogeneous of order 3 , because: f (t · x, t · y ) =

√ 3

t·x +

√ 3

t·y =

√ 3

t·

√ 3

x +

√ 3

y



1

= t 3 · f (x , y ).

Theorem 6.3.3 (Euler) Let K ⊂ Rn be an open cone and f : K → R be a function differentiable and homogeneuos of order p.Then we have: x1 ·

∂ f ∂ f (x) + · · · + xn · (x) = p · f (x) , ∀ x = (x1 , . . . , xn ) ∈ K . ∂ x1 ∂ xn

Proof By the hypothesis we have: f (t · x1 , . . . , t · xn ) = t p · f (x1 , . . . , xn ) , ∀ x = (x1 , . . . , xn ) ∈ K , t ∈ R∗ .

(6.12)

We notice that the function from left member can be seen as a composed function, if we denote by: u 1 (t) = t · x1 , . . . , u n (t) = t · xn and h(t) = f (u 1 (t) , . . . , u n (t)). Taking into account the rules of derivation (6.11) it results: h  (t) =

∂ f ∂ f (t · x) · u 1 (t) + · · · + (t · x) · u n (t) = p · t p − 1 f (x) ∂ u1 ∂ un

and further: x1 ·

∂ f ∂ f (t · x) + · · · + xn · (t · x) = p · t p − 1 · f (x). ∂ u1 ∂ un

In particular case t = 1 we obtain: x1 ·

∂ f ∂ f (x) + · · · + xn · (x) = p · f (x) , ∀ x = (x1 , . . . , xn ) ∈ K . ∂ x1 ∂ xn

6.4 The First Order Differenential and Its Invariance Form

125

Example 6.3.4 Let f : R2 → R be the function given by: f (x, y) =

√ 3

x+

√ 3

y.

From Example 6.3.3, we know that the function is f homogeneous of order R2 , hance satisfaces the following relation: x·

1 3

on

∂ f ∂ f 1 + y · = · f. ∂ x ∂ y 3

This also results from a direct computation: ∂ f 1 1 1 1 ∂ f ; = ·√ = · 3 3 2 ∂ x 3 ∂ y 3 y2 x ∂ f ∂ f 1 1 x· + y · = x · ·√ ∂ x ∂ y 3 3 x2 √ 1 √ 1 = · ( 3 x + 3 x) = · f (x, y). 3 3

+y ·

1 1 · 3 3 y2

6.4 The First Order Differenential and Its Invariance Form Definition 6.4.1 Let A ⊂ Rn be an open subset and let f : A → R be a differentiable function on A. The first order differential of the function f at the point a ∈ A is the linear map d f (a) : Rn → R, given by: d f (a)(h) =

∂ f ∂ f (a) · h 1 + · · · + (a) · h n , ∀ h = (h 1 , . . . , h n ) ∈ Rn . ∂ x1 ∂ xn

We consider now the projection functions pi : Rn → R, i = 1 , n, given by: pi (x) = pi (x1 , . . . , xi , . . . , xn ) = xi , ∀ x = (x1 , . . . , xn ) ∈ Rn . 

1 if i = j , it results that d pi (a)(h) = h i , ∀ i = 1 , n. 0 if i = j Therefore, the first order differentials of the functions pi do not depend on the point a. Hence we have: Since

∂ pi ∂ xj

=

d pi (h) = d xi (h) = h i , ∀ i = 1 , n. With this remark, the first-order differential of the function f at the point a is written:

126

6 Differential Calculus of Functions of Several Variables

d f (a)(h) =

∂f ∂f (a) · d x 1 (h) + · · · + (a) · d x n (h), ∀h ∈ Rn . ∂ x1 ∂ xn

(6.13)

If we write the previous relation as an equality of functions we have: d f (a) =

∂ f ∂ f (a) · d x1 + · · · + (a) · d xn ∂ x1 ∂ xn

where d xi is the first order differential of the project function pi , i = 1 , n. Remark 6.4.1 Let us consider the particular case n = 2. Let f : A ⊂ R2 → R be a function differentiable at the interior point (a , b) ∈ A. The first ordin differential of function f at the point (a , b) ∈ A is the linear map d f (a , b) : R2 → R, defined by: ∂ f ∂ f (a , b) · h + (a , b) · k ∂ x ∂ y ∂ f ∂ f = (a , b) · d x(h , k) + (a , b) · d y(h , k) , ∂ x ∂ y ∀ (h, k) ∈ R2

d f (a , b)(h , k) =

Therefore: d f (a , b) =

∂ f ∂ f (a , b) · d x + (a , b) · d y . ∂ x ∂ y

  Example 6.4.1 Let f : (x , y) ∈ R2 ; x > 0 , y > 0 ⊂ R2 → R, defined by: f (x, y) = x 2 · y 3 − 2 · y · ln x + y x . Compute d f (1 , 1) and d f (1 , 1)(2 , − 1). Taking into account to Example 6.1.1, we have: ∂ f y + y x · ln y (x, y) = 2 · x · y 3 − 2 · ∂ x x and ∂ f (x, y) = 3 · x 2 · y 2 − 2 · ln x + x · y x − 1 , ∂ y so; ∂ f ∂ f (1, 1) = 0; (1, 1) = 4. ∂ x ∂ y

6.4 The First Order Differenential and Its Invariance Form

127

Then d f (1 , 1) : R2 → R is: d f (1 , 1) =

∂ f ∂ f (1 , 1) · d x + (1 , 1) · d y = 4 · d y ∂ x ∂ y

and d f (1 , 1)(2 , − 1) = 0 · 2 + 4 · (−1) = −4. Let us consider two open subsets A , B ⊂ Rn , a vector function F = (u 1 , . . . , u n ) : A ⊂ Rn → B ⊂ Rn differentiable on A, a scalar function f : B ⊂ R2 → R differentiable on B and let h = f ◦ F : A ⊂ Rn → R be the composite function defined by: h(x) = h(x1 , . . . , xn ) = f (u 1 (x1 , . . . , xn ) , . . . , u n (x1 , . . . , xn )) , ∀ x = (x1 , . . . , xn ) ∈ A As f = f (u 1 , . . . , u n ) , (u 1 , . . . , u n ) ∈ B, it results: df =

∂f ∂f · du 1 + · · · + · du n . ∂u 1 ∂u n

(6.14)

On the other hand we have: dh =

∂h ∂h · d x1 + · · · + · d xn . ∂ x1 ∂ xn

Taking into account that: h(x) = h(x1 , . . . , xn ) = f (u 1 (x1 , . . . , xn ) , ... , u n (x1 , ..., xn )) , ∀ x = (x1 , ... , xn ) ∈ A from the derivation formulas (6.11), for composite function, it results:

∂ f ∂u 1 ∂ f ∂u n dh = · + ··· + · d x1 + · · · + ∂u 1 ∂ x1 ∂u n ∂ x1

∂ f ∂u n ∂ f ∂u 1 · + ··· + · d xn + ∂u 1 ∂ xn ∂u n ∂ xn

∂f ∂u 1 ∂u 1 ∂f = · · d x1 + · · · + · d xn + · · · + · Therefore, we have: ∂u 1 ∂ x1 ∂ xn ∂u n

∂u n ∂u n · d x1 + · · · + · d xn ∂ x1 ∂ xn ∂f ∂f · du 1 + · · · + · du n . = ∂u 1 ∂u n

128

6 Differential Calculus of Functions of Several Variables

dh =

∂ f ∂ f · d u1 + · · · + · d un ∂ u1 ∂ un

(6.15)

The formula (6.14) represents the expression of the first order differential of the function f seen as a function that depends on the variables u 1 , ... , u n . The formula (6.15) represents the expression of the first order differential of the function h regarded as the function that depends on the dependent variables: u i = u i (x1 , ... , xn ), i = 1 , n. This formal equality d h = d f is called the invariance property of the firstorder differential at a change of independent variables.   Example 6.4.2 Let f ∈ C 1 R2 be an arbitrary function and.   h(x, y) = f 2 · x + y 2 , 1 − y · cos x Compute d h(x , y) and verify the invariance property of the first-order differential. If we denote by u(x, y) = 2 · x + y 2 and by v(x, y) = 1 − y · cos x, then h(x, y) = f (u, v) and according to Example 6.3.2 we have: ∂h ∂h · dx + · dy ∂x ∂y

∂f ∂f ·2+ · y · sin x d x = ∂u ∂v

∂f ∂f ·2·y+ · (− cos x) dy + ∂u ∂v ∂f · (2 · d x + 2y · dy) =   ∂u 

dh(x, y) =

du

∂f + · (y · sin xd x − cos xdy)   ∂v  dv

∂f ∂f = · du + · dv = d f (u, v). ∂u ∂v

6.5 The Directional Derivative. The Differential Operators: Gradient, Divergence, Curl and Laplacian Let A ⊂ R3 be an open subset, let a = (a1 , a2 , a3 ) ∈ A be a fixed point and let − → − → − → − → l = cos α · i + cos β · j + cos γ · k be an unit vector i.e.:

6.5 The Directional Derivative. The Differential Operators …

129

 2  l  = cos2 α + cos2 β + cos2 γ = 1. Since a ∈ A is an interior point, it results that there exists r > 0 such that B(a, r ) ⊂ A. For any t ∈ (− r , r ), the point x = a + t · l ∈ B(a, r ) ⊂ A, because: x − a = a + t · l − a = |t| ·  l  = | t | < r. (here l = (cos α, cos β, cos γ )). Definition 6.5.1 We say that the function f : A ⊂ R3 → R is derivable at the − → point a ∈ A along the vector l if the following limit exists and is finite: lim

x →a x = a + t· l

f (x) − f (a) f (a + t · l) − f (a) = lim . t →0 x − a |t |

f This limit is denoted by ∂ − → (a) and is called the directional derivative of function ∂ l − → f along the vector l at the point a. From a geometric point of view, the set of points x = a + t · l , t ∈ R − → represents the line that passes through a and has the director vector l (Fig. 6.2). The positive half-axis coresponds to the values of the parameter t > 0, and the negative half-axis coresponds to values t < 0. Further we will note with:

f (a + t · l) − f (a) f (a + t · l) − f (a) ∂f = lim − → + (a) = t lim + t →0 →0 |t | t ∂ l and with: ∂f f (a + t · l) − f (a) f (a + t · l) − f (a) = lim − → − (a) = t lim − t →0 →0 |t | −t ∂ l ∂f =− − → (a). ∂ l + Teorema 6.5.1 If f : A ⊂ R3 → R is differentiable at the interior point a ∈ A, − → − → then f is derivable at the point a ∈ A along the vector l = cos α · i + cos β · − → − → j + cos γ · k and: Fig. 6.2 The line passing through a and has the director vector l

130

6 Differential Calculus of Functions of Several Variables

∂f ∂ f ∂ f ∂ f − → + (a) = ∂ x (a) · cos α + ∂ y (a) · cos β + ∂ z (a) · cos γ . ∂ l Proof Let r > 0 be such that B(a, r ) ⊂ A. If t ∈ (0 , r ), then a + t · l ∈ A, and taking into account that f is differentiable at the point a it results: f (a + t · l) − f (a) = d f (a)(t · l) + ϕ(t · l) |ϕ(h)| h →0 h 

where ϕ = o(h)(i.e., lim ϕ(h) = ϕ(0) = 0 and lim h →0

= 0).

Since  t · l  = t and d f (a)(t · l) = t · d f (a)( l), further we have: ϕ(t · l) f (a + t · l) − f (a) = d f (a)( l) + . t · l  t ϕ(t · l) t→ 0  t· l 

As lim

= 0,it results that there exists:

f (a + t · l) − f (a) = d f (a)( l) t ∂ f ∂ f ∂ f = (a) · cos α + (a) · cos β + (a) · cos γ . ∂x ∂y ∂z lim

t→ 0

hence: ∂ f ∂ f ∂f ∂ f − → + (a) = ∂ x (a) · cos α + ∂ y (a) · cos β + ∂ z (a) · cos γ . ∂ l Example 6.5.1 Let f : R3 → R be the function given by f (x, y, z) = x · y · z, − → − → − → the point a = (1, −1, 1) ∈ R3 and the unit vector l = √114 i − √314 j − − → f √2 k . Compute ∂ − → (1, −1, 1). 14 We have: ∂∂ ∂ f ∂ x (1,

f x

∂ l

= y · z,

−1, 1) = −1 ,

∂ ∂ ∂ ∂

f = y f y (1,

x · z,

∂ f ∂z

= x · y,

−1, 1) = 1 ,

∂ f ∂z

(1, −1, 1) = −1 and so:

3 2 2 ∂f 1 − √ + √ = − √ . − → (1, −1, 1) = − √ 14 14 14 14 ∂ l Remark 6.5.1 Dfinition 6.5.1 generalizes the definition of the partial derivative. − → − → Indeed, if l = i , namely l = (1, 0, 0), then: ∂f ∂ f ∂ f ∂ f − → + (a) = ∂ x (a) · 1 + ∂ y (a) · 0 + ∂ z (a) · 0, ∂ i hence

6.5 The Directional Derivative. The Differential Operators …

131

∂f ∂ f − →+ = ∂ x . ∂ i Similarly we have: ∂ f ∂ f ∂ f ∂ f − →+ = ∂ y ; − →+ = ∂ z , ∂ j ∂ k where j = (0, 1, 0) and k = (0, 0, 1). Definition 6.5.2 Let D ⊂ R3 be an open subset. The scalar field on D is a fancy name for any scalar function u : D ⊂ R3 → R. If in addition u ∈ C k (D), then we say that u is a scalar field of class C k on D. By vector field on D is meant any − → vectorial function V = (P , Q , R) : D ⊂ R3 → R3 . If P , Q , R ∈ C k (D),we − → say that V is a vector field of class C k on D. Examples of scalar fields are the temperature field, the pressure field, the density field and so on. A typical example of a vector field is the particle velocity field of a moving fluid. − → − → − → Further we consider a fix point O ∈ R3 and the standard basis i , j , k consisting of the unit vectors of the coordinate axes. If we identify any point −−→ − → M(x, y, z) ∈ R3 with its position vector O M, then the vector field V = 3 (P , Q , R) : D → R can be written in the form: − → − → − → V (x, y, z) = P(x, y, z) · i + Q(x, y, z) · j − → + R(x, y, z) · k , ∀ (x, y, z) ∈ D Definition 6.5.3 If u : D ⊂ R3 → R is a scalar field, u ∈ C 1 (D), then the vector · j + ∂∂ux · k is called the gradient of u.Thus we field on D, defined by ∂∂ux · i + ∂u ∂y have: grad u =

∂u  ∂u  ∂u  ·i + ·j + · k ∂x ∂y ∂z

− → − → − → Definition 6.5.4 A vector field V = P i + Q j + R k is said to be a potential field if there exists a scalar field u : D ⊂ R3 → R, u ∈ C 1 (D), such that V = grad u. This means that: P =

∂u ∂u ∂u , Q = , R = . ∂ x ∂y ∂z

Example 6.5.2 The vector field:    ∀(x, y, z) ∈ R3 V = x · z 2 · i + y · z 2 · j + z · x 2 + y 2 · k,

132

6 Differential Calculus of Functions of Several Variables

− → is a potential vector field, since V = grad u, where: u(x, y, z) =

1 · (x 2 + y 2 ) · z 2 , (x, y, z) ∈ R3 . 2

− → Definition 6.5.5 Let V = (P , Q , R) : D ⊂ R3 → R3 be a vector field of C 1 − class on D. Is called the divergence of the vector field V the following scalar field: ∂P ∂Q ∂R − → div V = + + . ∂ x ∂y ∂z A vector field V whose divergence is identically equal to zero is said to be solenoidal. Therefore V is solenoidal if div V = 0 on D. Example 6.5.3 The vector field:  V : R3 → R3 , V = x 2 · y · z · i + x · y 2 · z · j − 2 · x · y · z 2 · k, is solenoidal on R3 , because div V = 2 · x · y · z + 2· x · y · z − 4· x · y · z = 0. Definition 6.5.6 Let V = (P , Q , R) : D ⊂ R3 → R3 be a vector field of C 1 − class on D. − → Is called the curl of the vector field V the following vector fied:

∂Q ∂R − ∂y ∂z

− → · i +



∂P ∂R − ∂z ∂x

− → · j +



∂Q ∂P − ∂x ∂y

− → · k.

− → − → The curl of the vector field V is denoted by curl V .Therefore we have: − → V curl V =



∂Q ∂R − ∂y ∂z

− → · i +



∂R ∂P − ∂z ∂x

− → · j +



∂P ∂Q − ∂x ∂y

− → · k.

− → The expression of the curl of a vector field V = (P , Q , R) can be conveniently written as the symbolic determinant:





i j k



∂ ∂ ∂

∂ x ∂ y ∂z ,



P Q R

where i, j and k are the unit vectors in the directions of the coordinate axes and the ∂ are understood in the sense that the multiplication of such a symbols ∂∂x , ∂∂y and ∂z symbol by a function means the derivation with respect to the corresponding variable, i.e., ∂∂x · Q = ∂∂ Qx . Indeed, expanding (formally) the previous determinant in minors of the first row we obtain:

6.5 The Directional Derivative. The Differential Operators …

133



i j k







∂ ∂ ∂

∂R ∂Q ∂R − ∂P ∂P ∂Q − → → − →

= − · i + − · j + − · k.

∂ x ∂ y ∂z

∂ y ∂ z ∂ z ∂ x ∂ x ∂ y



PQR

−  → − → − → − → − → Example 6.5.4 If V : R3 → R3 , V = x · z 2 · i + y· z 2 · j + z· x 2 + y 2 k , then: − → − → − → − → − → curl V = (2 · y · z − 2 · y · z) · i + (2 · x · z − 2 · x z) · j + 0 · k = 0 .

Definition 6.5.7 Is called the Hamiltonian (nabla) operator the following symbolic operator: ∇ =

∂ ∂ ∂ − → − → − → · i + · j + · k. ∂x ∂y ∂z

Remark 6.5.2 Let u : D ⊂ R3 → R be a scalar field of C 1 − class on D and − → let V = (P , Q , R) : D → R3 be a vector field also of C 1 − class on D.The application of the nabla operator ∇ to the scalar function u can be performed as formal multiplication of the “vector” ∇ by the scalar u:

∂ − ∂ − ∂ − ∂u − → → → → i + j + k u = · i ∂x ∂y ∂z ∂x ∂u − ∂u − → → · j + · k = grad u, + ∂y ∂z

∇u =

− → and the divergence of a vector field V is the formal scalar product of the symbolic − → vector ∇ by the vector V .

∂ ∂ − → i + ∂x ∂y ∂Q ∂P + + = ∂x ∂y

− → div V =

 ∂ − → − → − → − → k · P i + Q j + R k ∂z ∂R − → = ∇ · V ∂z

− → j +

Similarly we have:

∂ − ∂ − ∂ − → → → i + j + k ∂x ∂y ∂z

− − →− →

→ i j k



 −  → − → − →



× P i + Q j + R k = ∂∂x ∂∂y ∂∂z



P Q R

− → curl F = ∇ × V =

134

6 Differential Calculus of Functions of Several Variables

− → i.e.,the curl of the vector field V is the formal vector product of the symbolic vector − → ∇ by the vector V . Definition 6.5.8 Is called the Laplacian operator and denoted by the following differential operator.

=

∂2 ∂2 ∂2 + + ∂ x2 ∂ y2 ∂ z2

∗)

Let D ⊂ R3 be a connected open subset and let u : D ⊂ R3 → R be a scalar field of C 2 class.Then: ∂ 2u ∂ 2u ∂ 2u + + 2 2 ∂ x ∂y ∂ z2

u =

(∗) for the partial derivatives of the second order see Definition 6.6.2 below). The name “laplacian” comes from the French mathematician Pierre Laplace (1749–1827). The connections between the differential operators , grad , div and curl are highlighted by the following remark: Remark 6.5.3 Let u : D ⊂ R3 → R be a scalar field of C 2 − class on the connected, − → open subset D, and let V = (P , Q , R) : D → R3 be a vector field also of C 2 − class on D.Then: 1. 2. 3.

div(grad u) = u − → curl(grad u) = 0 ;  − → div curl V = 0.

6.6 Partial Derivatives and Differentials of Higher Orders Definition 6.6.1 Let A ⊂ R2 be an open subset and let f : A ⊂ R2 → R be a real valued function of two variables. We assume that there exist the first partial derivatives ∂∂ xf : A → R and ∂∂ yf : A → R. Obviously, these derivatives are in turn real functions of two variables. If there exist the partial derivatives of the functions ∂∂ xf , ∂∂ yf , then they are called the second-order partial derivatives of the function f and are denoted as follows: f x2

  = f x x =

f yx =



f x

 y

=

∂ ∂ x ∂ ∂ y



∂ f ∂ x ∂ f ∂ x

=

∂2 f ; ∂ x2

=

∂2 f ∂y∂ x

6.6 Partial Derivatives and Differentials of Higher Orders

f xy

=

f y2 =





f y x

∂ f ∂2 f = ; ∂ y ∂x∂ y

∂ ∂ f ∂2 f = = ∂ y ∂ y ∂ y2

∂ = ∂ x

   fy y

135



Example 6.6.1 Let f (x, y) = x 2 · y 3 − 2 · y · ln x + y x , ∀x > 0, ∀y > 0. Compute. ∂2 f ∂2 f ∂2 f ∂2 f ; . ; ; ∂ x2 ∂y∂ x ∂ x ∂ y ∂ y2 We have successive: y ∂f ∂f = 2 · x · y3 − 2 · + y x · ln y, = 3 · x 2 · y2 ∂x x ∂y − 2 · ln x + x · y x − 1 .  ∂2 f ∂  y 3 x = − 2 · · ln y 2 · x · y + y ∂x2 ∂x x y 3 x = 2 · y + 2 · 2 + y · ln2 y. x  ∂2 f y ∂  = 2x · y 3 − 2 · + y x · ln y ∂y ∂x ∂y x 1 2 + x · y x − 1 · ln y + y x − 1 =6·x ·y − 2 · x  ∂2 f ∂  = 3 · x 2 y 2 − 2 · ln x + x · y x − 1 2 ∂y ∂y = 6 · x 2 · y + x · (x − 1) · y x − 2  ∂2 f ∂  = 3 · x 2 · y 2 − 2 · ln x + x · y x − 1 ∂x ∂y ∂x 1 + y x − 1 + x · y x − 1 · ln y. = 6 · x · y2 − 2 · x Remark 6.6.1 A real function of two variables has four second-order partial deriva2 2 2 2 2 2 tives i.e.: ∂∂ xf2 ; ∂ ∂y ∂f x ; ∂ ∂x ∂f y and ∂∂ yf2 . Two of these, namely ∂ ∂x ∂f y , ∂ ∂y ∂f x are called th mixed partial derivative of the second order and are generally not equal as shown in the following example: Example 6.6.2 Consider the function:  f : R → R , f (x, y) = 2

x· y

x 2 − y2 x 2 + y2

0

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

136

6 Differential Calculus of Functions of Several Variables

After easy-to-follow computations we obtain:

⎧ ⎨ x 2 − y2 4 ·x 2 · y 2 ∂ f y · x 2 + y2 + , if (x, y) = (0, 0) ( x 2 + y 2 )2 . (x , y) = ⎩ ∂ x 0 , if (x, y) = (0, 0) and further:

⎧ ⎨ x 2 − y2 4 ·x 2 · y 2 ∂ f , if (x, y) = (0, 0) x · x 2 + y2 − ( x 2 + y 2 )2 . (x , y) = ⎩ ∂ y 0 , if (x, y) = (0, 0) Then we have: ∂ f ∂2 f ∂2 f (0 , y) = − y; (0 , y) = − 1; (0 , 0) = − 1 ∂ x ∂y∂ x ∂y∂ x ∂ f ∂2 f ∂2 f (x , 0) = x; (x , 0) = 1; (0 , 0) = 1. ∂ y ∂x∂ y ∂x∂ y Therefore,

∂2 f , ∂ y ∂ x (0

0) =

∂2 f , ∂ x ∂ y (0

0).

Definition 6.6.2 Let A ⊂ Rn be an open subset and let f : A ⊂ Rn → R be a real valued function of n variables. We assume that there are ∂∂ xfi : A → R , ∀ i = 1 , n. Obvious, ∂∂ xfi is his turn a real valued function of n variables for any i = 1 , n.   If there exists ∂ ∂x j ∂∂ xfi then, this is called the partial derivative of second-order of the function f and is denoted by f x2 i

instead of mixed.

∂2 f ∂ xi ∂ xi

∂2 f ∂ x j ∂ xi

or f xj xi . If i = j we will note

∂2 f ∂ xi2

or

, and if i = j, the partial second-order derivatives are called

Remark 6.6.2 A real valued function of n variables has n 2 partial second-order derivatives. A sufficient condition for mixed second-order derivatives to be equal is given by the following theorem: Teorema 6.6.1 (Schwarz). Let A ⊂ R2 be an open subset, (a , b) ∈ A and f : A ⊂ 2 2 R2 → R. If there exist the mixed second-order derivatives ∂ ∂x ∂f y and ∂ ∂y ∂f x on an open neighborhood V of the point (a , b),V ⊂ A, and they are continuous at the point (a , b),then: ∂2 f ∂2 f (a , b) = (a , b). ∂x∂ y ∂y∂ x Proof Since A is an open subset, there exists a r > 0 such that :

6.6 Partial Derivatives and Differentials of Higher Orders

V =



(x, y) ∈ R2 ; (x − a)2 + (y − b)2 < r 2

137



⊂ A.

Let R(x, y) = f (x, y) − f (x, b) − f (a, y) + f (a, b) , ∀ (x, y) ∈ V. For any fixed point (x, y) ∈ V we denote by I (respectively J ) the closed interval of ends a and x (respectively b and y). If we denote by: g(t) = f (t, y) − f (t, b) , ∀ t ∈ I, then R(x, y) = g(x) − g(a). From Lagrange’s theorem it results that there exists a point ξ between a and x such that: R(x, y) = g  (ξ )(x − a) =



 f x (ξ, y) − f x (ξ, b) (x − a).

Applying again Lagrange’s theorem to the function f x on J , it results that there exists a point η betweeen b and y such that:  R(x, y) = f yx (ξ, η) (x − a)(y − b).

(6.16)

Similarly, if we denote by h(t) = f (x, t) − f (a, t) , t ∈ J , then: R(x, y) = h(y) − h(b). Applying Lagrange’s theorem twice, it results that there exists η between b and y and ξ  between a and x such that:   R(x, y) = f xy ξ  , η (x − a)(y − b).

(6.17)

Therefore, from (6.16) and (6.17) we have:    f yx (ξ, η) = f xy ξ  , η .

(6.18)

 Taking into account that f yx and f xy are continuous at the point (a , b) it follows:

    f yx lim f yx lim f xy ξ  , η = f xy (a, b). (a, b) = x→a (ξ, η) = x→a y→b

y→b

Corollary 6.6.1 Let A ⊂ R2 be an open subset and f : A ⊂ R2 → R. If f xy and f yx are continuous on A, then f xy (a) = f yx (a), ∀a ∈ A. Remark 6.6.3 Theorem 6.6.1 can be generalized to real functions of n variables as follows: if f : A ⊂ Rn → R, where A ⊂ Rn is an open subset and the mixed 2 2 second-order derivatives ∂ x∂i ∂f x j and ∂ x∂ j ∂f xi there exist on an open neighborhood V of the point a ∈ V , V ⊂ A, and they are continuous at the point a, then:

138

6 Differential Calculus of Functions of Several Variables

∂2 f ∂2 f (a) = (a). ∂ xi ∂ x j ∂ x j ∂ xi The definition of partial than two is obvious. E.g, if  derivatives of order greater ∂2 f ∂3 f ∂ there exists ∂ xk ∂ xi ∂ x j , then it is denoted by ∂ xk ∂ xi ∂ x j . Definition 6.6.3 Any element k = (k1 , ..., kn ) ∈ Nn is called multi-index. ∂ |k| f We will denote by |k| = k1 + ... + kn and by D k f = k1 kn , where ∂ x1 ... ∂ xn

f : A ⊂ Rn → R, and A ⊂ Rn is an open subset.The function f is said to be of C p − class onC p − A, if there exist D k f and are continuous on A, for any multi-index k with |k| ≤ p.We will use the notation f ∈ C p (A). Example 6.6.3 If k = (2 , 1 , 0 , 3), then |k| = 6 and: ∂6 f . ∂ x12 ∂ x2 ∂ x43

Dk f =

The function f ∈ C 5 (A), where A ⊂ R4 is an open subset, if there exist D k f and are continuous on A, for any multi-index k ∈ N4 with |k| ≤ 5. For mixed derivatives of order greater than 2, the possibility of permuting the derivation order is proved by applying Theorem 6.6.1 several times. Example 6.6.4 Let A ⊂ R3 be an open subset and f ∈ C 4 (A).Then we have: ∂4 f ∂2 = ∂z∂x∂ z∂ y ∂ z∂ x ∂ = ∂ x ∂ ∂ = ∂ x ∂



∂2 f ∂z∂ y ∂2 ∂ z∂ y ∂ ∂2 ∂ y∂z ∂

2

∂2 ∂ f = ∂ x∂z ∂ y∂z

f z

f ∂4 f . = z ∂ x ∂ y ∂ z2

Definition 6.6.4 Let A ⊂ Rn be an open subset, a ∈ A and f ∈ C 2 (A). Is called the second-order differential of function f at the point a, and is denoted by d 2 f (a) the following quadratic form on Rn : d 2 f (a)(h) =

n n ! ! i =1 j =1

∂2 f (a) · h i · h j , ∀ h = (h 1 , ..., h n ) ∈ Rn . ∂ xi ∂ x j

Notation is also used: d 2 f (a) =

n n ! ! i =1 j =1

∂2 f (a) · d xi d x j . ∂ xi ∂ x j

6.6 Partial Derivatives and Differentials of Higher Orders

The symmetrical matrix H f (a) =



∂2 f ∂ xi ∂ x j

(a)

matrix attached to the function f at the point a.

139

 1 ≤ i ≤n 1≤ j ≤n

is called the Hessian

Remark 6.6.4 Consider the particular n = 2. Let f : A ⊂ R2 → R be a function of C 2 − class on the open subset A and (a , b) ∈ A. The second-order differential of the function f at the point (a , b) ∈ A is the following quadratic form on R2 : ∂2 f ∂2 f 2 + 2 · b) · h (a, b) · h · k (a, ∂x2 ∂ x∂ y ∂2 f + 2 (a, b) · k 2 , ∂y ∀(h, k) ∈ R2 .

d 2 f (a, b)(h, k) =

or ∂2 f ∂2 f 2 + 2 · b) · d x (a, b) · d x · dy (a, ∂x2 ∂ x∂ y ∂2 f + 2 (a, b) · dy 2 . ∂y

d 2 f (a, b) =

  Example 6.6.5 Let f : (x , y) ∈ R2 ; x > 0 , y > 0 ⊂ R2 → R be the function given by f (x, y) = x 2 · y 3 − 2 · y · ln x + y x . Compute d 2 f (1 , 1) and d 2 f (1 , 1)(2 , − 1). From Example 6.1.1 we deduce: ∂2 f ∂2 f ∂2 f ∂2 f , 1) = , 1) = 4; , 1) = 6 ; (1 (1 , 1) = 5. (1 (1 ∂y∂ x ∂y∂ x ∂ x2 ∂ y2

Therefore we have: d 2 f (1 , 1) = 4 · d x 2 + 10 · d x · d y + 6 ·d y 2 and d 2 f (1 , 1)(2 , − 1) = 2. Remark 6.6.5 Let A ⊂ R3 be an open subset and let f : A ⊂ R3 → R be a function of three variables, f ∈ C 2 (A).Then the differential of second-order of f at the point (a , b , c) ∈ A has the following form: ∂2 f ∂2 f (a, b, c) · d x 2 + 2 (a, b, c) · dy 2 2 ∂x ∂y 2 ∂ f + 2 (a, b, c) · dz 2 ∂z ∂2 f +2· (a, b, c) · d x · dy ∂ x∂ y

d 2 f (a, b, c) =

140

6 Differential Calculus of Functions of Several Variables

∂2 f (a, b, c) · d x · dz ∂ x∂z ∂2 f +2· (a, b, c) · dy · dz. ∂ y∂z +2·

Returning to the general case of real functions of n variables and of C 2 − class, we find that the second order differential of funtion f at the point a is the following quadratic form on Rn : d 2 f (a)(h) =

n n ! ! i =1 j

If we denote by ai j = d 2 f (a)(h) =

∂2 f (a) · h i · h j , ∀ h = (h 1 , ..., h n ) ∈ Rn . ∂ x ∂ x i j =1 ∂2 f ∂ xi ∂ x j

n n ! !

(a), then ai j = a j i , ∀ i , j = 1 , n, and:

ai j · h i · h j , ∀ h = (h 1 , ..., h n ) ∈ Rn .

i =1 j =1

We recall the following definitions and results of linear algebra: the second 2 2 order  2 differential d f (a) is called npositive (negative) definite if d f (a)(h) > 0 n d f (a)(h) < 0 for any h ∈ R , h = 0R . If there exist h = k, such that d 2 f (a)(h) · d 2 f (a)(k) < 0 then, the quadratic form d 2 f (a) is called alternating. Further we will use the notations:



a11 a12 a13







a11 a12

1 = a11 ; 2 = a a ; 3 = a12 a22 a23 ; . .. ; n = det H f (a) . 12 22

a13 a23 a33

Theorem 6.6.2 (Sylvester). The necessary and sufficient condition that d 2 f (a) to be positive (negative) definite is:

i > 0 , ∀ i = 1 , n (respectively (− 1)i i > 0 , ∀ i = 1 , n). Next, we will introduce the following notation convention:  (2) 2 ∂ f instead of ∂∂ xf2 (a , b). , b) (a ∂ x ∂ f ∂ x (a

, b) · ∂∂ yf (a , b) instead of ∂ ∂x ∂f y (a , b).  (2) 2 ∂ f instead of ∂∂ yf2 (a , b). , b) ∂ y (a With this convention, for functions of two variables we have: 2

∂2 f ∂2 f (a , b) · d x · d y (a , b) · d x 2 + 2 · 2 ∂ x ∂x∂ y ∂2 f + (a , b) · d y 2 ∂ y2

d 2 f (a , b) =

6.6 Partial Derivatives and Differentials of Higher Orders

=

141

∂ f ∂ f (a , b) · d x + (a , b) · d y ∂ x ∂ y

(2)

.

To define the differential of the m− order, we will naturally extend the previous convention, namely, we will note with:  (m) m ∂ f instead of ∂∂ x mf (a , b) , b) (a ∂ x ∂m − k f (a ∂ xm − k

, b) ·

∂k f (a ∂ yk

, b) instead of

∂m f ∂ x m − k ∂ yk

(a , b) and so on.

Definition 6.6.5 Let A ⊂ R2 be an open subset, f : A ⊂ R2 → R and (a , b) ∈ A. If f ∈ C m (A), then the mth− order differential of the function f at the point (a , b) is defined as follows: d m f (a , b) = =

m ! k=0

= Cm0 ·

Cmk ·

∂ f ∂ f (a , b) · d x + (a , b) · d y ∂ x ∂ y

∂m f ∂ x m − k ∂ yk

(m)

(a , b) · d x m − k · d y k

∂m f ∂m f m 1 + C · , b) · d x (a , b) · d x m − 1 · d y + · · · + (a m ∂ xm ∂ x m−1 ∂ y +Cmm ·

∂m f (a , b) · d y m . ∂ ym

The following formula that generalizes Newton’s binomial can be proved by mathematical induction: " m! a k1 ... ankn . (a1 + ... + an )m = k1 ! ... kn ! 1 k1 + ...+ kn = m ki ∈N

Using this formula and the previous conventions, we can define the mth− order differential of the function f at the point a as follows: Let A ⊂ Rn be an open subset, f : A ⊂ Rn → R and a = (a1 , ... , an ) ∈ A. If f ∈ C m (A), then the mth− order differential of the function f at the point a = (a1 , ... , an ) ∈ A is: d f (a) = m

=

! k1 + ...+ kn = m ki ∈N

(m) ∂ f ∂ f (a) · d x1 + ... + (a) · d xn ∂ x1 ∂ xn m! ∂m f · (a) · d x1k1 · ... · d xnkn . kn k 1 k1 ! ... kn ! ∂ x1 ... ∂ xn

142

6 Differential Calculus of Functions of Several Variables

6.7 Second-Order Partial Derivatives of Functions Composed of Two Variables Theorem 6.7.1 Let A , B ⊂ R2 be two open subsets, F = (u, v) : A ⊂ R2 → B ⊂ R2 , f : B ⊂ R2 → R and h = f ◦ F : A ⊂ R2 → R, h(x, y) = f (u(x, y), v(x, y)), x, y ∈ A. If F ∈ C 2 (A) and f ∈ C 2 (B), then h ∈ C 2 (A) and we have the following formulas: ∂2 f ∂ 2h = · 2 ∂x ∂u 2



∂u ∂x

2 +2·

∂ 2 f ∂u ∂v · · ∂u∂v ∂ x ∂ x

2 ∂v ∂ f ∂ 2u ∂ f ∂ 2v ∂2 f + . · 2+ · + 2 · ∂v ∂x ∂u ∂ x ∂v ∂ x 2

∂ 2h ∂ 2 f ∂u ∂u ∂ 2 f ∂u ∂v ∂u ∂v · = · + · + · ∂ x∂ y ∂u 2 ∂ x ∂ y ∂u∂v ∂ x ∂ y ∂y ∂x ∂ 2 f ∂v ∂v + 2 · · ∂v ∂ x ∂ y ∂f ∂ 2u ∂f ∂ 2v + · + · . ∂u ∂ x∂ y ∂v ∂ x∂ y 2 ∂2 f ∂u ∂ 2h = · 2 2 ∂y ∂u ∂y 2 2 ∂ f ∂v ∂u ∂v ∂2 f +2 · · · · + ∂u ∂v ∂y ∂y ∂v 2 ∂y 2 2 ∂f ∂f ∂ u ∂ v + + . · · ∂u ∂ y 2 ∂v ∂ y 2 Proof According to (6.10) we have: ∂h ∂ f ∂u ∂ f ∂v = · + · ∂ x ∂u ∂ x ∂v ∂ x ∂h ∂ f ∂u ∂ f ∂v = · + · . ∂ y ∂u ∂ y ∂v ∂ y

(6.19)

We will examine the existence of the derivative ∂∂ xh2 and its computation method. From Theorem 6.3.1 it results that the function (x, y) → ∂ f y) , v(x, y)) has the first order partial derivaties on A(with respect (u(x, ∂ u to x respectively to y), namely: 2

6.7 Second-Order Partial Derivatives of Functions Composed of Two Variables

∂ ∂x



∂f ∂u

143

∂ ∂f ∂u ∂ ∂f ∂v + · · ∂u ∂u ∂x ∂v ∂u ∂x ∂ 2 f ∂u ∂2 f ∂v = · + · ∂u 2 ∂ x ∂v ∂u ∂ x =

(6.20)

Similarly we have: ∂ ∂ x



∂ f ∂v

=

∂2 f ∂u ∂2 f ∂ v · · + . ∂u∂ v ∂ x ∂ v2 ∂ x

(6.21)

From (6.19), (6.20) and (6.21) it results:



∂2 h ∂ ∂h ∂ ∂ f ∂u ∂ f ∂v = = · + · ∂ x2 ∂ x ∂ x ∂ x ∂u ∂ x ∂v ∂ x



2 ∂ ∂ f ∂ ∂ f ∂u ∂ f ∂ u ∂v ∂ f ∂2 v = + · + · · + · ∂ x ∂u ∂ x ∂ u ∂ x2 ∂ x ∂v ∂ x ∂ v ∂ x2

∂2 f ∂u 2 ∂2 f ∂u ∂v ∂ f ∂2 u = · + + · · + · ∂ u2 ∂ x ∂v∂ u ∂ x ∂ x ∂ u ∂ x2

∂2 f ∂v 2 ∂ f ∂2 v ∂u ∂v ∂2 f + · + . · · + · ∂u∂ v ∂ x ∂ x ∂ v2 ∂ x ∂ v ∂ x2 Because in our conditions

∂2 f ∂u∂ v

2 ∂ 2h ∂2 f ∂u = · + 2· 2 2 ∂x ∂u ∂x ∂f ∂f ∂ 2u + + · · 2 ∂u ∂ x ∂v

=

∂2 f , ∂v∂ u

finally we get:

∂2 f ∂u ∂v ∂2 f · · · + ∂u ∂v ∂x ∂x ∂v 2 ∂ 2v . ∂x2



∂v ∂x

2

Similarly, the other two formulas in the theorem statement are proved.     Example 6.7.1 Let f ∈ C 2 R2 , h(x, y) = f 2 · x + y 2 , 1 − y · cos x . 2 2 2 ∀ (x, y) ∈ R2 . Compute ∂∂ xh2 (x, y) , ∂ ∂x ∂h y (x, y) and ∂∂ yh2 (x, y). If we denote by u(x, y) = 2 · x + y 2 , v(x, y) = 1 − y · cos x, then we have: ∂h ∂ f ∂u ∂ f ∂v ∂ f ∂ f = · + · = · 2 + · y · sin x. ∂ x ∂u ∂ x ∂v ∂ x ∂u ∂v ∂h ∂ f ∂u ∂ f ∂v ∂ f ∂ f = · + · = · 2· y + · (− cos x). ∂ y ∂u ∂ y ∂v ∂ y ∂u ∂v ∂ 2h ∂2 f ∂2 f ∂2 f = · 4 + 4 · · y 2 · sin2 x · y · sin x + ∂x2 ∂u 2 ∂u ∂v ∂v 2

144

6 Differential Calculus of Functions of Several Variables

∂f · y · cos x. ∂v   ∂2 f ∂2 f ∂ 2h · 4 · y + = · − 2 · cos x + 2 y 2 · sin x ∂x ∂y ∂u 2 ∂u ∂v ∂2 f ∂f + 2 · (− y · sin x · cos x) + · sin x. ∂v ∂v +

∂ 2h ∂2 f ∂2 f = · 4 · y2 + 4 · · (− y · cos x) 2 2 ∂y ∂u ∂u ∂v ∂2 f ∂f + 2 · cos2 x + · 2. ∂v ∂u Remark 6.7.1 If we will use the following formal notations:  (2) 2 ∂ f instead of ∂∂ uf2 . ∂ u 2 ∂ f · ∂∂ vf instead of ∂ ∂u ∂f v . ∂ u  (2) 2 ∂ f instead of ∂∂ vf2 . ∂ v

then the expressions of the second order partial derivatives of the composed functions are more easily retained in the form:

∂2 h ∂ h (2) = + ∂ x2 ∂ x

∂ 2h ∂h = · ∂x∂ y ∂ x

∂2 h ∂ h (2) = + ∂ y2 ∂ y where by

 ∂h (2) ∂x

∂ f ∂ f ∂2 u ∂2 v + , · · ∂u ∂ x2 ∂v ∂ x2

∂h ∂ f ∂ 2u ∂ f ∂ 2v + · + · , ∂ y ∂u ∂x∂ y ∂v ∂x∂ y ∂ f ∂ f ∂2 u ∂2 v + . · · 2 ∂u ∂ y ∂v ∂ y2

we understand:

∂u ∂ f ∂ v (2) ∂ h (2) ∂ f · + · = ∂ x ∂ u ∂ x ∂ v ∂ x

∂u ∂v ∂2 f ∂2 f ∂2 f ∂u 2 ∂v 2 · · + = · + 2 · , and so on. ∂ x ∂u∂ v ∂ x ∂ x ∂ x ∂ u2 ∂ v2

In the case of Example 6.7.1 we have:

(2) ∂ f ∂ f ∂ h (2) = = · 2 + · y sin x ∂ x ∂u ∂v ∂2 f ∂2 f ∂2 f = · 4 + 4 · y 2 · sin2 x , and so on. · y · sin x + ∂ u2 ∂u∂ v ∂ v2



6.8 Change of Variables

145

6.8 Change of Variables Let I and J two open intervals of R and u : I → J with the properties: (i) (ii) (iii)

u ∈ C k (I ) , k > 1; u  (x) = 0 , ∀ x ∈ I ; u is bijection.

If y : I → R is such that y ∈ C 2 (I ), then the composed function: y = y ◦ u− 1 : J → R is also a function of C 2 − class on J . As y(x) = y (u(x)) , ∀ x ∈ I , it results: y  (x) = y  (u(x)) · u  (x).

(6.22)

If we denote by t = u(x) ∈ J then, the formula (6.22) becomes: dy dt dy = · . dx dt dx

(6.23)

Formula (6.23) shows the connection between the derivation operator with respect to x and the derivation operator with respect to t, namely: d dt d d = · = u  (x) · . dx dx dt dt

(6.24)

Further we have:



d dy dt d y d2 t d    d dy dt y (x) = · = · + · dx dx dt dx dx dt dx d t d x2

dt dt d y d2 t d dy d2 y d y d2 t dt 2 · · + · · = = · + . dt dt dx dx d t d x2 dx d t d x2 d t2

y  (x) =

(6.25) Similarly,if we assume that u and y are functions of C 3 − class, we get the formula: d3 y · y (x) = d t3 



dt dx

3 + 3

d2 y d t d2 t d y d3 t · + . · · d t2 d x d x2 d t d x3

(6.26)

Following the change of the independent variable x  → t : I  J , any form expression E = E x , y , y  , y  , ... becomes E

E t, y,

dy dt

,

d2 y d t2

, ... .

→ =

146

6 Differential Calculus of Functions of Several Variables

  Example 6.8.1 What becomes the following differential equation 1 − x 2 · y  − x · y  + y = 0, if we make the change of variable:  π ? x = cos t , t ∈ 0 , 2 By inversing the function, we get t = cos−1 x , x ∈ (0 , 1) and further: dt 1 1 = − =− √ 2 dx sin t 1 − x



2 d t d dt d 1 dt = = − · d x2 dx dx dt sin t dx

1 d 1 cos t =− · − = − . sin t d t sin t sin3 t From (6.23) and (6.25) we deduce: y = −

1 1 cos t d y d2 y dy − · · · and y  = . 2 2 sin t d t dt sin t d t sin3 t

Substituting in the initial differential equation we obtain: cos t d y cos t d y d2 y − · + · + y = 0, d t2 sin t dt sin t dt or: d2 y + y = 0. d t2   Therefore, following the change of variable x = cos t , t ∈ 0 , π2 , the   2 differential equation 1 − x 2 y  − x y  + y = 0, becomes dd t 2y + y = 0. In formulas (6.23) and (6.25) we noted the new function also with y instead of y, because in an equation it does not matter how the unknown function is noted. Let x → y(x) : I → J be a bijective function, y ∈ C k (I ), k > 1. Further we will show how the derivatives of the inverse function y → x(y) : J → I are computed. According to inverse function therem (Teorema 6.11.1 below), it follows: x  (y) =

1 dx =  dy y (x)

from which we deduce that: y  (x) =

1 . x  (y)

(6.27)

6.8 Change of Variables

147

Next we have: d    y  (x) = y (x) = dx =

−

x  (y) (x  (y))2 x  (y)

= −

d dy



1 x  (y)



dx dy

x  (y) . (x  (y))3

 3 Example. 6.8.2 What becomes the differential equation y · y  + y  = 0 following the inversion of the variables y(x) ↔ x(y)? Taking into account (6.27) and (6.28) the differential equation becomes: y·

1 x  (y) − = 0 (x  (y))3 (x  (y))3

or, after simplification: x  (y) − y = 0. Further we consider the case of the functions of two variables. Let  , D ⊂ R2 be two open and connected subsets. The function: F = (u , v) :  → D, (F(x, y) = (u(x, y) , v(x, y)) , ∀ (x, y) ∈ ), is said to be a change of variables if has the properties: (i) (ii) (iii)

F ∈ C 1 (); F is bijection; The Jacobian of this transformation i.e.: D(u, v) (x , y) = 0 , ∀ (x , y) ∈ . D(x , y)

Written on components the change of variables is given in the form: 

u = u(x , y) , ∀ (x , y) ∈ . v = v(x , y)

Let z :  → R be such that f ∈ C 2 () and let be the composed function: z = z ◦ F − 1 : D → R. Further, if we assume that F ∈ C 2 (), it results that z ∈ C 2 (D) and z = z ◦ F :  → R. As z(x, y) = z (u(x, y) , v(x, y)) , ∀ (x, y) ∈ , and taking into account the derivation rules (6.10) it results:

148

6 Differential Calculus of Functions of Several Variables

∂z ∂z ∂u ∂z ∂v = · + · . ∂ x ∂u ∂ x ∂v ∂ x ∂z ∂z ∂u ∂z ∂v = · + · . ∂ y ∂u ∂ y ∂v ∂ y

(6.28)

And:

∂u 2 ∂2z ∂u ∂v + 2 · · ∂x ∂u ∂v ∂ x ∂ x 2 ∂v ∂z ∂ 2 u ∂z ∂2z · + + · + 2 ∂v ∂x ∂u ∂ x 2 ∂v ∂2z ∂ 2 z ∂u ∂u = 2 · · ∂x ∂y ∂u ∂x ∂y

∂2z ∂u ∂v ∂u ∂v + · · + · + ∂u ∂v ∂x ∂y ∂y ∂x ∂z ∂ 2u ∂z ∂ 2v + · + · ∂u ∂ x ∂ y ∂v ∂ x ∂ y

∂2z ∂u 2 ∂2z ∂2z ∂u ∂v = · + 2 · · ∂ y2 ∂u 2 ∂y ∂u ∂v ∂ y ∂ y ∂z ∂ 2 u ∂z ∂ 2 v + · 2 + · ∂u ∂ y ∂v ∂ y 2 ∂2z ∂2z = · ∂x2 ∂u 2

·

∂ 2v ∂x2

∂ 2 z ∂v ∂v · · + ∂v 2 ∂ x ∂ y (6.29) ∂2z · + ∂v 2



∂v ∂y

2

Let us consider the expression:

∂ z ∂ z ∂2 z ∂2z ∂2 z E = E x, y, z, , , , , (x, y) ∈ . ∂ x ∂ y ∂ x 2 ∂ x ∂ y ∂ y2 Following the change of variables F, the expression E becomes: E = E



∂z ∂2z ∂ z ∂2 z ∂2 z u, v, z, , , , , , where (u , v) ∈ D ∂ u ∂ v ∂ u 2 ∂ u ∂ v ∂ v2

Example 6.8.3 1.

What becomes the equation:

x 2 ∂∂ xz2 − 2 x y variables: 2

∂2z ∂x∂ y

+ y 2 ∂∂ 

2

z y2

+ 2x

∂z ∂ x

= 0 following the change of

u= x· y , y = 0 ? v= y

Taking into account the formulas (6.29), (6.30) we get successively:

6.8 Change of Variables

149

∂z ∂z = · y ∂ x ∂u ∂z ∂z ∂z = · x + ∂ y ∂u ∂v ∂2 z ∂2 z = · y2 ∂ x2 ∂ u2 ∂2z ∂2z ∂2 z ∂z · xy + = · y + 2 ∂x∂ y ∂u ∂u∂ v ∂u ∂2 z ∂2 z ∂2z ∂2 z 2 = · x + 2 . · x + ∂ y2 ∂ u2 ∂u∂ v ∂ v2 Substituting these partial derivatives into the given equation is found: v2 · 2.

∂2 z ∂2 z = 0 or = 0. ∂ v2 ∂ v2

Find the expression of the Laplacian operator in polar coordinates. In other 2 2 words, what becomes the expression z = ∂∂ xz2 + ∂∂ yz2 after the change of variables:  x = ρ · cos θ , ρ > 0 (6.30) y = ρ · sin θ

To begin with we inverse the transformation (6.31) to obtain the new coordinates (ρ , θ) according to the old coordinates (x , y), as the theory is presented. We immediately notice that: 

ρ= θ =



x 2 + y2 , x · y = 0. tan−1 xy

We have successively: ∂ρ x y ∂ρ , . =  =  ∂ x x 2 + y2 ∂ y x 2 + y2 ∂ 2ρ y2 ∂ 2ρ x2 = , = 3    3 . 2 ∂ x2 x 2 + y2 2 ∂ y x 2 + y2 2 ∂θ ∂θ −y x , . = 2 = 2 ∂ x x + y2 ∂ y x + y2

(6.31)

150

6 Differential Calculus of Functions of Several Variables

∂ 2θ 2x y ∂ 2θ −2x y = , =    2 . 2 2 2 2 2 ∂ x2 ∂ y x + y x + y2 ∂z ∂z x −y ∂z + . = ·  · 2 ∂ x ∂ρ ∂θ x + y2 x 2 + y2 ∂z ∂z ∂z y x + . = ·  · 2 2 2 ∂ y ∂ρ ∂ θ x + y2 x + y ∂2z ∂2z x2 ∂2z −x · y = · + 2 · ·  3 2 2 2 2 ∂x ∂ρ x +y ∂ρ ∂θ x 2 + y2 2 +

∂2z y2 ∂z ∂z y2 2x · y ·  ·  2 + 2 .  23 + ∂θ ·  2 2 2 2 ∂θ ∂ρ 2 2 x +y x + y2 x +y

x· y ∂2 z y2 ∂2z ∂2 z x2 ∂2 z ·  = · 2 + 2· + ·  2 3  2 2 2 2 ∂ρ∂ θ ∂ y ∂ ρ x + y ∂ θ x 2 + y2 x 2 + y2 2 +

x2 ∂z ·  3 ∂ ρ x 2 + y2 2

z =

+

−2 · x · y ∂z ·  2 . ∂ θ x 2 + y2

∂2 z ∂2 z 1 ∂z 1 + · 2 + . ·  2 ∂ ρ2 ∂ θ2 x + y2 ∂ρ x + y2

Finally, taking into account (6.32) we obtain the expression of the Laplacian in polar coordinates:

z =

∂2 z 1 ∂z 1 ∂2 z + · 2 + · . 2 2 ∂ρ ∂θ ρ ∂ρ ρ

6.9 Taylor’s Formula for Functions of Several Variables Forn any a, b ∈ Rn , we denote by [a , b] {(1 − t) · a + t · b ; t ∈ [0, 1]}. The subset [a , b] ⊂ Rn is called the closed segment of heads a and b. The open segment of heads a and b is denoted by. (a , b) = {(1 − t) · a + t · b ; t ∈ (0, 1)}.

=

Definition 6.9.1 A subset A ⊂ Rn is called convex if for any a , b ∈ A it results that [a , b] ⊂ A. Remark 6.9.1 Any open (closed) n-dimentional ball in Rn is a convex subset. Indeed, if x , y ∈ B(a, r ),then x − a < r and y − a < r . For any t ∈ [0, 1] we have:

6.9 Taylor’s Formula for Functions of Several Variables

151

(1 − t) · x + t · y − a = (1 − t) · (x − a) + t · (y − a) ≤ ≤ (1 − t) · x − a + t · y − a < (1 − t) · r + t · r = r , hence [x, y] ⊂ B(a, r ). From Remark 6.9.1, we deduce that any interval in R, any disk (square)in R2 , any sphere (cube) in R3 are convex subsets. Theorem 6.9.1 (Taylor’s formula for functions of n variables). Let f : A ⊂ Rn → R,a ∈ A be an interior point and let r > 0 be such that. V = B(a, r ) ⊂ A. If f ∈ C m + 1 (V ) then, for any x ∈ V there exists ξ ∈ (a , x) such that: 1 2 1 d f (a)(h) + d f (a)(h) + · · · + 1! 2! 1 m 1 + d f (a)(h) + · d m+1 f (ξ )(h), m! (m + 1)!

f (a + h) = f (a) +

where h = x − a. Proof To simplify the writing, we prove the theorem in the particular case n = 2..Let a = (a1 , a2 ) ∈ A be an interior point, x = (x1 , x2 ) ∈ V a fix point and h = (h 1 , h 2 ) = = (x1 − a1 , x2 − a2 ) = x − a. For any t ∈ [0 , 1] we denote by: u 1 (t) = a1 + t · h 1 = a1 + t · (x1 − a1 ) u 2 (t) = a2 + t · h 2 = a2 + t · (x2 − a2 ) and we consider the composed function: g : [0, 1] → R , g(t) = f (u 1 (t), u 2 (t)). Obvious, g is a function of C m + 1 − class on [0, 1], and according to Maclaurin’s formula for function of one variable with Lagrange’s remainder (Cap. 4, Sect. 4.4), there exists 0 < θ < 1 such that: g(1) = g(0) + +

g  (0) g  (0) g (m) (0) + + ... + + 1! 2! m!

g (m+1) (θ ) . (m + 1)!

(6.32)

We remark that: g(1) = f (x) = f (a + h) and g(0) = f (a). On the other hand, we have:

(6.33)

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6 Differential Calculus of Functions of Several Variables

g  (t) =

∂ f ∂ f (u 1 (t), u 2 (t)) · u 1 (t) + (u 1 (t), u 2 (t)) · u 2 (t) ∂ x1 ∂ x2

or: g  (t) =

∂ f ∂ f (u 1 (t), u 2 (t)) · h 1 + (u 1 (t), u 2 (t)) · h 2 , ∂ x1 ∂ x2

hence: g  (0) =

∂ f ∂ f (a) · h 1 + (a) · h 2 = d f (a)(h) ∂ x1 ∂ x2

(6.34)

Also,from Theorem 6.7.1, it results: g  (t) =

 2 ∂2 f (u 1 (t), u 2 (t)) · u 1 (t) 2 ∂ x1

∂2 f (u 1 (t), u 2 (t)) · u 1 (t) · u 2 (t) ∂ x1 ∂ x2  2 ∂f ∂2 f + (u 1 (t), u 2 (t)) · u 2 (t) + (u 1 (t), u 2 (t)) · u 1 (t) 2 ∂ x1 ∂ x2

+ 2

+

∂ f (u 1 (t), u 2 (t)) · u 2 (t) . ∂ x2

Since u 1 (t) = u 2 (t) = 0, it follows: g  (0) =

∂2 f ∂2 f ∂2 f 2 + 2 · h + · h · h (a) (a) (a) · h 22 1· 2 1 ∂ x1 ∂ x2 ∂ x12 ∂ x22

(6.35)

= d 2 f (a)(h). It can be shown that: g (k) (0) = d k f (a)(h) , ∀ k ∈ N∗

(6.36)

If we denote by ξ = a + h · θ and taking into account (6.34)–(6.37) in (6.33) we get: 1 1 d f (a)(h) + d 2 f (a)(h) + · · · + 1! 2! 1 m 1 + d f (a)(h) + + d m+1 f (ξ )(h). m! (m + 1)!

f (x) = f (a) +

where ξ ∈ (a , x), that is, we obtained the formula from the statement of the theorem. Remark 6.9.2 Let us consider the particular case n = 2.

6.9 Taylor’s Formula for Functions of Several Variables

153

Let f : A ⊂ R2 → R, (a , b) ∈ A an interior point and let V be a neighborhood of the point (a , b). If f ∈ C m + 1 (V ), then for any (x , y) ∈ V , there exists a point (ξ , η) ∈ V , with ξ ∈ (a , x) and η ∈ (b , y), such that: $ ∂ f ∂ f (a , b) · (x − a) + (a , b) · (y − b) ∂ x ∂ y # $(m) ∂ f 1 ∂ f + ··· + (a , b) · (x − a) + (a , b) · (y − b) m! ∂ x ∂ y $(m + 1) # ∂ f 1 ∂ f + . (ξ , η) · (x − a) + (ξ , η) · (y − b) ∂ y (m + 1)! ∂ x

f (x , y) = f (a , b) +

1 1!

#

Remark 6.9.3 In the hypotheses of Theorem 6.9.1 and Remark 6.9.2, if a = 0, then the Taylor’s formula is called the Maclaurin’s formula. Example 6.9.1 Write the third degree Maclaurin’s formula for the function: f : R2 → R , f (x , y) = e y · cos x We have successive: f x = − e y · sin x , f x (0 , 0) = 0 , f y = e y · cos x , f y (0 , 0) = 1 f x2 = − e y · cos x , f x2 (0 , 0) = −1 , f y2 = e y · cos x , f y2 (0 , 0) = 1 f xy = − e y · sin x , f xy (0 , 0) = 0 , f (0 , 0) = 1 f x3 = e y · sin x , f x3 (ξ , η) = eη · sin ξ , f x2 y = − e y · cos x , f x2 y (ξ , η) = − eη · cos ξ f y3 = e y · cos x , f y3 (ξ , η) = eη · cos ξ , f xy 2 = − e y · sin x , f xy 2 (ξ , η) = − eη · sin ξ . The third degree Maclaurin’s formula is: & 1 %  · f x (0 , 0) · x + f y (0 , 0) · y f (x , y) = f (0 , 0) + 1! & 1 %  2 · f x 2 (0 , 0) · x + 2 · f x y (0 , 0) · x · y + f y2 (0 , 0) · y 2 + 2! ⎡  ⎤ 3  2  2 1 ⎣ f x 3 (ξ , η) · x + 3 · f x 2 y (ξ , η) · x · y + 3 · f x y 2 (ξ , η) · x · y + ⎦ · + 3! + f y3 (ξ , η) · y 3

or: 1 1 1 · x2 + · y2 + · (eη · sin ξ · x 3 2 2 6 −3 · eη · cos ξ · x 2 · y − 3 · eη · sin ξ · x · y 2 + eη · cos ξ · y 3 ) f (x , y) = 1 + y −

154

6 Differential Calculus of Functions of Several Variables

Theorem 6.9.2 (Lagrange’s Theorem). Let f : A ⊂ Rn → R be a function of n variables, a = (a1 , ..., an ) ∈ A an interior point, and V an open and convex neighborhood of a, V ⊂ A. If f ∈ C 1 (V ), then for any x = (x1 , ..., xn ) ∈ V there exists ξ ∈ (a , x) such that: f (x) − f (a) =

∂ f ∂ f (ξ ) · (x1 − a1 ) + ... + (ξ ) · (xn − an ). ∂ x1 ∂ xn

Proof The proof follows immediately from Theorem 6.9.1 for the particular case m = 0. If n = 2 we get: ∂ f (ξ1 , ξ2 ) · (x1 − a1 ) ∂ x1 ∂ f + (ξ1 , ξ2 ) · (x2 − a2 ). ∂ x2

f (x1 , x2 ) − f (a1 , a2 ) =

6.10 Local Extrema of a Function of Several Variables Definition 6.10.1 We shall say that a ∈ A is a local maximum (minimum) point of the function f : A ⊂ Rn → R, if there exists a neighborhood V of the point a, such that: f (x) ≤ f (a) (respectively, f (x) ≥ f (a)), for any x ∈ V ∩ A. As with the functions of one variable, a local maximum (minimum) point is called a local extreme point. Definition 6.10.2 Let f : A ⊂ Rn → R be a function of n variables, differentiable at the interior point a ∈ A. If d f (a) = 0, then a is called critical (stationary) point for the function f . Observatia 6.10.1. The critical points of a differentiable function f : A ⊂ Rn → R are determined by solving the following system: ⎧ ∂ f ⎪ ⎨ ∂ x1 (x1 , ..., xn ) = 0 ...................... ... ... ⎪ ⎩ ∂ f (x , ..., x ) = 0 . 1 n ∂ xn Theorem 6.10.1 (Fermat’s Theorem). Let f : A ⊂ Rn → R be a function of n variables, differentiable at the interior point a ∈ A. If the point x = a is a local extreme point for f , then x = a is a critical point for f , i.e. d f (a) = 0. Proof Let us denote by Ai = {t ∈ R ; (a1 , ..., ai and let us consider the functions of one variable:

− 1,

t, ai

+ 1,

..., an ) ∈ A}

6.10 Local Extrema of a Function of Several Variables

155

  ϕi : Ai → R, ϕi (t) = f a1 , ..., ai − 1 , t, ai + 1 , ..., an , t ∈ Ai , i = 1 , n

Obvious, ϕi is differentiable at the point t = ai , and this point is a local extreme point for ϕi . According to Fermat’s Theorem for functions of one variable it results that ϕi (ai ) = 0, hence ∂∂ xfi (a) = 0, for any i = 1 , n.Therefore d f (a) = 0, and so x = a is a critical point for f . As with the functions of one variable, not every critical point is an extreme local point. The following theorem establishes sufficient conditions for a critical point to be an extreme local point. Theorem 6.10.2 Let f : A ⊂ Rn → R, a ∈ A an interior point, and let r > 0 be such that V = B(a, r ) ⊂ A.We assume also that f ∈ C 2 (V ) and that x = a is a critical point for f .Then we have: (1) (2) (3)

If d 2 f (a) is positive definite, then x = a is a local minimum point for f. If d 2 f (a) is negative definite, then x = a is a local maximum point for f . If d 2 f (a) is alternating on any neighborhood V of the point a, then the point a does not is a local extreme point of f .

Proof Let x ∈ V be an arbitrary fixed point and h = x − a. From Theorem 6.9.1, it follows that there exists ξ ∈ (a , x) such that: f (x) = f (a) + and so: f (x) − f (a) = If we denote by αi =

1 2

1 2 1 d f (a)(h) + d f (ξ )(h), 1! 2!

· d 2 f (ξ )(h) =

hi  h 2 ,

then

n " i =1

1 2

·

n n " " i =1 j =1

∂2 f ∂ xi ∂ x j

(ξ ) · h i · h j .

αi2 = 1, and:

n n  h 2 ! ! ∂ 2 f f (x) − f (a) = · (ξ ) · αi · α j . 2 ∂ xi ∂ x j i =1 j =1

(6.37)

If we denote also by: ω(x) =

n # n ! ! i =1 j =1

$ ∂2 f ∂2 f (ξ ) − (a) · αi · α j ∂ xi ∂ x j ∂ xi ∂ x j

and taking into account that f ∈ C 2 (V ), it results that lim ω(x) = 0. The formula x →a (6.38) becomes: f (x) − f (a) =

  h 2  2 d f (a)(α) + ω(x) . 2

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6 Differential Calculus of Functions of Several Variables

Let:  S = x = (x1 , ..., xn ) ∈ R ; n

n !

, xi2

= 1 ⊂ Rn .

i =1

Obvious S is a closed and bounded subset, hence a compact subset. The application α → d 2 f (a)(α) : S → R is continuous (being a polynomial function), so it is bounded and attains its bounds on S, according to Theorem 5.6.2. (1)

Let us suppose that d 2 f (a) is positive definite, and let α0 ∈ S be such that: m = inf d 2 f (a)(α) = d 2 f (a)(α0 ) > 0. α∈S

As lim ω(x) = 0, it follows that there exists a neighborhood V1 of the point a, x →a

V1 ⊂ V , such that |ω(x)|
 h2 · m − m2 ≥ 0. 2 It results that f (x) ≥ f (a) , ∀ x ∈ V1 , hence a is a local minimum point for 2

f. (2)

(3)

If d 2 f (a) is negative definite, it is proved as in (1) that there is a neighborhood V2 ⊂ V of the point a, such that f (x) ≤ f (a) , ∀ x ∈ V2 , where results that the point a is a local maximum point for f . If d 2 f (a) is alternating, the difference f (x) − f (a) does not keep a constant sign on any neighborhood of the point a, so the point a is not an extreme local point for f . In the particular case of real functions of two variables, the nature of the local extreme points can be determined with the following theorem:

Theorem 6.10.3 Let f : A ⊂ R2 → R, and (a , b) ∈ A an interior point. We assume also that (a, b) is a critical point for f and that there exists a neighborhood V of the point (a , b) such that f ∈ C 2 (V ). Then: (1)

If

∂2 f ∂ x 2 (a

, b) > 0 and ∂2 f ∂2 f , b) · (a (a , b) − ∂ x2 ∂ y2

(2)

∂2 f (a , b) ∂ x∂ y

$2 > 0,

then (a , b) is a local minimum point for f . 2 If ∂∂ xf2 (a , b) < 0 and ∂2 f ∂2 f , b) · (a (a , b) − ∂ x2 ∂ y2

(3)

#

#

∂2 f (a , b) ∂ x∂ y

then (a , b) is a local maximum point for f . % 2 &2 2 2 If ∂∂ xf2 (a , b) · ∂∂ yf2 (a , b) − ∂ ∂x ∂f y (a , b) < 0,

$2 > 0,

6.10 Local Extrema of a Function of Several Variables

157

then (a , b) does not is a local extreme point for f . Remark 6.10.2 In the hypotheses of Theorem 6.10.3, if: ∂2 f ∂2 f , b) · (a (a , b) − ∂ x2 ∂ y2

#

∂2 f (a , b) ∂ x∂ y

$2 = 0,

we cannot say anything about the point (a , b); there are cases where it is the extreme point of the function, there are cases when it is not. In the particular case of real functions of three real variables, the nature of the extreme points can be determined with the following theorem: Theorem 6.10.4 Let f : A ⊂ R3 → R, and a ∈ A an interior point. We assume also that a is a critical point for f and that there exists a neighborhood V of this point„ V ⊂ A such that f ∈ C 2 (V ). If we denote by:

2

∂ f

2

1 =, 2 = ∂∂2x f

∂ x∂ y

∂2 f ∂ x∂ y ∂2 f ∂ y2

∂2 f

2

∂2x





, 3 = ∂∂x∂fy

∂2 f



∂2 f ∂2 f ∂ x∂ y ∂ x∂z ∂2 f ∂2 f ∂ y 2 ∂ y∂z ∂2 f ∂2 f ∂ x∂z ∂ y∂z ∂z 2











(where the all partial derivatives of the function f are computed at the point a). then: (1) (2)

If 1 > 0 , 2 > 0 , 3 > 0, then a is a local minimum point for f . If 1 < 0 , 2 > 0 , 3 < 0, then a is a local maximum point for f .

Example 6.10.1 Find the local extreme points of the follloing functions: 1.

f : R2 → R , f (x , y) = x 3 + y 3 + 21 x y + 36 x + 36 y − 3. By solving the system: ⎧ ∂ ⎪ ⎪ ⎨ ∂ ∂ ⎪ ⎪ ⎩ ∂

f (x, y) = x 2 + 7 · y + 12 = 0 x f (x, y) = y 2 + 7 · x + 12 = 0 y

we obtain two critical points i.e., M1 (− 4, − 4) and M2 (− 3, − 3). Further we have: d 2 f (x , y) = 6 · x · d x 2 + 42 · d x · d y + 6 · y · d y 2 . d 2 f (− 4, − 4) = − 24 · d x 2 + 42 · d x · d y − 24 · y · d y 2 .

158

6 Differential Calculus of Functions of Several Variables

1



−24 21



= 135 > 0. = −24 < 0 , 2 =

21 −24

As d 2 f (− 4, − 4) is negative definite we deduce that the point M1 (− 4, − 4) is a local maximun point for f . On the other hand we have: d 2 f (− 3, − 3) = − 18 · d x 2 + 42 · d x · d y − 18 · y · d y 2 .

1



−18 21



= − 117 < 0. = −18 < 0 , 2 =

21 −18

As d 2 f (− 3, − 3) is altrenating it results that the point M2 (− 3, − 3) does not is a local extreme point for f . 2.

f : R2 → R , f (x , y) = x 4 + y 4 − 2 · x 2 − 2 · y 2 + 4 · x · y. We have successively: ∂ f ∂ f = 4 · x3 − 4 · x + 4 · y ; = 4 · y 3 + 4 · x − 4 · y, ∂ x ∂ y ∂2 f ∂2 f ∂2 f = 12 · x 2 − 4 ; = 12 · y 2 − 4 ; = 4. 2 2 ∂ x ∂ y ∂ x∂ y The critical point are M1 d2 f

√

2, −

√  √ √  √  2, − 2 ; M2 − 2, 2 and M3 (0 , 0).

√  2 = 20 · d x 2 + 8 · d x · d y + 20 · d y 2 .



√

20 4

√ 

= 384 > 0, so M1 2, − 2 is a local

1 = 20 > 0 , 2 =

4 20

minimum point.  √ √  √ √  d 2 f − 2, 2 = 20 · d x 2 + 8 · d x · d y + 20 · d y 2 = d 2 f 2, − 2 ,

 √ √  It results that M2 − 2, 2 is also a local minimum point. d 2 f (0, 0) = −4 · d x 2 + 8 · d x · d y − 4 · d y 2 .

6.11 Local Inversion Theorem

1

159



−4 4



= 0. = − 4 < 0 , 2 =

4 −4

Because, 2 = 0, we cannot apply Theorem 6.10.3 to the point M3 (0 , 0). We can figure out whether or not the point is an extreme point starting from  the definition. Indeed, we notice that f (0, 0) = 0, f (x, 0) = x 2 · x 2 − 2 and f (x, x) = 2 · x 4 . It results that in any neighborhood of the point (0, 0) there are points where f > 0, and points where f < 0, while f (0, 0) = 0. Therefore, M3 (0 , 0) does not is a local extreme point for f . f : R3 → R , f (x , y , z) = x 2 + y 2 + z 2 − x · y + x − z.

3.

By solving the system: .  we find only a critical point, namely M − 23 , − Further we have:

1 3

,

1 2

 .

∂2 f ∂2 f ∂2 f ∂2 f ∂2 f ∂2 f = = = 2 , = −1 , = = 0. ∂ x2 ∂ y2 ∂ z2 ∂ x∂ y ∂ x∂ z ∂ y∂ z   d 2 f (− 1, − 2 , 2) = 2 · d x 2 + d y 2 + d z 2 − d x d y .

1





2 −1 0



2 −1



= 3 > 0 , 3 = −1 2 0 = 6 > 0. = 2 > 0 , 2 =



−1 2

0 02

  As, d 2 f − 23 , − 13 , 21 is positive definite, it results that:   M − 23 , − 13 , 21 is a local minimum point.

6.11 Local Inversion Theorem Definition 6.11.1 Let A , B ⊂ Rn be two open subsets. A function F : A → B is called diffeomorphism if it has the properties: (a) (b) (c)

F is bijective, F ∈ C 1 (A), F − 1 ∈ C 1 (B).

Such a definition is necessary because there are functions that satisfy conditions (a) and (b) but do not meet condition (c). Example 6.11.1 The function f diffeomorphism.

:

R → R , f (x)

=

x 3 does not is

160

6 Differential Calculus of Functions of Several Variables

Obvious, f is bijective and f ∈ C 1 (R), but f is not diffeomorphism because its √ inverse function f − 1 : R → R , f −1 (y) = 3 y, is not derivable at −1 1 ∈ / C (R). y = 0, therefore f Remark 6.11.1 Let A , B and D three open subsets of Rn and F : A → B, G : B → D two diffeomorphisms. Then the composed function H = G ◦ F : A → D is also diffeomorphism. The statement results from the fact that a composition of C 1 − class functions is also a function of C 1 − class, and by the ramark that: (G ◦ F)− 1 = F − 1 ◦ G − 1 . Example 6.11.2 The simplest example of diffeomorphism is the translation operator. Let y ∈ Rn be an arbitraty fixed point, and let τ y : Rn → Rn defined by: τ y (x) = x + y , ∀ x ∈ Rn . Obvious τ y ∈ C 1 (Rn ), τ y is bijective and τ y− 1 = τ−y ∈ C 1 (Rn ). The following result is known as the local inversion theorem for functions of one variable. Teorema 6.11.1 Let A ⊂ R be an open subset, a ∈ A and f : A ⊂ R → R. If f ∈ C 1 (A) and f  (a) = 0, then there exists an open neighborhood U of the point a, U ⊂ A, such that f  (x) = 0, ∀ x ∈ U , f : U → V = f (U ) is a diffeomorphism and: 

 f −1 (y) =

1 , ∀ y = f (x) ∈ V. f  (x)

Proof Let us suppose that f  (a) > 0. As f  is continuous at the point a, it results that there exists an open interval U containing the point a, U ⊂ A such that f  (x) > 0 , ∀ x ∈ U ( See Theorem 5.5.4). If we denote by V = f (U ), then V is an open interval and f : U → V is it is known that if f  = 0 bijective because is strictly increasing. On the other hand  −1  −1 : V → U is derivable on V ,and f on U , then f (y) = f 1(x) , ∀ y ∈ V, y = f (x).Obvious f −1 ∈ C 1 (V ), therefore f is a diffeomorphism. Next we present without proof the local inversion theorem for vector functions. Theorem 6.11.2 (Inverse function theorem). Let A ⊂ Rn be an open subset, a ∈ A and F = ( f 1 , ... , f n ) : A ⊂ Rn → Rn . D( f 1 , ..., f n ) If F ∈ C 1 (A) and the jacobian det JF (a) = D(x (a) = 0, then 1 , ..., x n ) there exists an open neighborhood U of the point a with the properties: U ⊂ A, D( f 1 , ..., f n ) = 0 , ∀ x ∈ U and F : U → V = F(U ) is diffeomorphism. D(x1 , ..., xn ) (x) More, if F − 1 = G = (g1 , ... , gn ) : V → U , then: D(g1 , ..., gn ) (y) = D(y1 , ..., yn )

1 D( f 1 , ..., f n ) D(x1 , ..., xn ) (x)

, ∀ y ∈ V , y = F(x).

6.12 Regular Transformations

161

6.12 Regular Transformations Definition 6.12.1 Let F = ( f 1 , ... , f n ) : A ⊂ Rn → Rn be a vector function. We say that F is a regular transformation at the point a ∈ A,if there exists an open neighborhood U of the point a, U ⊂ A with the properties: F ∈ C 1 (U ) and det JF (a) =

D( f 1 , ..., f n ) (a) = 0. D(x1 , ..., xn )

We say that F is a regular transformation on the set A if F is a regular transformation at any point of A. Proposition 6.12.1 If A ⊂ Rn is an open subset and F : A ⊂ Rn → Rn is a regular transformation at the point a ∈ A, then F is continuous at the point a. Proof If F is a regular transformation at the point a, then, according to Theorem 6.1.3, F is differentiable at a, hence F is continuous at a, as it results from Proposition 6.1.4. Proposition 6.12.2 Let A , B ⊂ Rn be two open subsets. If F : A ⊂ Rn → B ⊂ Rn is a regular transformation at the point a ∈ A and G : B ⊂ Rn → Rn is a regular transformation at the point b = F(a) ∈ B, then the composed function H = G ◦ F : A → Rn is a regular transformation at the point a ∈ A. D( f 1 , ..., f n ) Proof From the hypothesis it results that D(x (a) = 0, where f 1 , ... , f n 1 , ..., x n ) are the scalar components of the vector function F, and that there is an open neighborhood U of the point a,U ⊂ A, such that F ∈ C 1 (U ). Also there exists an open neighborhood V of the point b = F(a) ∈ B, V ⊂ B such that G ∈ C 1 (V ) and D(g1 , ..., gn ) = 0, where g1 , ... , gn are the scalar components of G. D(y1 , ..., yn ) (b) Since F is continuous at the point a, it follows that there is an open neighborhood U1 of a, U1 ⊂ U such that F(U1 ) ⊂ V . Obvious H = G ◦ F is of C 1 − class on U1 . On the other hand, we have:

det J H (a) = det JG (b) · det JF (a) = 0, hence H is a regular transformation at the point a ∈ A. If we denote by h 1 , ... , h n the scalar components of the vector function H, then we get the equality: D(h 1 , ..., h n ) D(g1 , ..., gn ) D( f 1 , ..., f n ) (a) = (b) · (a). D(x1 , ..., xn ) D(y1 , ..., yn ) D(x1 , ..., xn ) The following theorem highlights a remarkable property of regular transformations, namely that the direct image of an open set by a regular transformation is also an open set.

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6 Differential Calculus of Functions of Several Variables

Theorem 6.12.1 Let F : Rn → Rn be a regular transformation on Rn . If A ⊂ Rn is an open subset then the subset F(A) ⊂ Rn is also an open subset. Proof Let b ∈ F(A) arbitrary, and a ∈ A such that b = F(a). From Theorem 6.11.2 it results that there exists an open neighborhood U of the point a, U ⊂ A and an open neighborhood V = F(U ) of the point b such that F : U → V is diffeomorphism. Obvious, V ⊂ F(A), hence b is an interior point of the set F(A). As the point b was arbitrary it results that F(A) is an open subset. Definition 6.12.2 An open and connected subset of Rn is called domainRn . Proposition 6.12.3 Let F = ( f 1 , ... , f n ) : Rn → Rn be a regular transformation on Rn . If D ⊂ Rn is a domain, then F(D) is also a domain and the jacobian D( f 1 , ..., f n ) keeps constant sign on D. D(x1 , ..., xn ) Proof The fact that F(D) is an open subset of Rn follows from Theorem 6.12.1. On the other hand, from Proposition 6.12.1 it results that F is continuous on Rn . As a continuous function transforms a connected set into a connected set (Theorem 5.6.5), it follows that F(D) is connected, so F(D) is a domain. D( f 1 , ..., f n ) : Rn → R is continuous and D is a connected Since the function D(x 1 , ..., x n ) D( f 1 , ..., f n ) subset it results that the set D(x (D) is a connected subset of R, so an interval 1 , ..., x n ) (Theorem 5.6.2). If we suppose that there are two points u , v ∈ D such that D( f 1 , ..., f n ) D( f 1 , ..., f n ) < 0 and D(x (v) > 0 then, from Corollary 5.6.2. it results D(x1 , ..., xn ) (u) 1 , ..., x n ) that there is a point w ∈ D such that such that:

D( f 1 , ..., f n ) (w) = 0, D(x1 , ..., xn ) which contradicts the hypothesis that F is regular.

6.13 Implicit Functions Let us consider the rectangle D = [a , b] × [c , d] ⊂ R2 and the equation: F(x , y) = 0 , (x , y) ∈ D.

(6.38)

We wonder if for any x ∈ [a , b] there is only one value y ∈ [c , d] so that the pair (x , y) verifies the Eq. (6.39). If this happens, we will note by y(x) the value y corresponding to value x. The function x → y(x) : [a , b] → [c , d] it is said to be implicitly defined by Eq. (6.39). Obviously we have: F(x , y(x)) = 0 , ∀ x ∈ [a , b].

6.13 Implicit Functions

163

Example 6.13.1 Let us consider the equation: x 2 + y2 − 1 = 0

(6.39)

The set of points in the plane that verifies Eq. (6.40) represents, the circle C(0 ; 1), with the center in origin and radius 1 (Fig. 6.3). Let D1 = [a1 , b1 ] × [c1 , d1 ]. We remark that for any √ x ∈ [a1 , b1 ], there 1 − x 2 such that the is a unique value y = y(x) ∈ [c1 , d1 ], namely y(x) = pair (x , y) verifies the Eq. (6.40), i.e.,the point (x , y) belongs to circle C(0 ; 1). It results that on the rectangle D1 the Eq. (6.40) defines an implicit function. On the other hand, we notice that on the rectangle D2 , Eq. (6.40) does not define any implicit function of the form y = y(x), because for any x ∈ [a2 , − 1), there is no value y such that the pair (x √ , y) ∈ C(0 ; 1), and√for any x ∈ [− 1 , b2 ] there are two values, namely y1 = − 1 − x 2 and y2 = 1 − x 2 such that the points (x , y1 ) and (x , y2 ) belong to the circle C(0 ; 1). The following theorem establishes sufficient conditions for the existence of implicit functions. Theorem 6.13.1 (Implicit functions defined of equations of two variables). Let A ⊂ R2 be an open subset, (a , b) ∈ A and F : A ⊂ R2 → R with the properties: (a) (b) (c)

F ∈ C 1 (A); F(a , b) = 0; ∂ F , b) = 0. ∂ y (a

Then there exists an open neighborhood U of the point a, and an open neighborhood V of the point b such that U × V ⊂ A and an unique function f : U → V such that: (1)

F(x , f (x)) = 0 , ∀ x ∈ U ;

Fig. 6.3 The circle with center in origin and radious one = C(O; 1)

164

(2) (3)

6 Differential Calculus of Functions of Several Variables

f (a) = b; f ∈ C 1 (U ) and f  (x) = −

∂ ∂ ∂ ∂

F x (x F y (x

, f (x)) , f (x))

, ∀ x ∈ U.

More, if F ∈ C p (A), then f ∈ C p (U ). Proof Let us consider the vector function  = (ϕ1 , ϕ2 ) : A ⊂ R2 → R2 defined by: (x, y) = (x , F(x, y)) , (x, y) ∈ A. Obvious we have: ϕ1 (x, y) = x and ϕ2 (x, y) = F(x, y) , (x, y) ∈ A, (a , b) = (a , 0). and





D(ϕ1 , ϕ2 ) 1 0



det J (a , b) = (a , b) = ∂ F

∂ x (a , b) ∂∂Fy (a , b)

D(x , y) =

∂F (a , b) = 0. ∂y

According to Theorem 6.11.2, there exists an open neighborhood U × V of the point (a , b) and an open neighborhood U × W of the point (a , 0) such that  : U × V → U × W is diffeomorphism. More, det J (x , y) =

∂F D(ϕ1 , ϕ2 ) (x, y) = (x, y) = 0 , ∀ (x, y) ∈ U × V. D(x , y) ∂ y

If we denote by = (ψ1 , ψ2 ) = −1 : U × W → U × V , then ∈ C 1 (U × W ). Now we define f (x) = ψ2 (x, 0) , ∀ x ∈ U . Obvious, f : U → V , f ∈ C 1 (U ) and f (a) = = ψ2 (a, 0) = b. Further, for any x ∈ U we have: (x, 0) = ( (x, 0)) = (ψ1 (x, 0) , ψ2 (x, 0)) = (x , f (x)) = (x , F(x, f (x))) whence it results: F(x , f (x)) = 0 , (∀) x ∈ U. By deriving the relationship (6.41) we get: ∂F ∂F (x , f (x)) + (x , f (x)) · f  (x) = 0 ∂ x ∂ y whence it results:

(6.40)

6.13 Implicit Functions

165

f  (x) = −

∂ ∂ ∂ ∂

F x (x F y (x

, f (x)) , f (x))

, ∀ x ∈ U.

Example 6.13.2 1.

Show that the equation x 4 + y 4 − 2 · x 2 · y − 2 · x 2 − y = 0 defines in a neighborhood of the point (2 , 1) the implicit function y = y(x), and compute y  (2) and y  (2).

We denote by F(x , y) = x 4 + y 4 − 2 · x 2 · y − 2 · x 2 − y = 0 and we check the conditions in Theorem 6.13.1: (a) (b) (c)

F is a polynomial function, so indefinitely derivable on R2 ; F(2 , 1) = 0; ∂ F , y) = 4 · y 3 − 2 · x 2 − 1 and ∂∂ Fy (2 , 1) = − 5 = 0. ∂ y (x

From Teorema 6.13.1, it results that there exists an open neighborhood U of the point 2, an open neighborhood V of the point 1 and an unique implicit function x → y(x) : U → V . with the properties: 1. 2. 3.

F(x , y(x)) = 0 , ∀ x ∈ U ; y(2) = 1; ∂F 3 (x,y(x)) y ∈ C 1 (U ) and y  (x) = − ∂∂Fx (x,y(x)) = − 4·x4·y−4·x·y−4·x 3 −2·x 2 −1 , ∀x ∈ U, ∂y



where we deduce that y (2) = 16 . 5 More,    12 · x 2 − 4 · y − 4 − 4 · x · y  4 · y 3 − 2 · x 2 − 1 y  (x) = −  2 4 · y3 − 2 · x 2 − 1    4 · x 3 − 4 · x · y − 4 · x 12 · y 2 · y  − 4 · x −  2 4 · y3 − 2 · x 2 − 1 hence: y  (2) = 2.

2792 . 125

Parametric representation of the ellipse.

We recall that the most commonly used parametric equations of the circle with the center at origin and radius R of the equation x 2 + y 2 = R 2 are: 

x = R · cos t , t ∈ [0 , 2 π] y = R · sin t

(6.41)

166

6 Differential Calculus of Functions of Several Variables

Note that in this representation the parameter t represents the angle at the center −−→ between the position vector O M and the axis O x, where M(x , y) is a current point on the circle. By analogy with this representation, the parametric representation is 2 2 used for the ellipse of the equation ax 2 + by2 = 1 is: 

x = a · cos t , t ∈ [0 , 2 π] y = b · sin t

(6.42)

There is a fundamental difference between the two representations. In the case of the ellipse, the parameter t does not have the same geometric significance as in the case of the circle, t is no more the angle at the center formed by the position −−→ vector O M with the axis O x. We realize this, for example, when the point M on the ellipse is on the first bisector. Indeed, then x = y, whence it follows that a · cos t = b · sin t, and further t = arctg ab = π4 , which contradicts the fact that M is on the first bisector. In the following we present another parametric representation of the ellipse used especially in Geodesy. 2 2 Let M(x , y) , y = 0 be a current point on the ellipse ax 2 + by2 = 1. It is chosen as a parameter the angle B which the normal in M at the ellipse does with the axis O x(Fig. 6.4). Since the angle that the tangent in M to the ellipse makes with   the axis O x is B + π2 it results that the slope of the tangent is m = tg B + π2 = − ctg B. On the other hand, from the theorem of implicit functions, it results that locally, in a neighborhood of the point M, the ellipse is the graph of the implicit function y = y(x) defined by the equation F(x , y) = 0, where: F(x , y) =

x2 y2 + − 1. a2 b2

It follows that the slope of the tangent is equal to: Fig. 6.4 The geometric representation of the ellipse

6.13 Implicit Functions

167

y  (x) = −

∂ ∂ ∂ ∂

F x F y

= −

x · b2 . y · a2

By equalizing those two expressions of the slope of the tangent we obtain: x · b2 cos B = y · a2 sin B . and further: y =

b2 sin B · · x. a 2 cos B

(6.43)

Substituting (6.44) in the equation of the ellipse results:

b2 sin2 B x2 · 1 + 2 · = a2. a cos2 B  2 Taking into account that 1 − ab2 = e ∈ (0 , 1). represents the first eccentricity of the ellipse, we find: x · 1 + 2



1 − e

2



sin2 B · cos2 B

= a2

or: x2 ·

1 − e2 sin2 B = a2, cos2 B

from which we deduce that: a · cos B x =  . 1 − e2 · sin2 B a · 1 − e2 )·sin B Taking into account (6.44), it results that y = √( . 1 − e2 · sin2 B  Finally, if we note with w = 1 − e2 · sin2 B, the parametric representation of the ellipse used in Geodesy is obtained:



x= y=

a · cos B w a · (1 − e2 )· sin B w

(6.44)

where B is the geodesic latitude (the angle between the normal at M to the ellipse and axis O x).

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6 Differential Calculus of Functions of Several Variables

Remark 6.13.1 In the particular case of the circle with center in origin and radius R = a, the eccentricity e = 0 and if we replace in Eqs. (6.45) we obtain the parametric Eqs. (6.42) of the circle, where parameter B is the angle at the center −−→ between the position vector O M and the axis O x, because in this case the normal passes through origin. Next we present, without proof, two important generalizations of Theorem 6.13.1. Theorem 6.13.2 (Implicit functions defined of equations of n + 1 variables). Let A ⊂ Rn + 1 be an open subset, n ≥ 2, (a , b) = (a1 , ... , an , b) ∈ A and F : A → R. with the properties: F ∈ C 1 (A); F(a1 , ... , an , b) = 0; ∂ F , ... , an , b) = 0; ∂ y (a1

(a) (b) (c)

Then there are an open neighborhood U of the point a = (a1 , ... , an ), an open neighborhood V of the point b such that U × V ⊂ A and an unique function f : U → V with the properties: F(x1 , ... , xn , f (x1 , ... , xn )) = 0 , ∀ x = (x1 , ... , xn ) ∈ U ; f (a1 , ... , an ) = b; f ∈ C 1 (U );

1. 2. 3.

and ∀ x = (x1 , ... , xn ) ∈ U , ∀ i = 1 , n we have: ∂ f (x) = − ∂ xi

∂ ∂ ∂ ∂

F , xi (x F , y (x

f (x)) f (x))

.

More, if F ∈ C p (A), then f ∈ C p (U ). Example 6.13.3 Show that the equation x + y + z = e z defines on the some neighborhood of the point (− 1 , e , 1) an implicit function z = f (x , y), and compute d f (− 1 , e).   If we denote by F(x , y , z) = x + y + z − e z = 0, then F ∈ C 1 R3 , F(− 1 , e , 1) = 0 and ∂∂ Fz (− 1 , e , 1) = 1 − e = 0 We remark that the conditions of Theorem 6.13.2 are met, so there is an open neighborhood U of the point (− 1 , e), an open neighborhood V of the point 1 and an unique function z = f (x , y) : U → V with the properties (1), (2) and (3) from this theorem: We have: ∂ f (− 1 , e) = − ∂ x and

∂ ∂ ∂ ∂

F 1, x (− F 1, z (−

e , 1) e , 1)

=

1 . e − 1

6.13 Implicit Functions

169

∂ f (− 1 , e) = − ∂ y

∂ ∂ ∂ ∂

F 1, y (− F 1, z (−

e , 1) e , 1)

=

1 . e − 1

Therefore: d f (− 1 , e) =

1 (d x + d y). e − 1

Theorem 6.13.3 (Implicit functions defined by a system of equations). 2, (a , b) = Let A ⊂ Rn + m be an open subset, m ≥ (a1 , ... , an , b1 , ... , bm ) ∈ A and the vector function F = (F1 , ... , Fm ) : A ⊂ Rn + m → Rm with the properties: (a) (b) (c)

F ∈ C 1 (A); ⎧ ⎨ F1 (a1 , ... , an , b1 , ... , bm ) = 0 ............................................. ... ... ; ⎩ Fm (a1 , ... , an , b1 , ... , bm ) = 0 D(F1 , ..., Fm ) , b) = 0. D(y1 , ..., ym ) (a

Then there exists an open neighborhood U of the point a = (a1 , ... , an ) and an open neighborhood V = V1 × ... × Vm of the point b = (b1 , ... , bm ) such that U × V ⊂ A and m uniquely determined functions f i : U → Vi , i = 1 , m, with the properties: ⎧ =0 ⎨ F1 (x1 , ... , xn , f 1 (x1 , ... , xn ) , ... , f m (x1 , ... , xn )) (1) ............................................................................................. ... ... ; ⎩ =0 Fm (x1 , ... , xn , f 1 (x1 , ... , xn ) , ... , f m (x1 , ... , xn )) ∀ x = (x1 , ... , xn ) ∈ U ; (2) f 1 (a1 , ... , an ) = b1 , ... , f m (a1 , ... , an ) = bm ; (3) f i ∈ C 1 (U ) , ∀ i = 1 , m and for ∀ j = 1 , n, ∀ x = (x1 , ... , xn ) ∈ U , we have: ∂ f1 (x) = − ∂ xj

D(F1 , F2 , ..., Fm ) (x D (x j , y2 , ..., ym )

D(F1 , F2 , ..., Fm ) D(y1 , y2 , ..., ym ) (x

, f 1 (x) , ... , f m (x) ) , f 1 (x) , ... , f m (x) )

................................................................................................ ∂ fm (x) = − ∂ xj

D(F1 , ..., Fm−1 , Fm ) (x D ( y1 , ..., ym − 1 , x j ) D(F1 , ..., Fm−1 , Fm ) D(y1 , ..., ym−1 , ym ) (x

, f 1 (x) , ... , f m (x) ) , f 1 (x) , ... , f m (x) )

Example 6.13.4 Show that the nonlinear system: 

x 3 + 3 · y2 − z2 + x − y − 8 = 0 =0 x 2 − y2 − 3 · z − 3

.

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6 Differential Calculus of Functions of Several Variables

defines on a some neighborhood of the point (1 , 2 , − 2) two implicit functions y = f (x) and z = g(x) and compute f  (1) and g  (1). We have met the conditions of Theorem 6.13.3,i.e.: 

  F(x , y , z) = x 3 + 3 y 2 − z 2 + x − y − 8 = 0 , F , G ∈ C 1 R3 G(x , y , z) = x 2 − y 2 − 3 z − 3 = 0 F(1 , 2 , − 2) = 0, G(1 , 2 , − 2) = 0

D(F , G) (1 , 2 , − 2) = D(y , z)







∂ ∂ ∂ ∂

F y G y

∂ F ∂ z ∂G ∂ z







11 4

(1 , 2 , − 2) = − 4 − 3 = − 17  = 0.

From Teoremei 6.13.3, it follows that there are an open neighborhood U of the point 1, an open neighborhood V × W of the point (2 , − 2) and two functions y = f (x) : U → V , z = g(x) : U → W with the properties (1), (2), (3) of this theorem. Further we have:



4 4

D(F , G)

= − 20 (1 , 2 , − 2) =

2 −3

D(x , z) and:

11 D(F , G) (1 , 2 , − 2) =

−4 D(y , x)

4

= 38. 2

whence it results: f  (1) = −

20 − 20 =− − 17 17

g  (1) = −

38 38 = . − 17 17

and

Example 6.13.5 Find the locally extreme points of the function y = y(x) defined implicitly by the equation x 3 + y 3 − 3 x 2 · y − 3 = 0. We shall denote by F(x , y) = x 3 + y 3 − 3 x 2 · y − 3 . The critical points of the implicit function y = y(x) belong to the curve, i.e. verifies the equation F(x , y) = 0, and satisfy the equation y  (x) = 0, so they are the solutions of the system: ∂

=0 , y) = 3 · x 2 − 6 · x · y F(x , y) = x 3 + y 3 − 3 · x 2 · y − 3 = 0 . F ∂ x (x

6.14 Local Conditional Extremum

171

Solving this system we obtain the two critical points:  √ M1 0 , 3 3 and M2 (− 2 , − 1). Now we check if the condition ∂∂ Fy = 0 is fulfilled at the points M1 and M2 . As ∂∂ Fy (x , y) = 3 y 2 − 3 x 2 , it results that:  √  √ ∂ F 0 , 3 3 = 3 3 9 = 0 and ∂∂ Fy (−2 , −1) = −9 = 0. ∂ y To decide whether M1 and M2 are locally extreme points, as well as their nature, we shall compute y  .We have successively: y = and y



x2 − 2 x y x 2 − y2 

=

     2 x − 2 y − 2 x y  x 2 − y2 − x 2 − 2 x y 2 x − 2 y y .  2 x 2 − y2

 √   √   √  Since y  0 , 3 3 = 0, it results y  0 , 3 3 = √32 3 > 0, so M1 0 , 3 3 is a locally minimum point for the implicit function y = y(x) and since y  (− 2 , − 1) = 0 and y  (− 2 , − 1) = − 23 < 0, we deduce that M2 (− 2 , − 1) is a locally maximum point for this function.

6.14 Local Conditional Extremum In applications, the problem of determining the extreme values of a function of several variables often occurs in situations where the variables are subject to certain restrictions (satisfy certain connection relations). Example 6.14.1 Find the extreme values of the function: f : R2 → R , f (x , y) = x 2 + y 2 with the constraint: x + y − 2 = 0 The case being very simple, the problem is immediately reduced to an extremely free problem. Indeed, by substituting y = 2 − x in the expression of the function f we obtain the function g : R → R , g(x) = 2 · x 2 − 4 · x + 4, x ∈ R. Since g  (x) = 4 · x − 4 and g  (x) = 4 > 0, it results that the function g has a minimum equal with 2 at the point x = 1, hence the function f (x , y) = x 2 + y 2 , with the constraint x + y − 2 = 0, has a minimum equal with 2 at the point (1, 1).

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6 Differential Calculus of Functions of Several Variables

Note that the function f (x , y) = x 2 + y 2 admits the minimum value 0 at the point (0 , 0) if the variables are not subject to any constraint. Let A ⊂ Rn be an open subset and let f , F1 , ... , Fm : A ⊂ Rn → R be m + 1 functions, 1 ≤ m < n. Let S ⊂ Rn be the set of all system solutions: ⎧ ⎨ F1 (x1 , ... , xn ) = 0 ......................... ... ... ⎩ Fm (x1 , ... , xn ) = 0 .

(6.45)

Definition 6.14.1 We say that the point a ∈ A ∩ S is a local conditional maximum (minimum) point for the function f with the constraints (6.46) if there exists an open neighborhood U of the point x = a such that f (x) ≤ f (a). ( f (x) ≥ f (a)), for any x ∈ A ∩ S ∩ U . For a better understanding, consider the particular case n = 3 , m = 1. Let f : A ⊂ R3 → R be a funtion of C 1 − class and let the constraint: F(x1 , x2 , x3 ) = 0.

(6.46)

From a geometric point of view, the set of solutions of Eq. (6.47) represents a surface. We will denote by S this surface.The problem consists on determining the maximum and minimum value of the function f on the surface S. The following theorem establishes sufficient conditions for the existence of the local onditional extreme points of a real function with three variable and with one constraint. Theorem 6.14.1 Let A ⊂ R3 be an open subset, f , F ∈ C 1 (A) and let a ∈ A ∩ S be a local conditional point for f with the constraint (6.47). If ∂∂ xF3 (a) = 0,then there exists an unique number λ ∈ R such that: ⎧∂ f ⎪ ⎨ ∂ x1 (a) + λ ∂ f +λ ∂ x2 (a) ⎪ ⎩ ∂ f (a) + λ ∂ x3

∂ ∂ ∂ ∂ ∂ ∂

F x1 (a) F x2 (a) F x3 (a)

=0 =0 = 0.

(6.47)

(λ is called Lagrange’s multiplier). Proof To fix the ideas, suppose that a = (a1 , a2 , a3 ) is a local conditional minimum point for f . Then there is an open neighborhood U of the point a, such that: f (a) ≤ f (x) , ∀ x = (x1 , x2 , x3 ) ∈ A ∩ S ∩ U. As a is an interior point of the subset A we can assume that U ⊂ A. Therefore: f (a) ≤ f (x) , ∀ x ∈ S ∩ U.

(6.48)

6.14 Local Conditional Extremum

173

Since ∂∂ xF3 (a) = 0, from Theorem 6.13.2, it result that there exist ε > 0 and an implicit function ϕ : U1 × U2 → U3 , where U1 = (a1 − ε , a1 + ε),U2 = (a2 − ε , a2 + ε),U3 = (a3 − ε , a3 + ε), with the properties: (i) (ii) (iii)

F(x1 , x2 , ϕ(x1 , x2 )) = 0 , ∀ (x1 , x2 ) ∈ U1 × U2 ; ϕ(a1 , a2 ) = a3 ; ϕ ∈ C 1 (U1 × U2 ) and: ∂ϕ (a1 , a2 ) = − ∂ x1

∂ ∂ ∂ ∂

F x1 (a) F x3 (a)

∂ϕ ; (a1 , a2 ) = − ∂ x2

∂ ∂ ∂ ∂

F x2 (a) . F x3 (a)

Obviously we can choose ε > 0 so that the cartesian product U1 × U2 × U3 ⊂ U . We remark that if (x1 , x2 ) ∈ U1 × U2 ,then: (x1 , x2 , ϕ(x1 , x2 )) ∈ U1 × U2 × U3 ⊂ U. On the other hand, if (x1 , x2 ) ∈ U1 × U2 ,then, from the property (i) it results: (x1 , x2 , ϕ(x1 , x2 )) ∈ S. Therefore, if (x1 , x2 ) ∈ U1 × U2 ,then: (x1 , x2 , ϕ(x1 , x2 )) ∈ S ∩ U

(6.49)

Let us denote by g(x1 , x2 ) = f (x1 , x2 , ϕ(x1 , x2 )), Taking into account (ii), (6.49) and (6.50) it results:

(x1 , x2 ) ∈ U1 × U2 .

g(a1 , a2 ) = f (a1 , a2 , a3 ) = f (a) ≤ f (x1 , x2 , ϕ(x1 , x2 )) = g(x1 , x2 ), ∀(x1 , x2 ) ∈ U1 × U2 . Therefore, (a1 , a2 ) is a free local minimum point for the function g. From Fermat’s Theorem (6.10.1) we deduce that: ∂g ∂g (a1 , a2 ) = 0 and (a1 , a2 ) = 0. ∂ x1 ∂ x2 According to the rule of derivation of composed functions we obtain: ∂ f ∂ f ∂ϕ ∂g = + · ; ∂ x1 ∂ x1 ∂ x3 ∂ x1 ∂g ∂ f ∂ f ∂ϕ = + · . ∂ x2 ∂ x2 ∂ x3 ∂ x2

174

6 Differential Calculus of Functions of Several Variables

Taking into account (iii), further we have: − ∂∂ xF (a) ∂g ∂ f ∂ f = 0 (a1 , a2 ) = (a) + (a) · ∂ F 1 ∂ x1 ∂ x1 ∂ x3 ∂ x (a)

(6.50)

− ∂∂ xF (a) ∂ f ∂ f ∂g = 0. (a1 , a2 ) = (a) + (a) · ∂ F 2 ∂ x2 ∂ x2 ∂ x3 ∂ x (a)

(6.51)

3

3

We will analyze the following cases: Case 1.

∂F (a) = 0 ∂ x1

and ∂∂ xF2 (a) = 0. Then, from (6.51)and (6.52) it results: ∂ f ∂ x1 (a) ∂ F ∂ x1 (a)

=

∂ f ∂ x2 (a) ∂ F ∂ x2 (a)

=

∂ f ∂ x3 (a) ∂ F ∂ x3 (a)

= − λ,

where we immediately obtain the relations (6.48) from the theorem statement.

Case 2.

∂F (a) = 0. ∂ x1

and ∂∂ xF2 (a) = 0. From (6.51) and (6.52) we deduce that: ∂ f (a) = 0. ∂ x1 ∂ f ∂ x2 (a) ∂ F ∂ x2 (a)

=

∂ f ∂ x3 (a) ∂ F ∂ x3 (a)

(6.52)

= − λ.

(6.53)

It follows from (6.53) that the first relation in (6.48) is verified for any λ, and from (6.54) that the last two relations in (6.48) are also true.

Case 3.

∂F (a) = 0 ∂ x1

and ∂∂ xF2 (a) = 0. In this case, from (6.51) and (6.52) we deduce that ∂∂ xf1 (a) = ∂∂ xf2 (a) = 0 and it is obvious that the first two relations in (6.46) are true for any λ ∈ R.

6.14 Local Conditional Extremum

If we denote by

∂ f ∂ x3 (a) ∂ F ∂ x3 (a)

175

= − λ, we also obtain the third relation in (6.48) and

with this theorem is proved. Next, consider the following auxiliary function: φ(x1 , x2 , x3 , λ) = f (x1 , x2 , x3 ) + λ · F(x1 , x2 , x3 ).

(6.54)

We remark that: ⎧ ∂φ = ∂∂ xf1 + λ · ⎪ ⎪ ∂ x1 ⎪ ⎨ ∂φ = ∂ f +λ · ∂ x2 ∂ x2 ∂ f ∂φ ⎪ = + λ· ⎪ ∂ x ∂ x3 3 ⎪ ⎩∂φ = F . ∂ λ

∂ ∂ ∂ ∂ ∂ ∂

F x1 F x2 F x3

(6.55)

From Theorem 6.14.1 and from (6.56), it results that if a = (a1 , a2 , a3 ) is a point for f ,with the constraint (6.47), and λ verifies the local conditional extreme  system (6.48), then a1 , a2 , a3 , λ is a critical point for the auxiliary function φ defined in (6.55). It follows that the local conditional points of the function f with the constraint (6.47), are find among the points (x1 , x2 , x3 ) with the property that (x1 , x2 , x3 , λ) are critical  points of the auxiliary function given by (6.55). Let a1 , a2 , a3 , λ be a critical point for φ and let: φ(x) = f (x) + λ F(x) , ∀ x = (x1 , x2 , x3 ) ∈ A. If x ∈ U ∩ S, then F(x) = 0 and we have: f (x) − f (a) = φ (x) − φ (a) = d φ (a) (x − a) +

1 2 d φ(ξ )(x − a) 2!

As, d φ (a) = 0, it results that: f (x) − f (a) =

1 2 1 2 d φ(ξ )(x − a) = d φ(a)(x − a) + ω(x) 2! 2! (6.56)

where ω is an infinitely small function i.e. ω = 0(x − a). By differentiating the constraint (6.47) we get: ∂F ∂F ∂F (a) · d x1 + (a) · d x2 + (a) · d x3 = 0 ∂ x1 ∂ x2 ∂ x3

(6.57)

From the relation (6.58) we can solving, for example, d x3 in relation to d x1 and d x2 and replace it in d 2 φ(a). So, a quadratic form is obtained in two variables.

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6 Differential Calculus of Functions of Several Variables

If this quadratic form is positive (negative) definite, then the point a is a local conditional minimum ( maximum) point for f with the constraint (6.47). Example 6.14.2 Find the local conditional points of the function: f : R3 \{(0, 0, 0)} → R, f (x , y , z) = x 2 · y 3 · z 4 with the constraint: 2 · x + 3 · y + 4 · z = 9. We consider the auxiliary function: φ(x , y , z) = x 2 · y 3 · z 4 + λ · (2 · x + 3 · y + 4 · z − 9). The critical points of this function are obtained by solving the system: ⎧∂φ ⎪ = 2 · x · y3 · z4 ⎪ ∂ x ⎪ ⎨ ∂ φ = 3 · x 2 · y2 · z4 ∂ y ∂φ ⎪ = 4 · x 2 · y3 · z3 ⎪ ⎪ ⎩ ∂∂ φz = 2· x + 3 ·y ∂ λ

+ + + +

2· λ =0 3· λ =0 4· λ =0 4 · z − 9 = 0.

The system has only one solution, namely: x = y = z = 1, λ = − 1. If we denote by: φ(x , y , z) = x 2 · y 3 · z 4 − (2 · x + 3 · y + 4 · z − 9) then we have: d 2 φ(1 , 1 , 1) = 2 · d x 2 + 6 · d y 2 + 12 · d z 2 + 12 · d x · d y . + 16 · d x · d z + 24 · d y · d z By differentiating the the constraint F(x, y , z) = 2 ·x + 3 · y + 4· z − 9 = 0 we get: ∂F ∂F ∂F · dx + · dy + · d z = 2 · d x + 3 · d y + 4 · d z = 0, ∂ x ∂ y ∂z whence it results: (2 · d x + 3 · dy + 4 · dz)2 = 4 · d x 2 + 9 · d y 2 + 16 · d z 2 + 12 · d x · d y + 16 · d x · d z + 24 · d y · d z = 0

6.14 Local Conditional Extremum

177

and further: 12 · d x · d y + 16 · d x · d z + 24 · d y · d z = −4 · d x 2 − 9 · d y 2 − 16 · d z 2 If we take into account the last relation in the expression of d 2 φ(1 , 1 , 1) we get:   d 2 φ(1 , 1 , 1) = − 2 d x 2 + 3 d y 2 + 4 d z 2 . This quadratic form is negative

definite because:

−2 0

= 6 > 0;

1 = − 2 < 0 ; 2 =

0 −3



.

−2 0 0



3 =

0 − 3 0

= − 24 < 0

0 0 −4

Therefore, the point (1 , 1 , 1) is a local conditional maximum point for f with the constraint F(x, y , z) = 2 · x + 3 · y + 4 · z − 9 = 0. Example 6.14.3 Find the local conditional points of the function: f (x, y, z) = x · y + y · z + z · x with the constraint: x · y · z = 1. We consider the auxiliary function: φ(x, y, z) = x · y + y · z + z · x + λ · (x · y · z − 1) The critical points of this function are obtained by solving the system: ⎧ ∂φ ⎪ = y+z+λ·y·z =0 ⎪ ∂x ⎪ ⎨ ∂φ = x + z + λ · x · z = 0 ∂y ∂φ ⎪ = y+x +λ·x ·y =0 ⎪ ∂z ⎪ ⎩ ∂φ = x ·y·z−1=0 ∂λ The system has the unique solution: x = y = z = 1, λ = − 2. If we denote by:

178

6 Differential Calculus of Functions of Several Variables

φ(x , y , z) = x · y + y · z + z · x − 2 · (x · y · z − 1) then we have: d2 φ(x, y, z) = 2 · [(1 − 2 · z) · dx dy + (1 − 2y) · dx dz + (1 − 2x) · dy dz] and: d2 φ(1, 1, 1) = −2 · (dx dy + dx dz + dy dz). By differentiating the the constraint F(x, y , z) = x · y · z − 1 = 0 we get: ∂F ∂F ∂F dx + dy + dz = y · z dx + x · z dy + x · y dz = 0, ∂x ∂y ∂z For x = z = 1 we have dx + dy + dz = 0, whence it results dz = −(dx + dy). Replacing in d2 φ(1, 1, 1) we get:   d2 φ(1, 1, 1) = 2 · dx 2 − dx dy + dy 2 , This quadratic form is positive definite because:

1 = a11







a11 a12 1 − 1 3 2



= > 0.

= 1 > 0 and 2 =

= a12 a22 − 21 1 4

Therefore, the point (1, 1, 1) is a conditional minimum point for the function f (x, y, z) = = x · y + y · z + z · x with the constraint x · y · z = 1. Theorem 6.14.1 admits the following generalization. Theorem 6.14.2 Let A ⊂ Rn be an open subset, f , F1 , ... , Fm ∈ C 1 (A) and let a ∈ A ∩ S be a local extreme point for f with the constraints (6.46). 1 , ..., Fm ) = 0, then there are λ1 , ... , λm ∈ R, uniquely determined, If D(F D(x1 , ..., xm ) (a) such that: ⎧ ∂ f ∂ Fm ∂ F1 ⎪ ⎨ ∂ x1 (a) + λ1 · ∂ x1 (a) + ... + λm · ∂ x1 (a) = 0 (6.58) ... ... ... ... ... ... ... ... ... ⎪ ⎩ ∂ f (a) + λ · ∂ F1 (a) + ... + λ · ∂ Fm (a) = 0 . 1 m ∂ xn ∂ xn ∂ xn (λ1 , ... , λm are called Lagrange’s multipliers). Consider the auxiliary φ : A × Rm → R defined by: φ(x1 , ... , xn , λ1 , ... , λm ) = f (x) + λ1 · F1 (x) + ... + λm · Fm (x) , x∈A (6.59)

6.14 Local Conditional Extremum

179

and be the following system of n + m equations with the n + m unknowns. (x1 , ... , xn , λ1 , ... , λm ). Further we have: ⎧ ∂φ = ∂∂ xf1 ⎪ ∂ x1 ⎪ ⎪ ⎪ ⎪ ... ... ... ⎪ ⎪ ⎨ ∂φ = ∂ f ∂ xn ∂ xn ∂φ ⎪ = F 1 ⎪ ∂ λ 1 ⎪ ⎪ ⎪ ... ... ... ⎪ ⎪ ⎩ ∂φ = Fm ∂ λm

+ ... + = ... =

λ1 · ... λ1 · 0 ... 0.

∂ F1 ∂ x1 ∂ F1 ∂ xn

+ ... + λm · ... ... ... ... + ... + λm ·

∂ Fm ∂ x1 ∂ Fm ∂ xn

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

=0 ... ... =0

.

(6.60)

... ...

From Theorem 6.14.2 it results that, if a = (a1 , ... , an ) is a local conditional extreme point for f with  the constrains (6.46), and λ1 , ... , λm satisfies the system (6.60), then the point a1 , ... , an , λ1 , ... , λm is a critical point for the auxiliary function φ. It follows that the local conditional extreme points of f with the constraints (6.46) are find among the points (x1 , ... , xn ) with the property that points of the auxiliary function φ. (x1 , ..., xn , λ1 , ... , λm ) are critical  Let a1 , ... , an , λ1 , ... , λm be a critical point for the auxiliary function φ and let: φ(x) = f (x) + λ1 F1 (x) + ... + λm Fm (x) , ∀ x = (x1 , ... , xn ) ∈ A. If x ∈ U ∩ S, then Fi (x) = 0 , i = 1 , m and we have: f (x) − f (a) = φ (x) − φ (a) = d φ (a) (x − a) +

1 2 d φ(ξ )(x − a) 2!

As d φ (a) = 0,it results: f (x) − f (a) =

1 2 1 2 d φ(ξ )(x − a) = d φ(a)(x − a) + ω(x) 2! 2!

where ω = 0(x − a). By differentiating the constraints (6.46) we get: ⎧ ∂ F1 ∂ F1 ⎪ ⎨ ∂ x1 · d x1 + ... + ∂ xn · d xn = 0 ... ... ... ... ... ... ... ⎪ ⎩ ∂ Fm · d x + ... + ∂ Fm · d x = 0 . 1 n ∂ x1 ∂ xn

(6.61)

From the system (6.62) we can solving, for example, d x1 , ... , d xm in relation to d xm + 1 , ... , d xn and replace in the expression of d 2 φ(a). So, a quadratic form in n − m variables is thus obtained. If this quadratic form is positive (negative) define, then the point a is a local conditional minimum ( maximum) point for f with the constraint (6.46). As with the free local extreme points, if d 2 φ(a) is positive (negative) define, then a is a local conditional minimum (maximum) point for f with the constraints (6.46).

180

6 Differential Calculus of Functions of Several Variables

6.15 Dependent and Independent Functions Let A ⊂ Rn be an open subset and f 1 , ... , f m ∈ C 1 (A), m ≤ n. Definition 6.15.1 The functions f 1 , ... , f m are said to be dependent in the domain A if at last one of the functions,  say f m , is espressed in terms of the other functions i.e.,there exists a ϕ ∈ C 1 Rm − 1 such that: f m (x) = ϕ( f 1 (x), ... , f m − 1 (x)) , ∀ x ∈ A. Example 6.15.1 Consider the functions f i : R3 → R , i = 1 , 3 defined by: f 1 (x1 , x2 , x3 ) = x12 + x22 + x32 f 2 (x1 , x2 , x3 ) = x1 + x2 + x3 f 3 (x1 , x2 , x3 ) = 2 · x1 · x2 + 2 · x1 · x3 + 2 · x2 · x3 . It is immediately noticeable that: f 1 (x1 , x2 , x3 ) = ϕ( f 2 (x1 , x2 , x3 ) , f 3 (x1 , x2 , x3 )) where: ϕ(u , v) = u 2 − v , ∀ (u , v) ∈ R2 . Therefore, the functions f 1 , f 2 , f 3 are dependent in R3 . Remark 6.15.1 We recall that the functions f 1 , ... , f m are linearly dependent in the subset A if there are m real numbers λ1 , ... , λm ∈ R, not all null, such that: λ1 · f 1 (x) + ... + λm · f m (x) = 0 , ∀ x ∈ A. If we suppose, for example, that λm = 0, it results that: f m (x) =

−

λ1 λm

· f 1 (x) + ... +

λm − 1 · f m − 1 (x), ∀ x ∈ A. − λm

Therefore, the notion of linearly dependent functions is a particular case of Definition 6.15.1, namely the case when the function ϕ is linear. Theorem 6.15.1 Let A ⊂ Rn be an open subset, m ≤ n, and let F ∈ C 1 (A) be the vector function F = ( f 1 , ..., f m ) : A → Rm . If f 1 , ... , f m are dependent in A, then: rang JF (x) < m , ∀ x ∈ A.

6.15 Dependent and Independent Functions

181

  Proof let ϕ ∈ C 1 Rm − 1 be such that: f m (x) = ϕ( f 1 (x), ... , f m − 1 (x)), ∀x = (x1 , ..., xn ) ∈ A.

(6.62)

By differentiating the relation (6.63) we get: ∂ fm ∂ϕ ∂ f1 ∂ϕ ∂ fm − 1 = · + ... + · , j = 1 , n. ∂ xj ∂ f1 ∂ xj ∂ fm − 1 ∂ xj Since the Jacobian matrix is: ⎛

(6.63)

⎞ ... ∂∂ xfm1 (x) ... ∂∂ xf1n (x) ⎟ ⎜ JF (x) = ⎝ ... ... ... ... ... ⎠, ∂ fm ∂ fm ∂ fm ... ... (x) (x) (x) ∂ x1 ∂ xm ∂ xn ∂ f1 ∂ x1 (x)

from (6.64) it follows that the last line of the matrix JF (x) is a linear combination of the other lines, so any minor of the order m of the matrix JF (x) is null. Definition 6.15.2 Let A ⊂ Rn be an open subset. We say that the functions f 1 , ... , f m are independent in the point a ∈ A, if they are not dependent in any neighborhood V of the point a, V ⊂ A. We say that f 1 , ... , f m are independent in A if they are independent in every point of A. With this definition, from Theorem 6.15.1 results:. Corollary 6.15.1 If F = ( f 1 , ..., f m ) : A → Rm ,F ∈ C 1 (A) and. rang JF (x) = m , ∀x ∈ A, then the functions f 1 , ... , f m are independent in A. D( f 1 , ..., f m ) Corollary 6.15.2 If the Jacobian D(x (x) = 0 , ∀ x ∈ A, then the functions 1 , ..., x m ) f 1 , ... , f m are independent in A. In the following we present, without proof, the general theorem of functional dependence and independence:

Theorem 6.15.2 Let A ⊂ Rn be an open subset, m ≤ n, a ∈ A and the vector function F = ( f 1 , ..., f m ) : A ⊂ Rn → Rm , F ∈ C 1 (A). If rang JF (a) = s < m ≤ n, then there exists an open neighborhood U of the point a, U ⊂ A such that s functions between the functions f 1 , ... , f m are independent in U , and the other m − s functions depend on them in U . Example 6.15.2 If we return to Example 6.15.1 we find that: ⎞ 2 x1 2 x2 2 x3 ⎠. JF (x) = ⎝ 1 1 1 2(x2 + x3 ) 2(x1 + x3 ) 2(x1 + x2 ) ⎛

It is easy to verify that det (x1 , x2 , x3 ) ∈ R3 .  JF (x) = 0, ∀x = 3 Let us denote by M = (x1 , x2 , x3 ) ∈ R ; x1 = x2 = x3 .

182

6 Differential Calculus of Functions of Several Variables

and by A = R3 \M. If (x1 , x2 , x3 ) ∈ A, then at least one of the minors of order 2:



2 x1 2 x2



1 1 ,



2 x1 2 x3



1 1 ,



2 x2 2 x3



1 1

is different from zero. It results that the functions f 1 , f 2 are independent in A, and f 3 depend on f 1 and f 2 in A.

References

1. Ilyn, V.A., and E.G. Pozniak. 1982. Fundamentals of Mathematical Analysis. Moscow: Mir Publishers, Part 1 and Part 2, 1982. 2. Krasnov, M., A. Kiselev, G. Makarenko, and E. Shikin. 1990. Mathematical Analysis for Engineers. Moscow: Mir Publishers, Vol 1, 1989 and Vol 2, 1990. 3. Lange, S. 1987. Calculus of Several Variables, Third Edition. New York Inc.: Spinger. 4. Nikolsky, S.M. 1977. A Course of Mathematical Analysis. Moscow: Mir Publishers, Vol 1 and Vol 2, 1977. 5. Paltineanu, G., and M. Zamfir. 2015. Higher Mathematics 1 (Differential Calculus). Bucuresti: Conspress (In Romanian). 6. Rudin, W. 1964. Principles of Mathematical Analysis, McGraw-Hill, Inc. 7. Rudin, W. 1987. Real and Complex Analysis, Third Edition, McGraw-Hill, Inc. 8. Stanasila, O. 1981. Mathematical Analysis, E.D.P.Bucuresti (In Romanian). 9. Trench, W.F. 2003. Introduction to Real Analysis, Library of Congress Cataloging-in-Publication Data.

© The Editor(s) (if applicable) and The Author(s), under exclusive license to Springer Nature Singapore Pte Ltd. 2022 G. Paltineanu et al., Differential Calculus for Engineers, https://doi.org/10.1007/978-981-19-2553-5

183

Index

A Abel’s first theorem, 52 Abel’s second theorem, 56 Absolutely convergent series, 28 Accumulation point, 85 Adherent point, 81 Alternating second-order differential, 140 Alternating series, 25 Archimedes’ axiom, 2

B Binomial, logarithmic and geometric series, 70 Bolzano’s theorem, 101 Boundary of a set, 84 Bounded sets in Rn , 86

C Cantor’s axiom, 2 Cantor’s theorem, 100 Cauchy-Bolzano theorem, 91 Cauchy- Buniakovski-Schwarz’s inequality, 74 Cauchy-Hadamard theorem, 53 Cauchy’s criterion for series, 17 Cauchy’s root test, 20 Cauchy‘s test for sequences, 6 Cauchy’s theorem on the commutativity of absolutely convergent series, 30 Cauchy’s uniformly convergence criterion, 40 Change of variables, 145 Closed set, 81 Commutativity property, 28

Compact sets in Rn , 87 Conditionally convergent series, 28 Connected set, 101 Convergence interval of a power series, 52 Convergent and divergent number series, 13 Convergent real number sequences, 4 Conversion of a decimal fraction into a vulgar fraction, 15 Convex (concave) function on an interval, 64 Convex set, 151 Critical (stationary) point, 154 Curl of a vector field, 132

D D’Alembert’s test, 21 Darboux’s property, 101 Dependent and independent functions, 180 Derived set, 85 Diffeomorphism, 159 Differentiability of vector functions, 117 Differentials of higher orders, 141 Directional derivative, 128 Dirichlet-Abel’test, 24 Discrete set, 86 Divergence of a vector field, 132 Domain in Rn , 162

E Euler’s constant, 9 Euler’s formulas, 70 Euler’s identity, 124 Exponential function ez , 34

© The Editor(s) (if applicable) and The Author(s), under exclusive license to Springer Nature Singapore Pte Ltd. 2022 G. Paltineanu et al., Differential Calculus for Engineers, https://doi.org/10.1007/978-981-19-2553-5

185

186

Index

F Fermat’s Theorem, 154 First comparison test, 17 First order differential, 125 Functional series simply (uniformly) convergent on a set, 47 Function continuous at a point, 93 Function continuous at a point with respect to one of its variable, 95 Function homogeneous of order p, 123 Function implicitly defined, 162 Function of C p −class, 138 Function of several variables differentiable at a point, 112 Function uniformly continuous, 98 Fundamental (Cauchy) sequence, 5

Limit of a vector function, 88 Limit point of a sequence, 11 Limit superior (inferior) of a sequence, 11 Linear map, 103 Local conditional maximum (minimum) point with the constraints, 171 Local maximum (minimum) point, 154 Logarithmic series, 57

G Generalized harmonic series, 18 Geometric series, 14 Gradient of a scalar field, 131

N n–dimensional space Rn , 73 Neighborhood of the point, 79 Norms ∞ and 2 , 74 Number e, 7

H Hamiltonian (nabla) operator, 133 Harmonic series, 16 Hessian matrix, 139 Hyperbolic trigonometric functions, 71

I Implicit functions defined by a system of equations, 168 Infinitely small function, 112 Interior point, 80 Invariance property of the first-order differential, 128 Inverse function theorem, 160 Iterates limits of a function of several variable, 92

J Jacobian determinant, 119 Jacobian matrix, 119

L Lagrange’s multiplier, 172 Lagrange’s Theorem, 154 Laplacian operator, 134 Leibniz’s test, 26 Lemma Cesàro, 5

M Mac Laurin’s formula, 153 Mixed partial derivative of the second order, 135 Monotonic real number sequences, 4 Multi-index, 138

O Open cone, 123 Open (respectively closed) ball with center at a and radius r, 76 Open set, 80 Operations on convergent series, 31

P Partial derivatives of a function of several variables, 107 Partial sum of a number series, 13 Positive (negative) definited second-order differential, 140 Potential vector field, 131 Power series, 51

R Raabe’s test, 22 Radius of convergence of a power series, 52 Regular transformation, 161 Riemann‘s theorem, 29

S Scalar (inner) product, 73 Scalar (vector) field on a domain, 131 Schwarz’s theorem, 136

Index

187

Second comparison test, 19 Second –order differential, 138 Second-order partial derivatives, 134 Second-order partial derivatives of functions composed of two variables, 142 Sequence of functions simple (pointwise) convergent to a function on a set, 37 Sequence of functions uniformly convergent to a function on a set, 37 Sequence of vectors convergent in Rn , 75 Sequences of complex numbers convergent (fundamental), 33 Series of complex number convergent, 33 Series with arbitrary terms, 24 Series with positive terms, 17 Set of convergence of a power series, 52 Subsequence of a sequence, 4 Sum of a number series, 13 Supremum (infimum) of a set of real numbers, 3 Sylvester’s theorem, 140

Term-by-term differentiation theorem, 49 Term-by-term differentiation theorem of a power series, 55 Term-by-term integration theorem, 56 Theorem on the existence and differentiability of an implicit function, 163 Theorem regarding the continuity of the limit function, 42 Theorem regarding the continuity of the sum of a functional series, 49 Theorem regarding the derivability of the limit function, 43 Theorem regarding the integrability of the limit function, 45 Third comparison test, 19

T Taylor and Mac Laurin series, 65 Taylor formula, 59 Taylor’s formula for functions of n variables, 150 Taylor’s formula with Cauchy’s remainder, 60 Taylor’s formula with Lagrange’s remainder, 59

V Vector function, 87

U Upper (lower) bound of a set of real numbers, 3

W Weierstrass’ theorem for real number sequences, 7 Weierstrass’ uniform convergence criterion, 48