E1: CALCULUS - lecture notes

Transcription

1 E1: CALCULUS - lecture notes Ştefn Blint Ev Kslik, Simon Epure, Simin Mriş, Aureli Tomoiogă Contents I Introduction 9 1 The notions set, element of set, membership of n element in set re bsic notions of mthemtics 9 2 Symbols used in set theory 10 3 Opertions with sets 10 4 Reltions 11 5 Functions 14 6 Composite function. Inverse of function Logic symbols 16 8 Converse theorem nd contrry theorem 17 9 Necessity nd sufficiency 18 II Single vrible clculus Topology in R Sequences 20 1

2 12 Convergence Rules (for convergence of sequences) Limit points of sequence Series of rel numbers Rules (for convergence of series) Absolute convergent series Limit of function t point Rules for the limit of function One sided limits Infinite limits Limit points of function t point Continuity Rules for continuity Properties of continuous functions Sequence of functions. Set of convergence Continuity nd uniform convergence Equl continuous nd equl bounded sequence of functions Series of functions. Convergence nd uniform convergence Convergence criteri for series of functions Power Series 58 2

3 32 Arithmetics of power series Differentible functions Rules of differentibility Locl extremum Theorems concerning bsic properties of differentible functions Higher-order derivtives nd differentils Tylor polynomils Clssifiction theorem for locl extrem The Riemnn-Drboux integrl Properties of the Riemnn-Drboux integrl Clsses of Riemnn-Drboux integrble functions Men vlue theorem The fundmentl theorem of clculus Techniques to find primitives Improper integrls Fourier series Different forms of Fourier series 101 III Functions of severl vribles Topology in R n Limit of function t point 107 3

4 51 Continuity Importnt properties of continuous functions Differentition Bsic properties of differentible functions Higher order prtil differentibility Tylor s theorems Clssifiction theorem for locl extrem Conditionl extrem Jordn mesurble subsets of R The Riemnn-Drboux integrl of functions of two vribles Integrble functions Properties of the Riemnn-Drboux integrl Riemnn-Drboux integrl clculus when A is rectngulr Riemnn-Drboux integrl clculus when A is not rectngle Jordn mesurble subsets of R n The Riemnn-Drboux integrl of n vrible function Integrble functions of n vribles Properties of the Riemnn-Drboux integrl of n-vrible functions Riemnn-Drboux integrl clculus for n-vrible functions when A is hypercube Elementry curves nd elementry closed curves 143 4

5 71 Line integrl of first type Line integrls of second type Trnsformtion of double integrls into line integrls Elementry Surfces Surfce integrls of first type Surfce integrls of second type Properties of surfce integrls Differentition of n integrl contining prmeter 165 5

6 In which wy cn Clculus course be useful to first yer computer science student? This is frequently sked question of first yer students t the beginning of their Clculus course. It is difficult to give full nd convincing nswer to this question t the very beginning of the course, s we hve to tlk bout the utility of some concepts nd mthemticl instruments, tht re unknown to those who sk, in solving prcticl problems which re out of their rech t the moment. However, the question cnnot nd must not be voided. It is necessry to formulte prtil nswer showing the utility of this course in solving rel problems, tht future computer scientists could find interesting. We hve to emphsize here tht for mthemtics students, Clculus is bsic nd very importnt prt of their curriculum, nd its utility is usully not questioned outside the field of mthemtics. So let s get bck to giving prtil nswer to computer science students. We would like to point out tht in this course, bsic concepts nd instruments will be presented, used for nlyzing rel or vector functions of one ore more vribles. To illustrte the utility of some of these concepts nd instruments, we will consider the following prcticl problem: constructing trin schedule. Constructing trin schedule for rilwy network is rel nd complex problem. It is bsed on the knowledge of speed restrictions in the network, trin sttions, trnsport mteril, options concerning the stops of some trins in certin sttions, nd previous computtion tht gurntees tht in idel conditions, the trins will not collide. Some concepts of clculus prove to be useful in this computtion. To gurntee tht the trins will not collide, it is necessry to know, t every moment, the position of every trin nd to ssure tht these positions do not coincide t certin moment of time. Let s consider for exmple the Timişor-Buchrest rilwy which cn be represented s curve ÃB like in the following figure: nd trin tht circultes on this rilwy in the time rnge [t 0, t 0 +T ] will be represented by point P. If in the considered time rnge there re more trins circulting on this rilwy, we will hve to describe the motion of ech of them. In order to describe the motion of trin represented by the point P, we cn ssocite to ech moment of time t [t 0, t 0 + T ] the length of the rc of curve ÃP, where P is the position on the curve ÃB where the trin is t the moment t. Therefore, function f is obtined, which is defined for t [t 0, t 0 + T ] nd tkes its vlues in the set [0, l]: f : [t 0, t 0 + T ] [0, l]; l is the distnce from A to B, on the considered rilwy. 6

7 We must emphsize tht the object tht ppered in nturl wy in this problem of describing the position of trin on rilwy, is rel function of one rel vrible, mthemticl object tht belongs to the field of interests of this course. Our trin hs to rrive t given times to its sttions nd hs some speed restrictions long the wy, hence, the function f could be quite complicted. However, there re some chrcteristics of rel motion tht hve to be trnslted mthemticlly s properties of the function f. For exmple, the rel motion is continuous, mening tht the trin moves from the position P 1 to the position P 2 grdully, pssing through ll the intermedite positions nd not by jumping. This mens tht the function f, even if complicted, must hve to following property: for ny t 2 [t 0, t 0 + T ], if t 1 tends to t 2 then f(t 1 ) tends to f(t 2 ). A function with the bove property is sid to be continuous on the intervl [t 0, t 0 + T ]. The concept of continuity is studied in this course, reveling severl properties. Hence, continuous functions tht re studies in this course re useful, for exmple, for describing the motion of trin on rilwy. If our trin leves t the moment t 0 from sttion A nd moves off continuously from A without stopping until the moment t 1 t the first sttion S 1, then the function f which describes the motion of the trin hs the following property: for ny t, t [t 0, t 1 ], t < t it results tht f(t ) < f(t ). In this course, such function is sid to be incresing. The course presents severl properties of monotonous functions. In the cse of the considered motion, this concept is useful for expressing moving off or pproching. Due to speed restrictions nd stops t the sttions, the velocity of the trin depends on its position. More exctly, it depends on the moment of time t, s in the time rnge [t 0, t 0 + T ], the trin my pss through the sme plce couple of times. In order to find the velocity of the trin t the moment t 1, we consider the men velocity f(t) f(t 1) t t 1 (distnce over time) on short time rnge [t, t 1 ] nd the limit of this men velocity when t tends to t 1 represents the velocity of the trin t the moment t 1. In this course, this limit is clled the derivtive of the function f t t 1 nd is denoted by f (t 1 ). If the trin stys in sttion in the time rnge [t 1, t 2 ] then it s velocity is zero, f (t) = 0, for t [t 1, t 2 ]. If f (t) > 0, then the trin moves off A, nd if f (t) < 0 then the trin pproches A. If the trin moves with constnt velocity in the time rnge [t 1, t 2 ], then f (t) = const in the intervl [t 1, t 2 ]. These show the utility of the concept of derivtive for describing mechnic motion. Finlly, we point out tht strting from velocity profile v(t) (which results from speed restrictions nd previously ssigning the rrivl nd deprture times) the function f(t) which describes the motion cn be recovered using the integrl formul: f(t) = f(t 0 ) + t v(τ)dτ. presented in this course. t 0 We hope tht this extremely simple nd prtil resoning mnges to convince computer science students tht they will study t this course mthemticl objects nd results tht will be useful in their future creers. 7

8 The written course is presented in stndrd form, similr to the course presented to mthemtics students. However, the spoken course is full of comments nd exmples tht re ment to illustrte the utility nd pplicbility of the concepts nd results t solving rel problems. The uthors 8

9 Prt I Introduction 1 The notions set, element of set, membership of n element in set re bsic notions of mthemtics A strict mthemtics course requires precise definition of ll the notions used to present the mteril. A definition should precisely describe notion (A) using n other notion (B), which is ssumed to be known, or in ny event simpler thn (A). Notion (B) must lso be strictly defined, nd its definition will contin nother notion (C) simpler then (B), nd so on. For the construction of mthemticl theory with exct definitions, of ll the notions, it is necessry to hve collection of very simple notions to which the rest cn be reduced nd which re themselves not defined. We will cll such notions bsic notions. From the point of view of common sense, the bsic notions of mthemtics re so self evident tht they do not require definitions. The mening of bsic notions cn be described by exmples. The notions: set, n element of set, membership of n element in set, re bsic notions of mthemtics. We cnnot obtin n exct definition of the bove notions, but it is possible to clrify their mening, by exmples. Thus, let us consider the notion of set. We my spek of the set of dys in yer, points in plne, students in lecture-room, nd so on. In these cses, ech dy of yer, ech point in plne, ech student in lecture-room is n element of the set. When concrete set is considered, n essentil thing is to be ble to ffirm for ny element if it belongs or not to the set. Thus, for the set of dys in yer, the 3 rd of July, 20 th of My, 29 th of December re ll elements of the set, while Wednesdy, Fridy, holidy, dys in yer re not. In the second exmple, only the points in the given plne re elements of the set. If the point does not lie in the given plne, or the element is not point, then the point or the element is not n element of the set. In order to define concrete set it is necessry to describe clerly the elements belonging to it. Any fulty description my led to logicl contrdiction. 9

10 2 Symbols used in set theory If x is member (n element) of set A, then we write x A, otherwise we write, x / A ( is clled the membership symbol). Two sets A nd B tht hve precisely the sme elements re sid to be equl. Thus, with respect to sets, the equlity A = B mens tht the sme set is denoted by different letters, tht is, A nd B re two nmes for the sme set. The nottion A = {x, y, z,...} mens tht the set A consists of elements x, y, z,.... In this nottion, duplicted elements re regrded s one element. For instnce: {1, 2, 3, 4, 5} = {1, 1, 1, 2, 2, 3, 4, 5}. If set A consists of ll the elements x of set B tht posses given property, then we write A = {x B... } where the property is written fter the verticl line. For instnce, let nd b two rel numbers stisfying the condition < b; then the set of points of the closed intervl [, b], tht is the set of ll rel members x such tht x b, cn be written s: [, b] = {x R 1 x b} where R 1 mens the set of ll rel members. If every element in set A is lso n element of set B, then we sy tht A is subset of B nd write A B or B A. The first reltion reds set A is contined in set B, nd the second reltions reds B contins A. It is esy to prove tht if A B nd B A, then A = B. 3 Opertions with sets Definition 3.1. For ny two sets A nd B the set of elements belonging to A or B or to both sets is clled the union of A nd B, nd is written A B. Definition 3.2. For ny two sets A nd B the set of elements belonging to A nd B t the sme time is clled the intersection of A nd B nd is written A B. Definition 3.3. For ny two sets A nd B the set of elements of B tht re not elements of A is the difference B A written B \ A. If the set A is subset of B, then B \ A is clled the complement of A in B nd is denoted s C B A. Comment The notions of union nd intersection of sets cn be extended to three, four or ny number of sets. Nmely, the union of n sets A, B, C,... is the set of those elements which belong to t lest one of these sets. The intersection of n sets A, B, C,... is the set of those elements which belong simultneously to ech set. - It is possible tht two sets A nd B hve no elements in common. In such cse A B contins no elements. Nevertheless, it is still convenient to view A B s set (contining no elements). It is clled the empty (or null) set, nd is denoted by the symbol. 10

11 For ny set A we hve A A nd A ; thus A nd re subsets of A; they re clled improper subsets, ll other subsets being proper subsets. Sometimes, the union of sets is clled the sum of sets, nd the intersection of sets the product sets. Usully, the opertions of union nd intersection of sets re defined on the set of ll subsets of given set S. These opertions, for ny A, B, C S, stisfy the following properties: (A B) C = A (B C) ssocitivity of union; (A B) C = A (B C) ssocitivity of intersection; A B = B A commuttivity of union; A B = B A commuttivity of intersection; (A B) C = (A C) (B C) distributivity of intersection over union; (A B) C = (A C) (B C) distributivity of union over intersection; for A S there is unique B S such tht A B = S, A B = : this set is S \ A; the set S possesses the property A S = A for ny A S, the empty set possesses the property: A = for ny A. There re identities, known s rules of De Morgn, which relte the opertions of complementtion, tking unions, nd tking intersections. These rules re expressed by the formuls: C S (A B) = C S A C S B; C S (A B) = C S A C S B. Definition 3.4. For ny two sets A nd B, the set of ordered couples (, b) with A, b B is clled the crtesin product of A nd B nd it is denoted A B. The crtesin product hs the following properties: A (B C) = (A B) (A C); for ny sets A, B, C. A (B C) = (A B) (A C). 4 Reltions Definition 4.1. A binry reltion in the set A is subset R of the crtesin product A A : R A A. 11

12 Trditionlly, the membership (x, y) R is denoted by xry. The set R = {(x, y) R R : x 2 + y 2 1} is binry reltion in the set of ll rel numbers R. Definition 4.2. A binry reltion R in the set A is clled reflexive if for ny x A we hve xrx. The set R = {(x, y) R R : x y 0} is reflexive binry reltion in the set of ll rel numbers R. Definition 4.3. A binry reltion R in the set A is clled symmetric if xry yrx for ny x, y A The set R = {(x, y) R R : x 2 + y 2 1} is symmetric binry reltion in the set of ll rel numbers R. Definition 4.4. A binry reltion R in the set A is clled ntisymmetric if xry nd yrx x = y for ny x, y A The set R = {(x, y) R R : x y 0} is n ntisymmetric binry reltion in the set of ll rel numbers R. Definition 4.5. A binry reltion R in the set A is clled trnsitive if xry nd yrz xrz for ny x, y, z A. The set R = {(x, y) R R : x y 0} is trnsitive binry reltion in the set of ll rel numbers R. Definition 4.6. A binry reltion R in the set A is totl if for ny x, y A, t lest one of the following sttements is true: xry, yrx. The set R = {(x, y) R R : x y 0} is totl binry reltion in the set of ll rel numbers R. Definition 4.7. A binry reltion R in the set A is prtil of there exist x, y A such tht none of the following sttements is true: xry, yrx. The set R = {(x, y) R R : x 2 + y 2 1} is prtil binry reltion in the set of ll rel numbers R. Definition 4.8. A binry reltion R in the set A is reltion of prtil order if it stisfies the following properties: R is prtil reltion; R is reflexive; R is ntisymmetric; R is trnsitive. The inclusion of sets is reltion of prtil order in the set of ll prts of given set S. 12

13 Definition 4.9. A binry reltion R in the set A is reltion of totl order if it stisfies the following properties: R is totl reltion ; R is reflexive; R is ntisymmetric; R is trnsitive. The set R = {(x, y) R R : x y 0} is reltion of totl order in the set of ll rel numbers R. Definition A set A, together with reltion of prtil order R in A is clled prtilly ordered system nd it is denoted by (A, R). The set of ll prts of given set S, together with the reltion of inclusion is prtilly ordered system. Definition A set A together with reltion of totl order R in A is clled totlly ordered system nd it is lso denoted by (A, R). The set of rel numbers R, together with the binry reltion R = {(x, y) R R : x y 0} is totlly ordered system. Definition Let (A, R) be prtilly ordered system nd A subset of A : A A. An element A is n upper bound for the set A if verifies R for ny A. An upper bound for A is sid to be lest upper bound for A if verifies R for ny upper bound of A. If it exists, lest upper bound of A is denoted by sup A. Definition Let (A, R) be prtilly ordered system nd A subset of A : A A. An element A is lower bound for the set A if verifies R for ny A. A lower bound for A is sid to be gretest lower bound for A if verifies R for ny lower bound of A. If it exists, gretest lower bound of A is denoted by inf A. Definition Let (A, R) be prtilly ordered system. An element A is mximl if for ny A with the property R, one hs R. The fmily P(X) of ll subsets of set X ffords n illustrtion of this concepts. The inclusion reltion R = between the sets contined in X mkes the pir (P(X), ) prtilly ordered system. An upper bound for subfmily B P(X) in ny set contining B is the only lest upper bound of B. B nd B B B B Similrly, B is the only gretest lower bound of B. The only mximl element of P(X) B B is X. Definition A reltion R in set A is clled equivlency if possesses the following properties: R is reflexive, symmetric nd trnsitive. For instnce, the equlity of sets is n equivlency. For exmple, the equlity in the set of prts P (X) of given set X is n equivlency. The set R = {(x, y) Z Z : x y divisible by 5} is reltion of equivlency in the set of integers Z. 13

14 Definition A reltion R between the elements of set A nd the elements of set B is subset of the crtesin product A B; R A B. Trditionlly, (x, y) R is denoted xry. Definition A function (mpping) f of the set A into the set B written f : A B is reltion R between the elements of the sets A nd B (R A B) which posses the following properties: ) for every x A there exists y B such tht x R y; b) if (x, y 1 ), (x, y 2 ) R, then y 1 = y 2. Trditionlly, function f defined on the set A into the set B is denoted by f : A B. 5 Functions The notion of function plys n importnt role in mthemtics. It is not bsic notion, since we hve lredy seen tht it cn be defined in terms of sets. However for those strting mthemticl nlysis, it is esiest to consider mpping (function) s bsic notion clrifying it by exmples nd describing it in mnner tht stisfies common sense. If for every x A n element y B is chosen ccording to some rules, then we sy tht there is function (mpping) f of the set A into the set B, written f : A B. Thus, function is defined uniquely by the rule which mkes every x A correspond to y B. Wht does the bove description of function lck for it to be strict definition? Firstly, we must explin wht rule is; secondly, wht correspondence is. Intuitively it is cler wht rule nd correspondence re. In simple cses, these notions do not involve misunderstndings nd re sufficient for meningful mthemticl theory to be constructed on their bsis. Let us note once gin tht the rule defining the element y B is pplicble to every x A. The element x A is clled the rgument of the function f, the element y B is clled the vlue of the function f corresponding to the element x A, y = f(x), nd the function itself is rule which processes every x A into y = f(x). The set A is clled the domin of the function, nd the set of ll the elements y B for which there re x A such tht y = f(x) is clled the rnge of the function f. We shll consider functions which ssocite every rel number x A R 1 with number y = f(x) R. For this kind of functions, rule cn be given by n explicit lgebric expression; for instnce: y = x x; y = 1 x x + 2 ; y = x. 14

15 The right-hnd sides of the equlities contin the rule tht processes x into y. The rule in the first expression is: ech x should be squred nd dded to twice x. The rules in the second nd third expression cn be formulted in similr wy. The rule cn be given lso by the symbols exp, log, sin, cos, tn, cot nd lso combintions of the symbols nd lgebric opertions. For instnce, y = log sin x; y = 1 (tn x) x. The right-hnd sides of equlities define the rules for processing x into y. The rule cn be given by nother frequent method. Let f 1 nd f 2 be functions defined by expressions given bove, nd be number. We hve then set: { f1 (x) for x < f(x) = f 2 (x) for x The bove equlity cn be interpreted s rule ccording to which every x hs corresponding y. This rule cn be formulted thus: if n x is less thn, then the corresponding y is computed by rule f 1 ; but if x is greter thn or equl to, then the corresponding y is determined by rule f 2. 6 Composite function. Inverse of function. Definition 6.1. Let f : X U, g : U Y be, respectively, mppings of the set X into the set U nd of the set U into the set Y. For every x X the element g(f(x)) belongs to the set Y. The correspondence x g(f(x)) defines mpping of the set X into the set Y which is denoted by g f nd clled the composition of mppings. If X, U, Y re sets of numbers, then the composition of the mppings (functions) g f is clled the superposition of the functions or composite function. Comment 6.1. The rule ssociting the element x X with the element g(f(x)) is tht the mpping f is pplied first to x (s result, the element f(x) U is obtined), nd then the mpping g is pplied to the obtined element f(x) U; finlly we hve g(f(x)) Y. For instnce: - if y = u 2, u = sin x, then y = (sin x) 2 = sin 2 x; - if y = tn u, u = x 2, then y = tn (x 2 ); - if y = cos u, u = x ( x ) 2, then y = cos ; 2 re composite functions. Definition 6.2. A mpping (function) f : X Y is sid to be injective (n injection) if for different vlues of the rgument there re different vlues of the function. Definition 6.3. A mpping (function) f : X Y is sid to be surjective ( surjection) if every y Y is the imge of some x X, tht is, there is n x such tht f(x) = y. 15

16 Definition 6.4. A mpping (function) f : X Y is sid to be bijective ( bijection) if it is both injective nd surjective. Comment An injective mpping possesses the following property: different vlues of the function correspond to different vlues of the rgument. For instnce, the number functions: y = 5 x; y = e x ; y = rctn x re injective. 2. Surjective functions re lso clled onto mppings. For instnce, the number function y = sin x is surjective mpping of R 1 onto the set [ 1, 1] but is not surjective mpping of R 1 onto ll R 1 (there is no inverse imge of the point y = 2). 3. A bijective function is one-to-one mpping f : X Y. This mens tht every x X hs corresponding y Y, y = f(x), with different x X hving different corresponding y Y, nd every y Y hving corresponding x X (such tht y = f(x), different x corresponding to different y Y ). Definition 6.5. Let f : X Y be bijective mpping. Then for every y Y there exists unique x X such tht f(x) = y. The correspondence y x defines mpping Y X, which is clled the inverse of f nd is denoted by f 1. For the number sets X nd Y the mpping f 1 is clled the inverse of the function f (or n inverse function). Comment The rule in the Definition 6.5 implies the following property of n inverse mpping (inverse function): f(f 1 (y)) = y for ny y Y. 2. The functions (mppings) f nd f 1 re mutully inverse, tht is, (f 1 ) 1 = f. 3. To find the inverse of given number function y = f(x), we must express x in terms of y. Thus, for y = 3 x + 2 the inverse mpping is x = y 2 3 ; for y = x3 it is x = 3 y, for y = 10 x it is x = log y. 7 Logic symbols The expressions for ny element nd there exists re frequently used in mthemtics. They re designted in specil mnner: - the first is denoted by the symbol (the first letter of the word Any inverted); - the second by the symbol (the first letter of the word Exist reflected). We shll lso use the symbol to men follows. Thus, if A nd B re two sentences, then A B mens tht B follows from A. If A B nd B A, then the sentence A nd B re sid to be equivlent, written A B (A is equivlent to B). Using this nottion, the injectivity of mpping f : X Y cn be written in the form: x 1, x 2 X, x 1 x 2 f(x 1 ) f(x 2 ) 16

17 nd the surjectivity of the sme mpping in the form: y Y, x X f(x) = y the verticl line before f(x) = y is red such tht. The designtion A def B is used when we wnt to describe notion A using sentence B. It is red A is by definition B. For instnce the nottion: X Y def {( x)(x X) (x Y )} defines X s subset of set Y : the right-hnd side of this nottion is sentence nd it is red: ny element x of X is lso n element of the set Y. 8 Converse theorem nd contrry theorem Mny mthemticl sttements (including theorems) hve the following form: if A, then B, or, which is the sme, B follows from A, A B, where A is the condition, nd B is the conclusion of the theorem. For ny sttement A B we cn construct new sttement by interchnging A nd B, nmely, write B A, tht is if B, then A, A follows from B. The theorem (sttement) B A is the converse of the theorem (sttement) A B. It is obvious tht the converse of converse is the originl theorem, therefore the two theorems re sid to be mutully converse. If the direct theorem is true, its converse my be either true or flse. Exmple 8.1. The direct theorem (Pythgors theorem) is: if tringle is right-ngled, then the squre of the hypotenuse is equl to the sum of the squres of the other two sides. The converse is: if the squre of the biggest side equls the sum of the squres of the two smller sides, the tringle is right-ngled. In this cse, both the direct theorem nd the converse re true. Exmple 8.2. The direct theorem is: if two ngles re right ngles, they re equl. The converse is: if two ngle re equl, then they re right ngles. Here the direct theorem is true, but the converse is flse. For ny sttement A we denote A the proposition tht A is flse. Exmple 8.3. If A denotes the sttement 7 is n even number then A denotes the sttement 7 is not n even number. If A is the sttement It will rin tomorrow then A is the sttement It will not rin tomorrow. If A is the sttement All bullets will hit the trget, then A is the sttement At lest one bullet will not hit the trget. For the theorem if A, then B, the sttement if A, then B is clled the contrry theorem. The contrry of contrry theorem is the initil theorem. 17

18 Exmple 8.4. For the theorem If the sum of two opposite ngles in qudrilterl is equl to 180, then circle cn be circumscribed bout the qudrilterl the contrry theorem is If the sum of two opposite ngles in qudrilterl is not equl to 180, then circle cnnot be circumscribed bout the qudrilterl. In this cse, both the direct theorem nd its contrry re true. The contrry theorem is equivlent to the converse. This mens tht the contrry theorem is true if nd only if the converse theorem is true. 9 Necessity nd sufficiency Let the sttement if A, then B be true. In this cse the condition A is sid to be sufficient for B, nd the condition B to be necessry for A. Let lso the converse be true, tht is, if B, then A. In this cse B is the sufficient condition for A nd the condition A is necessry for B. Thus, the condition A is necessry nd sufficient for B (nd the condition B is necessry nd sufficient for A). In other words, conditions A nd B re equivlent: A occurs if nd only if B is true. Exmple 9.1. Bézout s theorem is: If α is root of polynomil P (x), then the polynomil P (x) is divisible by x α without reminder. The converse is: If polynomil P (x) is divisible by x α, then α is root of the polynomil P (x). We know tht both Bézout s theorem nd its converse re true. Therefore, the necessry nd sufficient condition for the number α to the root of polynomil P (x) is tht the polynomil P (x) is divisible by x α. The following sttement is lso true: for polynomil P (x) to be divisible by x α without reminder it is necessry nd sufficient tht the number α be root of the polynomil P (x). 18

19 Prt II Single vrible clculus 10 Topology in R 1 Definition A neighborhood of the point x R 1 is set V R 1 which contins n open intervl (, b) R 1 contining x; i.e x (, b) V. For instnce, ny open intervl contining x is neighborhood of the point x. Definition Let be A R 1. A point x R 1 is clled n interior point of the set A if there exists n open intervl (, b) such tht: x (, b) A. For instnce, point x of the open intervl (, b) is n interior point of the set (, b). Definition The interior of set A R 1 is the set of ll interior points of the set A. Usully, the interior of set A is denoted by Å or Int(A). For instnce, if A is n open intervl A = (, b), then Å = (, b) = A. Definition A set A R 1 is open, if A = Å. For instnce, ny open intervl is n open set. A set A R 1 is open if nd only if it contins neighborhood of ech of its points. The union of ny fmily of open sets is open. The set of ll rel numbers R 1 nd the empty set re open. The intersection of finite number of open sets is open. Definition A set A R 1 is sid to be closed if its complement is open. The intersection of ny fmily of closed sets is closed. The union of finite number of closed sets is closed. The set of ll rel numbers R 1 nd the empty set re closed. Any closed intervl [, b] is closed set. Definition If A is subset of R 1, then point x R 1 is limit point, or point of ccumultion, of A provided every neighborhood of x contins t lest one point y x, with y A. Definition The closure A of set A R 1 is the intersection of ll closed sets contining A. The set of points belonging to A nd not to the interior Å of A is clled the boundry of A, denoted usully by A. The closure opertion hs the following properties: ) A B = A B; b) A A; c) A = A; d) A = A if nd only if A is closed set; 19

20 e) x A if nd only if every neighborhood V (x) of x intersect A. Definition A set A R 1 is bounded if there exist m, M R 1 such tht m x M for every x A. Definition A set A R 1 is compct if it is both bounded nd closed. For instnce, ny closed intervl [, b] is compct. 11 Sequences Definition A function whose domin is the set of positive integers N = {1, 2,..., n,... } nd whose vlues belong to the set R 1 of rel numbers, is clled sequence of rel numbers. Comment the vlue of the function (defining sequence of rel numbers) corresponding to rgument 1 is denoted by 1, tht corresponding to the rgument 2 by 2,..., tht corresponding to the rgument n by n. Here, 1 is clled the first term of the sequence, 2 the second term,..., n the n-th term. The sequence 1, 2,..., n,... is denoted by ( n ). In order to define sequence the vlue of the first, second,..., nd n-th terms of the sequence must be indicted. In other words, rule must be given for evluting the n-th term of the sequence, given its plce in the sequence for n = 1, 2,.... Exmple Let n = q n 1, q 0 then 1 = 1, 2 = q, 3 = q 2,..., n = q n 1,.... Exmple Let n = 1 n then 1 = 1, 2 = 1 2, 3 = 1 3,..., n = 1 n,.... Exmple Let n = n 2 then 1 = 1, 2 = 4, 3 = 9,..., n = n 2,.... Exmple Let n = ( 1) n then 1 = 1, 2 = 1, 3 = 1,..., n = ( 1) n,.... Thus: { 1 for n odd n = 1 for n even Exmple Let n = 1 + ( 1)n, then 1 = 0, 2 = 1, 3 = 0, 4 = 1. Thus: 2 { 0 for n odd n = 1 for n even It my hppen tht s the number n increses, the terms n of the sequence increses too. Definition An incresing sequence ( n ) is one in which n n+1 for ll n N. Definition A decresing sequence ( n ) is one in which n n+1 for ll n N. 20

21 Definition A sequence which is either incresing or decresing is clled monotone sequence. Exmple If q > 1, then the sequence n = q n is incresing nd if 0 < q < 1, then the sequence n = q n is decresing. If q (0, + ) nd q 1, then the sequence n = q n is monotone. Definition A sequence ( n ) is clled bounded if there exists number M such tht n M for ll n. For instnce, if 0 < q < 1, then the sequence n = q n is bounded ( n < 1). The sequence n = ( 1) n is lso bounded ( n 1). Definition A sequence ( n ) is clled unbounded if it is not bounded. In other words, if for ny M > 0 there exists n M such tht nm > M. For instnce, if q > 1, then the sequence n = q n is unbounded. Definition If ( n ) is sequence, then ny sequence ( nk ), where (n k ) = n 1, n 2,... is strictly incresing sequence of positive integers, is clled subsequence of the sequence ( n ). Comment ny subsequence of n incresing sequence is incresing; ny subsequence of decresing sequence is decresing; ny subsequence of bounded sequence is bounded. 12 Convergence It my hppen tht s the number n increses without bound, the terms n of the sequence pproch closely certin number L. In this cse we rrive t n importnt mthemticl concept tht of the limit of sequence. Definition A number L is sid to be the limit of the sequence ( n ) if for ny number ε > 0 there is number N (dependent on ε) such tht ll the terms n of the sequence with subscript n exceeding N stisfy the condition: n L < ε. In this cse we write lim n = L nd red: s n tends to infinity, the limit of n equls L or n n n L nd red: s n tends to infinity n tends to L. If n n L, then the sequence ( n ) is sid to be convergent to L. Comment

22 If the sequence ( n ) converges to L, then ny subsequence ( nk ) of the sequence ( n ) converges to L. Indeed: for ny ε > 0 there exists N such tht for n > N we hve n L < ε. Hence for n k > N we hve nk L < ε. Not every sequence hs limit. For instnce, the sequence n = ( 1) n hs no limit. Tht is becuse the subsequence 2k = ( 1) 2k = 1 converges to 1 nd the subsequence 2k+1 = ( 1) 2k+1 = 1 converges to 1. The limit of sequence ( n ), if it exists, it is unique. Assuming the contrry, tht is ( n ) converges to L 1 nd L 2, L 1 L 2, we find N 1 nd N 2 such tht n L 1 < L 1 L 2 /2 for n > N 1 nd n L 2 < L 1 L 2 /2 for n > N 2. Since for n > mx{n 1, N 2 } we hve L 1 L 2 L 1 n + L 2 n < L 1 L 2 we obtin tht L 1 L 2 < L 1 L 2 wht is bsurd. If the sequence ( n ) converges to L, then it is bounded. Indeed, considering ε = 1 nd N 1 such tht n L < 1 for n > N 1 we hve n = n L + L n L + L < 1 + L for n > N 1. Therefore n mx{ 1, 2,... N1, 1 + L } Exmple Let us show tht lim n 1 n = 0 Indeed, let ε > 0. Consider the inequlity 1 0 n < ε 1 we hve < ε, 1 n n < ε2 tht is n > 1 ε. [ ] [ ] We set N = + 1 where is the integrl prt of the number 1. It is obvious tht ε 2 ε 2 ε2 if n > N, then n > 1 ε nd inequlity 1 n 0 2 < ε will be fulfilled. Note tht when proving the existence of limit we clculted the number N for the given ε in forml, textbook mnner. From now on, we shll compute limits using other, more simple nd convenient rules. In some cses the limit of sequence ( n ) is sid to be infinity. The mening of this concept is the following: Definition The limit of the sequence ( n ) is sid to be + if for ny M > 0 there is N M such tht n > M for n > N M. For instnce, the limit of the sequence n = n 2 is +. Definition The limit of the sequence ( n ) is sid to be if for ny M > 0 there is N M such tht n < M for n > N M. For instnce, the limit of the sequence n = n 2 is. 22

23 13 Rules (for convergence of sequences) Suppose tht ( n ) nd (b n ) re convergent sequences with limits nd b respectively, then the following rules pply: Sum rule: ( n + b n ) converges to + b. Proof. Consider the inequlity: ( n + b n ) ( + b) = ( n ) + (b n b) n + b n b. Given ε > 0, let ε = 1 2 ε. Then ε > 0 nd, since lim n = nd lim b n = b, there exist n n nturl numbers N 1 nd N 2 such tht n > N 1 n < ε nd n > N 2 b n b < ε. Let N be the mximum of N 1 nd N 2 nd so n > N n + b n b = 2 ε = ε In other words ( n + b n ) converges to + b. Product rule: ( n b n ) converges to b. Proof. Since lim n = b, there is M > 0 such tht b n M for ny n N. It follows tht: n b n b = n b n b n + b n b = b n ( n ) + (b n b) Given ε > 0, let ε 1 = b n n + b n b M n + b n b, for ll n N ε 2M nd ε 2 = ε 2( + 1). Since lim n n = nd lim n b n = b, there exist N 1 nd N 2 such tht : nd n > N 1 n < ε 1 n > N 2 b n b < ε 2. Let N 3 the mximum of the N 1 nd N 2 nd so conclude tht if n > N 3 then: n b n b < ε. In other words lim n n b n = b. Quotient rule: ( n /b n ) converges to /b provided tht b n 0 for ech n nd b 0. Proof. Firstly it is shown tht if lim b n = b 0 nd b n 0 for ll n, then lim b n = 1 n n b. It is cler tht we hve: 1 1 b n b = b n b b n b Since lim b n = b there exists n integer N 1 such tht b n b < 1 n 2 b for ll n > N 1. Let 2 M be the mximum of b, 1 b 1,..., 1 b N1. Then 1 < M for ll n. 23 b n

24 So, given ny ε > 0, let ε = ε b M. Then ε > 0 nd there exists n integer N 2 such tht b n b < ε for ll n > N 2. Hence, 1 < ε for ll n > N 3 where N 3 is the mximum 1 b n b 1 of N 1 nd N 2. In other words lim = 1. By the product rule then lim n b n b n n b n = b. Sclr product rule: (k n ) converges to k for every rel number k. The sclr product rule is specil cse of the product rule. Appliction Evlute lim n n 2 + 2n + 3 4n 2 + 5n + 6. Solution: The quotient rule cnnot be pplied direct since neither the numertor nor n 2 + 2n + 3 the denomintor of converges to finite limit. 4n 2 + 5n + 6 However, if the numertor nd denomintor re divided by the dominnt term n 2 the following is obtined: It is esy to prove tht 1 n x lim n n = 1 4 n = n + 3 n n + 6 n 2. 0 nd the constnt sequence (k) hs limit k. Hence freely using the sum, product, sclr product nd quotient rules. Squeeze rule: Let ( n ), (b n ), (c n ) be sequences stisfying n b n c n for ll n N. If ( n ) nd (c n ) both converge to the sme limit L, then (b n ) lso converges to L. Proof. If n b n c n, then n L b n L c n L. Hence b n L mx{ n L, c n L }. Given ε > 0, there exist nturl numbers N 1 nd N 2 such tht n > N 1 n L < ε nd n > N 2 c n L < ε. Let N be the mximum of N 1 nd N 2. Then for n > N it follows tht b n L < ε. In the other words lim n b n = L Appliction Show tht lim ( 1) n 1 n n = 0. 2 Solution: Note tht ( 1) n 1 < 1 n. 2 Now let n = 1 n, b 2 n = ( 1) n 1 n, c 2 n = 1 n. 2 Both ( n ) nd (b n ) converge to 0. By the squeeze rule (b n ) converges to 0. n 2 24

25 Principle of monotone sequences: A bounded monotone sequence is convergent. Proof. The sttement for bounded incresing sequence is proved, the proof being similr for decresing sequence. Let ( n ) be such tht n... nd n M for ll n N. Let M 0 = sup { n n N} the lest upper bound of the set of numbers ppering in the sequence. Given ε > 0, M 0 ε cnnot be n upper bound for { n n N}. Hence, there exists vlue n = N such tht N > M 0 ε. Furthermore n M 0 by the definition of M 0 nd hence, for n > N, n M 0 < ε. This proves tht lim n = M 0. n Appliction A sequence ( n ) is defined by 1 = 1 nd n+1 = n + 1 for n 1. Show tht lim n = n 2 Solution: First, it is shown by induction on n, tht ( n ) is n incresing sequence. Since 1 = 1 nd 2 = 2, it follows tht 1 2. Now n+1 n = n + 1 n = n n 1 n n nd since n n is positive if n 1 n then n n+1. So, by induction ( n ) is n incresing sequence. ( Now 2 n 2 n+1 = 2 n n 1 = n 1 ) 2 5 ( 2 4 nd since ( n) is incresing, n 1 ) This quickly leds to ( n ) being bounded bove by 1 2 (1 + 5). By the principle of monotone sequences ( n ) is convergent. Hence, suppose tht lim n = L. Since lim n+1 = L we obtin L = L + 1 nd so L 2 = L + 1. The n n qudrtic eqution L 2 = L + 1 hs two roots, nmely 1 2 (1 ± 5). Since n 1 for ll n N, the positive root is required. Hence L = 1 2 (1 + 5). Theorem 13.1 (Bolzno-Weierstrss theorem). Any bounded sequence ( n ) of rel numbers contins convergent subsequence. Proof. Let S N = { n n > N}. If every S N hs mximum element, then define subsequence of ( n ) s follows: b 1 = n1 is the mximum of S 1, b 2 = n2 is the mximum of the S n1, b 3 = n3 is the mximum of S n2 nd so on. Therefore (b n ) is monotone decresing subsequence of ( n ). Since ( n ) is bounded, then so is (b n ) too. It follows tht (b n ) is convergent subsequence of ( n ). On the other hnd if, for some M, S M does not hve mximum element, then for ny m with m > M there exists n n following m with n > m. Now let c 1 = M+1 nd c 2 the first term of ( n ) following c 1 for which c 2 > c 1. Now let c 3 the first term of ( n ) following c 2 for which c 3 > c 2 nd so on. Therefore (c n ) is monotone incresing subsequence of ( n ). Since (c n ) is bounded, it is convergent. It is intuitively cler tht if n n L, then ll the terms of the sequence with lrge subscripts will differ very little, ll of them being pproximtely equl to L. More precisely, we hve: 25

26 Theorem 13.2 (Cuchy s criterion for the convergence of sequence). A sequence ( n ) hs limit if nd only if for ny ε > 0 there exists N ε such tht ll the terms of the sequence with subscripts p, q > N ε stisfy p q < ε. Proof. Let ssume tht the sequence ( n ) hs limit L nd let ε > 0 be number. Consider the number ε ; by definition of limit there exists n integer N such tht 2 n L < ε 2 for ll n > N. Hence p L < ε 2, q L < ε for p, q > N nd it 2 follows tht p q p L + q L < ε for p, q > N. Let ssume now tht for ny ε > 0 there exists N such tht p q < ε for p, q > N 1. Considering ε = 1 nd N 1 such tht p q < 1 for p, q > N 1 we hve: Therefore: n = n N N1 +1 n N N N1 +1, for n > N 1 n mx{ 1, 2,..., N1, N } = M According to Bolzno-Weierstrss theorem the sequence ( n ) contins convergent subsequence ( nk ). Let be L = lim n n k k nd ε > 0 number. There exists N 1 such tht for n k > N 1 we hve nk L < ε 2 nd N 2 such tht p q < ε 2 for p, q > N 2. Considering N 3 = mx{n 1, N 2 } nd n > N 3 we hve: where n k > N 3 nd n k is fixed. n L n nk + nk L < ε 14 Limit points of sequence Definition The set of limit points of sequence ( n ) is the collection of points x R 1 for which there exists subsequence ( nk ) of the sequence ( n ) such tht lim n n k k = x. Usully the set of limit points of sequence ( n ) is denoted by L( n ). The sequence ( n ) converges nd lim n = L if nd only if L( n ) = {L}. n + Definition The limit superior of sequence ( n ) is sup L( n ). The limit superior of sequence ( n ) usully is denoted by lim sup n or by lim n. n n Definition The limit inferior of sequence ( n ) is inf L( n ). The limit inferior of sequence ( n ) usully is denoted by lim inf n or lim n. n n Exmple If n = ( 1) n then L( n ) = { 1, 1} nd lim n = 1, lim n = 1. n n The sequence ( n ) converges if nd only if lim n = lim n = L. n n 26

27 15 Series of rel numbers If sequence ( n ) is given, the finite sum s n = n for ech n N cn be formed. If the sequence (s n ) converges to some limit s, then s cn justifibly be clled the sum of the infinite series n = More precisely: Definition It is sid tht the symbol n is convergent series, with sum s, if the sequence (s n ) of n-th prtil sums converges to s. If (s n ) is divergent sequence then, irrespective of its precise behvior, divergent series. n is clled Rther regrettbly, not possess sum. n is still used to denote divergent series even through it does Exmple Show tht hs sum Solution: The n-th prtil sum of lim s n = 1 it cn be deduced tht n 1 2 n is s n = n = n. Since 1 converges nd hs sum 1. 2n Exmple Show tht is divergent series. Solution: The n-th prtil sum is s n = n = 1 2 n (n+1). Since s n is divergent sequence, n is divergent. Exmple Show tht 1 n 2 + n = 1 27

28 Solution: Since the n-th prtil sum of s n = ( 1 1 ) 2 1 n 2 + n = 1 n(n + 1) = 1 n 1 n n 2 + n + cn be written s ( ) ( n 1 ) = 1 1 n + 1 n + 1 Now lim s n = 1. n 1 Exmple is specil cse of n importnt clss of infinite series, nmely 2n the geometric series x n, where x is rel number. n=0 Notice tht the summtion here begins t n = 0, nd not t n = 1. For this series the sum of the first n terms is so nd by substrction s n = + x + x x n 1 x s n = x + x x n s n = (1 xn ) 1 x for x 1. Therefore, lim s n = for x < 1. n 1 x Since (s n ) diverges for x 1 the following result is obtined: Result: The geometric series x n = + x + x , 0 n=0 converges if nd only if x < 1. Moreover, its sum is then 1 x. Since the sum of convergent series is defined to be the limit of the sequence of n-th prtil sums of the series in question the rules concerning the convergence of sequence cn be used to estblish theorems concerning convergence of series. The first result provides useful test for the divergence of series. The vnishing condition If n is convergent, then lim n = 0. n Proof. Suppose tht (s n ) converges to some limit s. Hence, (s n 1 ) lso converges to s. But n = s n s n 1 nd so lim n n = 0. 28

29 Exmple Consider vnishing condition n n + 1. Since n = n n + 1 nd lim n n n n + 1 does not converge. In other words n n + 1 = 1 0, by the is divergent. It is importnt to note tht the converse of the vnishing condition is flse! In other words there re divergent series whose terms nevertheless tend to zero. Exmple Consider ( n n 1). The n-th prtil sum my be written s: s n = ( 1 0) + ( 2 1) + + n n 1 = n Clerly (s n ) is divergent sequence nd so ( n n 1) is divergent series. However n = n n 1 = ( n n 1)( n + n 1) ( n + n 1) = 1 ( n + n 1) 0 Cuchy s criterion for the convergence of series A series n converges if nd only if for ny ε > 0 there exists N such tht for n N nd p 1 the following inequlity holds: n+1 + n n+p < ε. Proof. Let be s n the n-th prtil sum of the series: The series s n = n. n converges if nd only if the sequence (s n ) converges. The sequence (s n ) converges if nd only if for ny ε > 0 there exists N ε such tht for q, r > N ε the following inequlity hold s q s r < ε. This is equivlent with the condition: for ny ε > 0 there exists N ε such tht for n N ε nd p 1 the following inequlity hold: n+1 + n n+p < ε. 16 Rules (for convergence of series) By considering the n-th prtil sums of the pproprite series nd the sum nd the sclr product rules for sequences the following elementry results cn be esily proved. 29

30 Sum rule: nd If n nd b n re convergent series, then ( n +b n ) is lso convergent ( n + b n ) = n + b n Sclr product rule: k R 1 nd If n is convergent, then (k n ) is convergent for ny (k n ) = k n. Rules will be estblished below which cn be used to test whether given series converges or not. Integrl test: n N. Let j n = Let f : R 1 + R 1 + be decresing function nd let n = f(n) for ech f(x) dx. The series n converges if nd only if j n converges. n 1 The proof of the sttement is given in the section where the Riemnn integrl is rigourously defined. 1 Appliction Estblish tht the p series converges if nd only if p > 1. np Solution: Consider the function f p : R 1 + R 1 + given by f p (x) = 1. When p > 0 this is xp decresing function of x nd n = 1 n = f 1 p(n) is the n-th term of the p series p n. p For p 1 n n 1 x1 p j n = dx = 1 xp 1 p = p (n1 p 1). So, for p > 1, lim j n = 1 n p 1, nd forp < 1 the sequence (j n) is divergent. For p = 1, j n = n 1 1 x dx = ln x n 1 = ln(n) nd so (j n ) gin diverges. 1 When p 0, diverges by the vnishing condition. np Exmple The hrmonic series 1 n diverges nd the series 1 n converges. 2 The p-series, together with geometric series, give fund of known convergent nd divergent series. 30

31 First comprison test: n convergent. If 0 n b n for ll n N, then b n convergent implies Proof. Let s n = n N. If n k=1 k nd t n = k=1 b k. From the given conditions 0 s n t n for ll b n converges, then lim n t n = t nd, since (t n ) is n incresing sequence, t n t for ll n N. Therefore, s n t for ll n N nd hence (s n ) is bounded nd incresing sequence. It follows tht (s n ) converges nd hence n converges. Exmple The series Solution: Let n = 1 + cos n is convergent. 3 n + 2 n3 1 + cos n 3 n + 2 n 3. Then n 0 since cos n 1. Also n 2 3 n + 2 n 3 since cos n 1. Therefore, since 3 n > 0, n < 2 2 n = 1 3 n. Let b 3 n = 1 n. Then 3 converges, being the p series with p = 3. Hence n lso converges. b n Second comprison test: Let n lim = L 0. n b n Then, n converges if nd only if Proof. Suppose tht n nd b n b n converges. b n is convergent nd let s n = n ; t n = b 1 + b b n. n Since lim = L for ε = 1 there is n N 1 such tht n b n n L b n < 1, for ll n > N 1 Hence, n b n = n b n be positive term series such tht = n L + L b n n L b n + L < 1 + L = k, n > N 1. 31

32 Now consider the positive term series k b N1 +n. α n nd Hence 0 α n β n for ll n N. Since nd hence test, β n where α n = N1 +n nd β n = b n converges, then so does n=n 1 +1 b n too β n converges by the sclr product rule for series. By the first comprison α n converges nd, since the ddition of finite number of terms to convergent series produces nother convergent series, convergent implies tht n lso converges. This proves tht n convergent. The converse of this sttement cn be proved by reversing the roles of n nd b n in the bove rgument nd observing tht b n n 1 L. Exmple Show tht the series 2 n n 2 5 n + 8 is divergent. 2 n Solution: Let n = n 2 5 n + 8 nd b n = 1 n. Then n 2 n 2 = b n n 2 5 n + 8 n diverges by comprison with the divergent hrmonic series. b n 2 0. Hence Rtio test: Let n be series of positive terms nd for ech n N let α n = n+1. n Suppose tht (α n ) converge to some limit L. If L > 1 then n diverges; if L < 1 then n converges; nd if L = 1 the test give no informtion. Proof. Suppose tht L < 1 nd let ε = 1 (1 L). 2 Now ε > 0 nd L + ε = k < 1. Since lim α n = L there is vlue N ε such tht n α n = α n L + L ε + L = k < 1 for ll n > N ε. Therefore n+1 k n for ll n > N ε. Let β n = Nε+n. Then β n+1 k β n for ll n N nd so (by induction on n) β n+1 k n β 1, for ll n N Now k n β 1 is convergent geometric series since k < 1. Hence n=0 n lso converges. β n converges nd 32

33 Suppose now tht L > 1 nd let ε = L 1. Now ε > 0 nd since lim α n = L there is n vlue N ε such tht α n > L ε = 1 for ll n > N ε. Hence n+1 > n for ll n > N ε nd so n > Nε for ll n > N ε. Since Nε 0, ( n ) does not tend to zero. By the vnishing condition n diverges. Exmple Determine those vlues of x for which Solution: Here n = n (4x 2 ) n, nd so Now α n 4x 2. Hence α n = (n + 1) (4x2 ) n+1 n (4x 2 ) n = 4x 2 n (4x 2 ) n is convergent. ( ). n n diverges if 4x 2 > 1 (in other words x > 1 2 ), n converges if x < 1 2, nd no informtion is gined if x = 1 2. If x = 1 2 then n = n diverges. Therefore n (4x 2 ) n converges if nd only if x < 1 2. The root test: Let n be series of positive terms. If there exists N nd k (0, 1) such tht n n k for n > N, then the series converges. If n n 1 for n infinity of terms of the series, then the series diverges. Proof. If there exists N nd k (0, 1) such tht n n k for n > N, then n k n for n > N. consequently, the series n cn be compred with the geometric series k n which converges since k < 1. This proves the first cse. If n n 1 for n infinity of terms of the series, then n 1 for n infinity of terms of the series. Hence the vnishing condition is not stisfied nd the series diverges. Appliction The series 1 n 1 2 for n 2. 1 n n n converges. Using the root test we hve 1 n = n Alternting series test (Leibnitz): If the sequence (b n ) is monotonic decresing sequence nd b n 0, then the lternting series ( 1) n 1 b n converges. n 33

34 Proof. Let s m = ( 1) n 1 b n. Then s 2m = b 1 (b 2 b 3 ) (b 2m 2 b 2m 1 ) b 2m b 1 nd hence the sequence (s 2m ) is bounded bove. Since s 2m = (b 1 b 2 ) + (b 3 b 4 ) + + (b 2m 1 b 2m ) nd (b n ) is decresing, (s 2m ) is incresing. Consequently, (s 2m ) converges nd let be s = lim s 2m. m Similrly, the sequence (s 2m+1 ) is decresing sequence which is bounded below by b 1 b 2 nd so (s 2m+1 ) converges to limit t = lim s 2m+1. 2m+1 Now t s = lim (s 2m+1 s 2m ) = lim b 2m+1 = 0 m m Finlly, it is shown tht lim s n = s. n If ε > 0, there re integers N 1 nd N 2 such tht s 2m s < ε for ll m > N 1 nd s 2m+1 s < ε for ll m > N 2. Let N be the mximum of 2N 1 nd 2N If n > N, then either n = 2 m ( nd m > N 1 ) or n = 2 m + 1 ( nd m > N 2 ). In either cse s n s < ε for ll n > N. In other words, the sequence (s n ) converges nd hence, the series ( 1) n 1 b n is convergent. Exmple The series with limit zero. ( 1) n 1 1 n is convergent since 1 n is decresing sequence 17 Absolute convergent series Definition The series bsolute vlues n converges. n is sid to be bsolute convergent if the series of A convergent series which is not bsolute convergent is clled conditionlly convergent. Absolute convergence implies convergence. If the series n converges, then the series n converges too. Proof. By tringle inequlity we hve s n s m = m n m n where n > m. Since n converges, for ny ε > 0, there exists n integer N such tht for n > m > N we hve m n < ε nd hence s n s m < ε. By the Cuchy criterion we obtin tht the series n converges. 34

35 ( 1) n Exmple The series converges, since it is bsolute convergent; 2 n n=0 2. n=0 1 2 n = ( 1) n The lternting hrmonic series is conditionlly convergent. It converges (by n 1 Leibnitz criterion) but it is not bsolutely convergent since n diverges. Comment Absolute convergence cn be tested by the convergence tests given bove for series with positive terms. Absolute convergence is importnt for the following reson: the sum of n bsolute convergent series n does not depend on the order in which the terms n re tken. It cn be shown tht for ny conditionlly convergent series S, we cn hve n = S by rerrnging the terms of n nd ny rel number n. For exmple rerrnging the terms of the lternting hrmonic series we cn hve it sum to 10 6 or to 10 6 or within ε to the number of toms in universe. Cuchy product of series: If n nd b n re bsolute convergent series nd c n = 1 b n + 2 b n n b 1 then c n is bsolute convergent nd ( ) ( ) c n = n b n Proof. Suppose first tht n nd b n re positive term series nd consider the rry: 1 b 1 1 b 2 1 b b 1 2 b 2 2 b b 1 3 b 2 3 b If w n is the sum of the terms in the rry tht lie in the n n squre with 1 b 1 t one corner, then w n = s n t n, where s n nd t n re the n-th prtil sums of n nd respectively. 35 b n

36 ( ) ( ) Hence lim w n = n b n. n Now c n is the sum of the terms in the rry summed by digonls nd so if u n is the prtil sum of c n then: By the squeeze rule we hve s required. lim u n = n w [ n 2 ] u n w n. ( ) ( ) u n b n For the generl cse the bove rgument cn be pplied to the series nd c n to deduce tht the series combintion of the series + n, c n is bsolute convergent. As n, b + n nd b n we hve: ( ) ( ) c n = n b n. n, b n, c n is liner 18 Limit of function t point An importnt concept in clculus is the limit of function t point. It is used in the study of the continuity, derivtives, integrls, nd other importnt topics in clculus. Consider function f : A R 1, whose domin A is subset of R 1. The behviour of the function f s x pproches fixed rel vlue shll now be investigted. It shll be ssumed tht f(x) is defined for ll x close to but not necessrily t. In other words, it is ssumed tht the domin of f contins set of the form ( r, ) (, + r) for some r > 0. Definition L is clled the limit of f(x) s x pproches, if for ny ε > 0, there exists number δ > 0 (depending on ε) such tht f(x) L < ε for ll x A, x nd x < δ. We denote this by lim f(x) = L or f(x) L for x. x 36

37 Comment Definition 18.1 does not depend on the vlue of f t (if this exists) s the point is excluded from considertion. If the vlue f() exist, my violte the inequlity. Given the function f nd the vlue L the inequlity f(x) L < ε mens L ε < f(x) < L + ε nd therefore, ε cn be regrded s the prescribed ccurcy of pproximting L, i.e how close to L one wnts to get. The number δ is not uniquely determined by ε. You cn lwys tke smller δ in the sense tht if, for given ε we hve: then, for ny 0 < δ < δ 1 we hve Exmple Let us show tht 0 < x < δ 1 f(x) L < ε 0 < x < δ f(x) L < ε. x 2 4 lim x 2 x 2 = 4. Let ε > 0 nd consider the inequlity x 2 4 x 2 4 < ε or 4 ε < x2 4 x 2 < 4 + ε is equivlent, for x 2 to 4 ε < x + 2 < 4 + ε or 2 ε < x < 2 + ε showing tht we cn tke δ = ε. Exmple Let us show tht t ny point > 0, the function f(x) = x hs the limit lim x =. x Indeed, if ε > 0, then the inequlity becomes, by squring x < ε or ε < x < + ε 2 ε + ε 2 < x < + 2 ε + ε 2 For the given nd ε we cn tke δ = 2 ε + ε 2. Exmple The limit lim sin 1 does not exist. x 0 x Recll tht rbitrry close to = 0 there exists x such tht f(x) = 1 s well f(x) = 1. Therefore for ny L there is ε > 0 such tht for ny δ > 0 there exist x such tht: x 0 nd x 0 < δ nd sin 1 x L > ε. The limit of f s x pproches is unique. Indeed: ssume tht lim f(x) = L 1 nd lim f(x) = L 2 nd L 1 L 2. For ε = L 1 L 2 x x 2 there exists δ 1 such tht f(x) L 1 < ε for 0 < x < δ 1, nd there exists δ 2 such tht f(x) L 2 < ε for 0 < x < δ 2. Hence, for 0 < x < min{δ 1, δ 2 } we hve L 1 L 2 L 1 f(x) + f(x) L 2 < L 1 L 2 which is bsurd. 37

38 Heine s criterion for the limit: The function f : A R 1 R 1 hs limit s x pproches if nd only if for ny sequence (x n ), x n A, x n, nd x n s n, the sequence (f(x n )) converges. Proof. Assume tht L is the limit of f(x) s x pproches nd consider sequence (x n ), x n A, x n nd x n s n. For ε > 0 there exists δ > 0 such tht 0 < x < δ f(x) L < ε. For δ > 0 there exists N such tht x n < δ for n > N. Hence: f(x n ) L < ε for n > N. Therefore, the sequence (f(x n )) converges. Assume now tht for ny sequence (x n ), x n A, x n nd x n s n the sequence (f(x n )) converges. Firstly, we show tht lim f(x n ) is independent on (x n ). n For tht ssume the contrry i.e there exist (x n), (x n); x n, x n A, x n, x n nd lim n x n = lim x n = for which lim f(x n n n) = L L = lim f(x n n). Consider the sequence (x n ) defined s { x k for n = 2 k x n = x k+1 for n = 2 k + 1 nd remrk tht x n A, x n nd lim x n =. Hence the sequence (f(x n )) converges. n Let be L = lim f(x n ) nd remrk tht lim f(x n n n) = L nd lim f(x n n) = L hs to be the sme L, i.e. L = L; L = L. It follows tht L = L wht is bsurd. Denote now by L the common vlue of lim f(x n ) nd show tht lim f(x) = L. For tht ssume the contrry. n x It follows tht there exists ε 0 > 0 such tht for ny n N there exists x n A, x n such tht x n < 1 n nd f(x n) L ε 0. Hence the sequence (f(x n )) does not converge to L even x n A, x n nd x n s n. Tht is bsurd. Cuchy-Bolzno s criterion for the limit: The function f : A R 1 R 1 hs limit s x pproches if nd only if for ny ε > 0 there exists δ > 0 such tht 0 < x < δ nd 0 < x < δ f(x ) f(x ) < ε. Proof. Assume tht L = lim x f(x) nd consider ε > 0. There is δ > 0 such tht 0 < x < δ f(x) L < ε 2. Hence, 0 < x < δ nd 0 < x < δ f(x ) f(x ) f(x ) L + f(x ) L < ε. Assume now tht for ny ε > 0 there exists δ > 0 such tht 0 < x < δ nd 0 < x < δ f(x ) f(x ) < ε nd consider sequence (x n ), x n A, x n nd x n n. For δ > 0 there is N such tht x n < δ for n > N. Hence, for n, m > N we hve f(x n ) f(x m ) < ε. This mens tht the sequence (f(x n )) converges. Applying Heine s criterion, the function f hs limit s x pproches to. 19 Rules for the limit of function ) If k is constnt, then lim x k = k. 38

39 b) If lim x f(x) = L nd lim x g(x) = M, then lim x (f(x) ± g(x)) = L ± M. c) If lim x f(x) = L nd lim x g(x) = M, then lim x f(x) g(x) = L M. f(x) d) If lim f(x) = L, g(x) 0 nd lim g(x) = M 0, then lim x x x g(x) = L M. Proof. We prove prt b) (the other prts re proved similrly, the proof is rther technicl nd cn be skipped t first reding). For ny ε > 0, there re positive δ 1 nd δ 2 such tht 0 < x < δ 1 f(x) L < ε 2 For 0 < x < min{δ 1, δ 2 } we hve 0 < x < δ 2 f(x) M < ε 2 f(x) ± g(x) (L ± M) f(x) L + g(x) M < ε. Pinching rule: Suppose tht the inequlity f(x) g(x) h(x) holds for ll x in some intervl round, except perhps t x =. If lim f(x) = L nd lim h(x) = L then lso x x lim g(x) = L. x Proof. Since f(x) g(x) h(x), we hve f(x) L g(x) L h(x) L. Hence g(x) L mx { f(x) L, h(x) L }. For ε > 0 there exist δ 1 > 0 nd δ 2 > 0 such tht 0 < x < δ 1 f(x) L < ε, 0 < x < δ 2 h(x) L < ε. Hence, for 0 < x < min {δ 1, δ 2 } we hve g(x) L mx { f(x) L, h(x) L } < ε. Exmple Show tht lim sin θ = 0 nd lim cos θ = 1. θ 0 θ 0 Let θ be mesured in rdins nd consider the ngle θ s centrl ngle in circle with rdius of 1. 39

40 The re of the circulr sector OAB is θ 2. The re of the tringle OAB is sin θ 2. sin θ Therefore 0 θ 2 2. Pinching sin θ between 0 nd θ, both of which pproch 0 s θ 0, proving the first limit. From the Pythgoren reltion sin 2 θ + cos 2 θ = 1 we get moreover lim θ 0 cos2 θ = lim(1 sin 2 θ) = 1 θ 0 But lim cos 2 θ = (lim cos θ) 2 nd therefore lim cos θ = +1 or 1. The negtive sign is θ 0 θ 0 θ 0 eliminted, since cos θ is positive ner θ = 0. Exmple We use the pinching rule to prove lim x sin 1 x 0 x = 0 The function x sin 1 x is bounded below by x nd bove by x, so tht x x sin 1 x x. As x 0, we lso hve x 0, nd therefore proving the climed limit. 0 lim x 0 x sin 1 x 0. We cn similrly show tht lim x 0 x 2 sin 1 x = 0 Exmple We cn show tht lim x 0 e x = 1 pinching the exponentil function e x ner 0 between the functions 1 + x nd 1 + x + x 2 i.e. 1 + x < e x < 1 + x + x 2 for x (, 1). Using the bove inequlities we cn similrly clculte e x 1 lim x 0 x = 1. The substitution rule: lim g(f(x)) = M. x Assume tht lim x f(x) = L nd lim y L g(y) = M. Then Proof. Let ε > 0 be given. Since g(y) M s y L there exists δ 1 > 0 such tht 0 < y L < δ 1 g(y) M < ε. Also since lim f(x) = L there exists δ 2 > 0 such tht x 0 < x < δ 2 f(x) L < δ 1. Therefore, 0 < x < δ 2 f(x) L = y L < δ 1 g(y) M = g(f(x)) M < ε proving tht lim g(f(x)) = M. x 40

41 20 One sided limits The limit lim x f(x) = L in Definition 18.1 is two-sided limit, since the vrible x pproches the point from both sides. We now nlyze one-sided limits, where the vrible x pproches the point on one side. This is necessry if the function is defined only on one side of the point in question, or if pproching the point from different sides gives different limits. We use the following terminology: x pproches from the right, lso x pproches from the bove, denoted by x + or x, mens < x < + δ for δ > 0 sufficiently smll. x pproches from the left, lso x pproches from the below, denoted by x or x, mens δ < x < for δ > 0 sufficiently smll. Definition (one-sided limits) ) L is clled the right limit of f t or the limit of f(x) s x pproches from the right (or from bove), denoted by lim f(x) = L or lim f(x) = L x + x if for ny ε > 0, there exists number δ > 0 such tht < x < + δ f(x) L < ε; b) L is clled the left limit of f t, or the limit of f(x) s x pproches from the left (or below), denoted by lim f(x) = L or lim f(x) = L x x if for ny ε > 0, there exists number δ > 0 such tht δ < x < f(x) L < ε. Remrk If the left limit nd the right limit exist nd re equl lim f(x) = lim = L x x + then the limit of f t exists nd equls the sme vlue L lim f(x) = L. x Exmple The function f(x) = x is defined only for x 0. As x pproches 0 from the right, the vlue of x tends to 0, lim x x = 0. Exmple The function sign(x) = x x = { 1 if x > 0 1 if x < 0 does not hve limit t = 0; however, the two one-sided limits exist lim sign(x) = 1 nd lim x sign(x) = 1. x 0

42 Exmple Step nd stircse function. The step function is defined s: 0 if x < 0 1 step(x) = if x = if x > 0 which, for x 0, cn be expressed s The step function hs one-sided limits t 0 step(x) = 1 (1 + sign(x)). 2 lim step(x) = 1 nd lim x 0 + step(x) = 1. x 0 The trnslted step function step(x ) hs its step t the point where it hs the two one-sided limits lim step(x ) = 1 nd lim x + step(x ) = 1. x A stircse function is function with severl steps, for exmple, S m (x) = m step(x n). n=0 At ech step point, the stircse function hs left limit nd right limit which re different,nd lso not equl to the vlue of the function S m t tht point. At ll other points the left limits nd the right limits coincide, nd therefore the two sided limits exist t x n. Definition A function f : A R 1 R 1 is incresing if x 1, x 2 A, x 1 < x 2 f(x 1 ) f(x 2 ). Definition A function f : A R 1 R 1 is decresing if x 1, x 2 A, x 1 < x 2 f(x 1 ) f(x 2 ). Definition A function f : A R 1 R 1 is monotone if it is incresing or decresing function. Monotone limits exist: If function f : (, b) R is monotone, then the one-sided limits lim f(x) nd lim f(x) exists for ny x 0 (, b). x x0 x x0 Proof. Consider x 0 (, b) nd the set S x0 = {f(x) x < x 0 } If f is incresing, then the set S x0 is bounded bove by f(x 0 ) nd if f is decresing, then the set S x0 is bounded below by f(x 0 ). If f is incresing, then the lest-upper-bound of S x0 i.e. sup S x0 is the left limit of f t x 0 nd if f is decresing the gretest-lower-bound of S x0, i.e. inf S x0 is the left limit of f 42

43 t x 0. In this wy it ws shown tht: for n incresing function f the left limit in x 0 exists nd lim x x 0 f(x) = sup S x0. For decresing function f the left limit in x 0 exists nd lim x x 0 f(x) = inf S x0. Considering the set R x0 increses, then nd if f decreses, the = {f(x) x > x 0 } it cn be proven, in similr wy, tht if f lim f(x) = inf R x0 x x + 0 lim f(x) = sup R x0 x x Infinite limits Infinity (± ) is mthemticl symbol nd not number which is subject to rithmetic opertions. Definition (infinite limits) The function f hs the right limit + t denoted by M > 0 there is δ > 0 such tht f(x) > M whenever < x < + δ. The function f hs the right limit t denoted by M > 0 there is δ > 0 such tht f(x) < M whenever < x < + δ. lim f(x) = + if for ny x + lim f(x) = if for ny x + The left limits: nd the two-sided limits lim f(x) = +, lim x f(x) = x lim f(x) = +, lim f(x) = x x re defined nlogously. Exmple Show tht 1 ) lim x 0 x = b) lim x 0 x = nd lim 1 x 0 + x = +. 43

44 Exmple A function cn hve, t point finite one-sided-limit s the point is pproched from one side, nd n infinite one-sided-limit s the point is pproched from the other side. For exmple the function { 0 if x 0 h(x) = 1 if x > 0 x lim h(x) = 0, lim x 0 h(x) = +. x 0 + Limits s x + or s x re clled limits t infinity, not to be confused with the infinite limits! Definition (limits t infinity) The number L is the limit of f(x) s x pproches +, denoted by lim x + f(x) = L if for ny ε > 0, there exists number M > 0 such tht The limit t, x > M f(x) L < ε lim f(x) = L is defined nlogously. x Exmple The function f(x) = lim f(x) = 1 nd lim x 1 x2 hs the following limits t infinity 1 + x + x2 f(x) = 1. x + 22 Limit points of function t point Definition (limit point t ) The number L is limit point of f(x) t if there exists sequence (x n ) such tht x n A, x n, lim x n = nd lim f(x n) = L. n + n + Usully, the set of limit points of f(x) t is denoted by L (f). inf L (f) is clled the inferior limit of f t nd it is denoted by lim f(x), x lim f(x) def = inf L (f). x sup L (f) is clled the superior limit of f t nd it is denoted by lim x f(x), def lim f(x) = sup L (f). x 44

45 The following sttement holds: The number L is the limit of f(x) s x pproches if nd only if lim f(x) = lim f(x) = L. x x Proof. Assume first tht lim f(x) = L nd consider sequence (x n ) such tht x n x A, x n nd lim x n =. For ε > 0 there exists δ > 0 such tht n 0 < x < δ f(x) L < ε. Since lim n x n = there is N such tht x n < δ for n > N. Hence f(x n ) L < ε for n > N. Therefore, lim n f(x n ) = L. We obtin in this wy tht L (f) = {L} nd consequently lim f(x) = lim f(x) = L. x x Assume now tht lim f(x) = lim f(x) = L nd suppose tht L is not the limit of f(x) x x s x pproches. Then, there exists ε 0 > 0 such tht for every n N there exists x n such tht x n < 1 n nd f(x n) L > ε 0. On the other hnd, the sequence (f(x n )) is bounded nd hs subsequence (f(x nk )) which hs limit. It is cler tht is different from L. Hence L (f) contins t lest two elements. Exmple If f(x) = sin 1 x for x R1 \ {0} then L 0 (f) = [ 1, 1]. lim f(x n n k k ) 23 Continuity Nively function f : A R 1 R 1 is continuous if its grph is continuous curve. In prticulr if the domin of f contins neighborhood of fixed rel number, then the grph of f cn be drwn through the point (, f()) without removing the pen from the pper. The desired behvior t (, f()) cn be rrnged by insisting tht, for ll vlues of x, sufficiently close to, f(x) is close to f(). If f : A R 1 R 1 is function whose domin contins neighborhood of, then the following definition holds. Definition The function f is continuous t if lim x f(x) = f(). Note tht this definition demnds three things: - firstly tht lim x f(x) exists, - secondly tht f() is defined, 45

46 - finlly tht the previous two vlues re equl. The ε, δ definition of lim x f(x) = L cn be esily dpted to give the following ε, δ definition of continuity t. Definition The function f is continuous t if nd only if for every ε > 0, there exists δ > 0 such tht: x < δ f(x) f() < ε. Exmple Use the ε, δ definition of continuity to prove tht f(x) = x 2 is continuous t = 0. Solution: For ny ε > 0, determine those x for which f(x) f(0) < ε. Now f(x) f(0) = x 2 0 = x 2 < ε provided x < ε. So let δ = ε. If x 0 < δ, then f(x) f(0) < ε. In other words, lim f(x) = 0 nd, since f(0) = 0, lim f(x) = f(0). x 0 x 0 Hence, f is continuous t 0. If function f is continuous for ll x in the rnge < x < b, then it cn be sid tht f is continuous on the intervl (, b). If f is continuous for ll x in its domin, it cn be simply sid tht f is continuous. Definition If lim f(x) exists nd equls f(), then f is clled right-continuous t x +. Definition If lim f(x) exists nd equls f(), then f is clled left-continuous t x. The ε, δ formultion of the lst two definitions re not difficult to write down. Moreover, the following result holds. A function f is continuous t if nd only if is both left-continuous nd right-continuous t. Note: If function f is only defined on the closed intervl [, b] nd it is climed tht f is continuous on [, b] wht is ment is tht f is continuous on (, b), right-continuous t nd left-continuous t b. Heine s criterion for continuity: The function f : A R 1 R 1 is continuous t A if nd only if for ny sequence (x n ), x n A, x n the sequence (f(x n)) n converges to f(). Proof. The result is obtined from Heine s criteri for the limit. Cuchy-Bolzno s criterion for continuity: The function f : A R 1 R 1 is continuous t A if nd only if for ny ε > 0 there exists δ > 0 such tht x < δ nd x < δ f(x ) f(x ) < ε. Proof. The result is obtined from Cuchy-Bolzno s criteri for the limit. 46

47 24 Rules for continuity Sum rule: If f nd g re continuous t, then f + g is continuous t. Product rule: If f nd g re continuous t, then f g is continuous t. Reciprocl rule: If f is continuous t nd f(x) 0, then 1 f is continuous t. Squeeze rule: Let f, g nd h be such tht h(x) f(x) g(x) for ll x in some neighborhood of nd such tht h() = f() = g(). If h nd g re continuous t, then so is f too. The proofs of the bove rules re left s exercises. Composite rule: Let f nd g be continuous t nd f(), respectively. Then g f is continuous t. Proof. Let f() = b. Since g is continuous t b for every ε > 0 there exists δ 1 > 0 such tht t b < δ 1 g(t) g(b) < ε. Since f is continuous t for δ 1 > 0, there exists δ 2 > 0 such tht Now we deduce tht x < δ 2 f(x) f() < δ 1. x < δ 2 g(f(x)) g(f()) < ε. Hence, for ny ε > 0, there exists δ = δ 2 > 0 such tht x < δ (g f)(x) (g f)() < ε. Exmple Given tht the identity function x x the constnt function x k nd the trigonometric functions sine nd cosine re ll continuous, the following re proved to be continuous functions: ) x x2 + 2x + 3 x 2 + x + 1, b) x x 3 cos x 2, { x sin 1 if x 0 c) x x 0 if x = 0 47

48 The extension by continuity: If f : A R 1 R 1 is not defined t the point but L = lim f(x) exists, then the function g : A {} R 1 R 1, defined by g(x) = f(x) for x x nd g() = L, is continuous t. The function g is clled the extension by continuity of the function f. Exmple The function g : R 1 R 1 defined by { sin x for x 0 g(x) = x 1 for x = 0 is the extension by continuity of the function f : R 1 \ {0} R 1, defined by: f(x) = sin x x 25 Properties of continuous functions The results in this section show tht the definition of continuity leds nturlly to the intuitive geometric interprettion which is used when sketching the grphs of continuous functions. The boundedness property: Let f be continuous on the intervl [, b]. Then (1) f is bounded on [, b]; (2) f ttins its bounds somewhere on [, b]. Comment Wht is being sid is tht, for (1) there exist numbers m nd M such tht m f(x) M for ll x [, b]. For (2) if m nd M re chosen to be the infimum nd supremum, respectively, of the set {f(x) x b}, then (2) clims tht there re numbers c nd d in [, b] such tht m = f(c) nd M = f(d). Proof of boundedness property. (1) Let B = {x x [, b] nd f is bounded on [, x]}. Clerly B nd B is bounded bove by b. By the completness property of R 1, B possesses t lest upper bound. Let c = sup B. Since f is right-continuous t, for ε = 1 there exists δ > 0, such tht < x < + δ f(x) f() < 1 f(x) < 1 + f() Hence, f is bounded on [, + δ] nd so c + δ >. We wnt to show tht = b. 2 Suppose tht c < b. Since c >, f is continuous t c. Then, for ε = 1 there exists δ > 0 such tht x c < δ f(x) 1 + f(c). 48

49 In the other words, f is bounded on [c, c + δ ]. But then c + δ B nd this contrdicts 2 c being the supremum of B. Thus c = b nd so f is bounded on [, b], s required. (2) Since f is bounded on [, b] A = f([, b]) = {f(x) x b} is set which is bounded both bove nd below. Let m = inf A nd M = sup A. Suppose 1 tht there is no x [, b] such tht f(x) = M nd define g(x) = for x [, b]. M f(x) Now g is continuous on [, b] by the sum nd reciprocl rules. By (1) g is bounded on [, b]. Let such bound be K. Then: g(x) K 1 M f(x) K 1 K M f(x) f(x) M 1 K This contrdicts the fct tht M is the lest upper bound for f on [, b], so the ssumption tht f never tkes the vlue M is flse. Hence f ttins its upper bound on [, b]. A similr rgument shows tht f ttins its lower bound somewhere on [, b]. The intermedite vlue property: Let f be continuous on [, b] nd suppose tht f() = α nd f(b) = β. For every rel number γ between α nd β there exists number c, < c < b with f(c) = γ. Comment Here it is being sid tht, if f tke the vlues α nd β somewhere on the intervl [, b], then f must tke ll posible vlues between α nd β. Proof of the intermedite vlue property. Suppose α < γ < β nd let S = {x x [, b] nd f(x) < γ}. The set S is non-empty, since it contins. Let c = sup S. It is cler tht < c < b. If f(c) < γ, then, for ε = γ f(c) > 0, there exists δ > 0 such tht x c < δ f(x) f(c) < ε. In prticulr (c f + δ ) 2 f(c) < ε nd so But then f ( c + δ ) < γ nd hence c + δ 2 2 f ( c + δ ) f(c) < γ f(c). 2 S, which contrdicts the fct tht c is the supremum of S. Hence, f(c) γ. If f(c) > γ then, for ε = f(c) γ > 0, there exists δ > 0 such tht x c < δ f(x) f(c) < ε. 49

50 Hence c δ < x c f(x) > γ nd so x does not lie in S. In other words, sup S c δ which in turn contrdicts the definition of c. It follows tht f(c) = γ. Hence, γ is vlue of f. The intermedite vlue property hs mny pplictions nd the following exmple illustrtes one of these. Exmple Any polynomil of odd degree hs t lest one rel root. Solution: Let P (X) = X + + n X n where n is odd, nd without loss of generlity, let n = 1. We know tht P is continuous. Define r(x) = P (x) x n 1, x 0 Now r(x) = P (x) 1 x n = n 1 x x + 0 n n 1 x n x x n 1 x n nd if M is the mximum of n 1,..., 0 then r(x) M n r=1 1 x < M 1 r x = r r=1 M 1 x 1 1 x = M, for x > 1. x 1 Hence, r(x) < 1 for x > 1 + M. In prticulr 1 + r(x) > 0 for x > 1 + M. Hence, P (x) = x n (1 + r(x)) hs the sme sign s x n for x > 1 + M. Since n in odd, there exist α, β R 1 with P (α) > 0 (choose α > 1 + M) nd P (β) < 0 ( choose β < (1 + M). By the intermedite vlue property P (γ) = 0 for some γ, γ < 1 + M. Incidentlly, this shows tht P hs zero in the intervl ( (1 + M), (1 + M)). In fct ll the rel zeros of P lie in this intervl. Theorem 25.1 (The intervl theorem). Let f be continuous on I = [, b]. Then, f(i) is closed bounded intervl. Comment The clim here is tht continuous functions mp intervls onto intervls. This mens tht the intuitive picture of continuous functions s one hving continuous grph is round. Proof of the intervl theorem. By the boundedness property there exist numbers c nd d in I such tht f(c) = m 0 nd f(d) = M 0 nd m 0 f(x) M 0, for ll x I. Suppose for simplicity, tht c d. Apply the intermedite vlue property to f on the subintervl [c, d] to deduce tht f tkes ll possible vlues between f(c) = m 0 nd f(d) = M 0. In other words f(i) = [m 0, M 0 ]. Theorem 25.2 (The fixed point theorem). Let f : [, b] [, b] be continuous function. Then there is t lest one number c which is fixed by f. Tht is f(c) = c. Comment This result sys tht if proceeding continuously from (, f()) to (b, f(b)), then the line y = x must be crossed. 50

51 Proof of the fixed point theorem. Let g : [, b] R 1 be defined by g(x) = f(x) x. Since the identity function x x is continuous on [, b] the function g is continuous on [, b]. If f() = or f(b) = b, then there is nothing to prove. So it is ssumed tht f() nd f(b) b. Since f mps onto [, b], g() > 0 nd g(b) < 0. The intermedite vlue property pplied to g on the intervl [, b] implies tht g(c) = 0 for some c, < c < b. Hence f(c) = c. Theorem 25.3 (The continuity of the inverse function). Suppose tht f : A B is bijection where A nd B re intervls. If f is continuous on A, then f 1 is continuous on B. Proof. Consider the continuous bijection f : A B where A nd B re intervls. First it is shown tht f is either strictly incresing or strictly decresing. If f is neither strictly incresing nor decresing, then without loss of generlity, there re numbers 1, 2 nd 3 such tht 1 < 2 < 3 nd f( 1 ) < f( 3 ) < f( 2 ). Apply the intermedite vlue theorem to f on the intervl [ 1, 2 ] to deduce tht f(c) = f( 3 ) for some c ( 1, 2 ). This contrdicts the fct tht f is bijection. For the rest of the proof it is ssumed tht f is strictly incresing, the proof for the strictly decresing cse being similr. Hence f 1 is strictly incresing. Let b B nd f 1 (b) =, so tht f() = b. For every ε > 0, f mps the intervl I = [ ε, + ε] onto some intervl f(i) = [m, M]. Since f is strictly incresing, m < b < M, so let δ be the minimum of b m nd M b. Clerly δ > 0. Now [b δ, b + δ] is subset of [m, M] = f(i) nd so f 1 mps [b δ, b + δ] into I = [ ε, + ε]. Thus given ny ε > 0, there exists δ > 0 such tht y b < δ f 1 (y) f 1 (b) < ε. If f : A B is strictly monotone surjection, where A nd B re intervls, then f nd f 1 re continuous. Let be f : A R 1 R 1 continuous function. Definition The function f is clled uniformly continuous on A if for ny ε > 0 there exists δ > 0 such tht x x < δ f(x ) f(x ) < ε for ny x, x A. Theorem 25.4 (Theorem of the uniform continuity). If f : [, b] R 1 R 1 is continuous, then f is uniformly continuous. Comment This result sys tht continuous function on closed intervl [, b] is uniformly continuous. 26 Sequence of functions. Set of convergence. Definition A sequence of rel vlued functions defined on A R is function F : N {f f : A R}. We write F (n) = f n nd the sequence of functions is denoted by (f n ). 51

52 Let be A R 1 nd f 1, f 2,..., f n,... sequence of rel vlued functions f n : A R 1. We will denote this sequence by (f n ). Definition An element of A is clled point of convergence of the sequence (f n ) if the sequence (f n ()) converges. The set of ll points of convergence is clled the set of convergence of the sequence (f n ). Let be B A the set of convergence of sequence (f n ). For x B we denote by f(x) the limit f(x) = lim n f n (x). We estblish in this wy correspondence from B to R 1 ; i.e. function f : B A R 1 R 1. The function f defined bove is clled the limit function, on the set B, of the sequence (f n ). We will sy tht the sequence (f n ) converges on B to f. Definition Let be (f n ) sequence of function defined on A i.e. f n : A R 1 R 1. A function f : A R is clled the limit function of sequence (f n ) if for ny x A nd ε > 0 there exists N(x, ε) such tht for n > N(x, ε) we hve written: f n n f on A. f n (x) f(x) < ε Comment If in the bove definition N depends only on ε nd does not depend on x, then we sy tht the sequence (f n ) converges uniformly to f on A. Definition The sequence (f n ) is uniformly convergent on A to f if for ny ε > 0, there exists N(ε) such tht for n > N(ε) nd x A we hve f n (x) f(x) < ε. If the sequence (f n ) is uniformly convergent to f we will write f n Exmple A = [0, 1], f n (x) = x n, f(x) = f n u / f. n u n f. { 1 for x = 1 0 for x [0, 1), f n n Exmple A = [0, 2 π], f n (x) = sin n x n, f(x) = 0, f u n f. n f but Uniform convergence criteri: Let be (f n ) sequence of functions f n : A R 1 R. First criterion (Cuchy): The sequence (f n ) converges uniformly to function f defined on A if nd only if for ny ε > 0 there exists N(ε) such tht, for ny n, m > N(ε) nd ny x A we hve: f n (x) f m (x) < ε. 52

53 Proof. Assume first tht f n hve We hve u f. For ε > 0 there exists N ε such tht for p N ε we n f p (x) f(x) < ε, for ny x A. 2 f n (x) f m (x) < f n (x) f(x) + f(x) f m (x) ε 2 + ε 2 = ε. for ny n, m > N ε. Assume now tht for ny ε > 0 there exists N ε, such tht for n, m > N ε nd x A we hve f n (x) f m (x) < ε nd show tht there exists f : A R 1 R 1 u such tht f n f. n From hypothesis we hve tht the sequence of the rel numbers (f n (x)) converges. Let be f(x) = lim f n (x). We obtin in this wy function f : A R 1 R 1. The sequence n of function (f n ) converges in ny point x A to f. Let now ε > 0 nd N ε such tht for n, m > N ε nd x A we hve We choose n 0 N ε nd since f n n f n (x) f m (x) < ε. f we hve f n f n0 n f f n 0 nd more f(x) f n0 (x) < ε for x A. Since n 0 N ε ws rbitrry chosen, it follows tht for ny n N ε nd x A we hve i.e. f n u n f. f n (x) f(x) < ε Second criterion: Let be (f n ) sequence of functions defined on A: f n : A R 1 R 1 nd f : A R 1 R 1. If there exists sequence ( n ) of positive rel numbers ( n > 0) which converges to 0 ( n 0), such tht f n (x) f(x) n, for ny n N nd ny u x A, then f n n f. Proof. Let be ε > 0. Since n 0, there exists N ε such tht for ny n N ε we hve n < ε. It follows tht f n (x) f(x) < ε for n N ε nd x A i.e. f n u n f. 27 Continuity nd uniform convergence The following sttement shows tht the uniform convergence conserves the continuity. Proposition Let be (f n ) sequence of functions f n : A R 1 R 1 which converges uniformly to f : A R 1 R 1 u ; f n f. If ll the functions f n re continuous t n point A, then f is continuous t. 53

54 Proof. Let be ε > 0. Since f n x A we hve u n f, there exists N ε such tht for ny n N ε nd f Nε (x) f(x) < ε 3. In prticulr, we hve f Nε () f() < ε 3. Since f Nε is continuous t, there exists δ ε > 0 such tht x < δ ε f Nε (x) f Nε () < ε 3. Therefore, for ny x with x < δ ε we hve f(x) f() f(x) f Nε (x) + f Nε (x) f Nε () + f Nε () f() < ε 3 + ε 3 + ε 3 = ε Tht mens tht f is continuous t. Corollry The limit of uniform convergent sequence of continuous functions is continuous function. 28 Equl continuous nd equl bounded sequence of functions Let be A R 1 nd (f n ) sequence of function defined on A; f n : A R 1. The expression: (f n ) is sequence of continuous functions mens: for ny n N, x A nd ε > 0 there exists δ = δ(n, x, ε) > 0 such tht: x x < δ f n (x ) f n (x) < ε. If the functions f n re uniformly continuous on A, then δ does not depend on x. Hence: (f n ) is sequence of uniformly continuous function on A mens: for ny n N nd ny ε > 0, there exists δ = δ(n, ε) > 0 such tht: x x < δ f n (x ) f n (x ) < ε for ny x, x A. It is possible tht δ does not depend on n, but depends on x nd ε. In this cse the sequence (f n ) is sequence of equl continuous functions. More precisely: The sequence (f n ) is sequence of equl continuous functions on A if for ny x A nd ε > 0 there exists δ = δ(x, ε) > 0 such tht for ny n : x x < δ f n (x ) f n (x) < ε. If δ is independent on x nd n, then the functions of the sequence (f n ) re uniformly continuous nd equl continuous too; they re equl uniformly continuous. More precisely: 54

55 (f n ) is sequence of functions equl uniformly continuous on A if for ny ε > 0 there is δ = δ(ε) > 0 such tht for ny n N nd x, x A. x x < δ(ε) f n (x ) f n (x ) < ε The sequence (f n ) is sequence of bounded functions on A if for ny n N there is M = M(n) > 0 such tht f n (x) < M for ny x A. If M is independent on n, then the functions of the sequence (f n ) re equl bounded. More precisely: (f n ) is sequence of equl bounded functions on A, if there is M > 0, such tht for n N, x A. f n (x) < M Theorem 28.1 (Arzel-Ascoli). Let be I = [, b] closed intervl nd (f n ) sequence of functions f n : I R 1. If (f n ) is sequence of equl continuous nd equl bounded functions, then (f n ) contins subsequence (f nk ) which is uniformly convergent on I. Proof. The proof is rther technicl nd it will omitted. 29 Series of functions. Convergence nd uniform convergence. Let be A R 1 nd (f n ) sequence of functions f n : A R 1. Definition It is sid tht the symbol f n is convergent series of functions t the point A, if the numericl series f n () is convergent. The symbol f n is divergent series of functions t the point A, if the numericl series f n () diverges. A point A is clled point of convergence of the series of functions converges t. f n if the series 55

56 The collection of ll the points of convergence of the series is clled the set of convergence of the series f n. Let be B A the set of convergence of the series f n. For x B we denote by S(x) the sum S(x) = f n (x). We estblish in this wy correspondence from B to R 1 i.e. function S : B A R 1 R 1 The function S defined bove is clled the sum function, on the set B, of the series f n. We will sy tht the series f n converges to S on B, nd we will write S = f n, for x B. Definition Let be series of functions defined on A, nd S function defined on B A. The series f n f n converges to S on B if for ny x B nd ny ε > 0 there exists N = N(x, ε) > 0 such tht for ny n > N we hve f 1 (x) + f 2 (x) + + f n (x) S(x) < ε. If the number N is independent on x, then the series is uniformly convergent on B to S. In this cse we hve the following: Definition The series f n converges uniformly to S if for ny ε > 0 there is N = N(ε) > 0 such tht f 1 (x) + f 2 (x) + + f n (x) S(x) < ε. for ny x B nd n > N(ε). Definition The series converges on B. f n converges bsolutely on B if the series f n If the series f n converges bsolutely on B, then it converges on B. 56

57 Exmple ) Consider the sequence of functions defined by nd the series f n (x). n=0 f n (x) = x 2 (1 + x 2 ) n, n 0 The set of convergence of this series is R 1 nd the sum of series is { 1 + x 2 for x 0 S(x) = 0 for x = 0 The series is bsolutely convergent on R 1. b) For n 1 consider f n defined on R 1 s nd the series f n. f n (x) = sinn x n 2 The series is bsolutely convergent on R 1. The series is uniformly convergent on R 1. c) For n 1 consider f n (x) = cos n x nd the series f n. The set of convergence is R 1 \ {k π} k Z. The series is bsolutely convergent on the set of convergence. en x d) consider for n 1 the functions f n (x) = n nd the series f n. The set of convergence of the series is empty. 30 Convergence criteri for series of functions Consider the series of functions f n defined on A, i.e. f n : A R 1 R 1. Definition The series of functions f n is clled the reminder of order k of the series f n. n=k+1 57

58 1 st Criterion: The series of the series converges. Proof. Consider nd nd remrk tht f n converges if nd only if the reminder of ny order k S k = f 1 + f f n σ p = f k+1 + f k f k+p S k+p = S k + σ p. Therefore, the sequence (S k+p ) converges s p if nd only if the sequence (σ p ) converges s p. 2 nd Criterion: The series reminders tends to 0. f n converges if nd only if the sequence of the sums of Proof. Obvious. 3 rd Criterion (Cuchy): The series f n converges uniformly on A if nd only if for ny ε > 0 there is N = N(ε) such tht for n N nd p 1 we hve f n+1 (x) + f n+2 (x) + + f n+p < ε for x A. Proof. An immedite consequence of the Cuchy criterion for sequences. 4 th Criterion: Let be n convergent series of positive numbers. If f n (x) n for x A nd n N then the series Proof. Obvious. f n is uniform convergent. 31 Power Series Definition A series of functions of the form Clerly, ny power series converges when x = n x n n=0 is clled power series.

59 Theorem 31.1 (The set of the convergence of power series. theorem). Abel-Cuchy-Hdmrd - The power series n x n is bsolutely convergent for x < R (R clled rdius of n=0 convergence) where R is given by R = 1 ω if 0 < ω + nd ω = lim n n n. R = + if ω = 0 - The series diverges for ny x with x > R. - For ny r (0, R) the series is uniformly convergent on the closed intervl [ r, r]. Proof. Consider x 0 R 1 nd the series n x 0 n. n=0 Apply the root test to this series nd obtin: If lim n n x 0 < 1, then the series n x n 0 is bsolutely convergent. In other words, n n=0 the series n x n 0 is bsolutely convergent for x 0 < R, where R = 1 if 0 < ω + ω n=0 n nd R = + if ω = 0 nd ω = n. lim n + Applying the sme test, it follows tht the series diverges for ny x 0 with x 0 > R. For r (0, R) the series n r n converges (x = r is point t which the series n=0 n x n converges bsolutely) nd for x [ r, r], we hve n=0 n x n n r n. According to the 4 th criterion of convergence the series, for x r. Exmple ) The series n x n is uniformly convergent x n converges bsolutely for x < 1 nd diverges for x > 1. The n=0 n=0 rdius of convergence is R = 1; the convergence set is ( 1, 1). b) For the series [ 1, 1). x n n the rdius of convergence R is R = 1. The convergence set is 59

60 c) The set of convergence of series d) The series ( 1) n xn n is ( 1, 1]. x n (α > 1) is bsolutely convergent on [ 1, 1]. nα Concerning the continuity of the sum of power series we hve: The sum S of the power series n x n is continuous function on ( R, R). n=0 Proof. Let be x 0 ( R, R). There exists r (0, R) such tht R < r < x 0 < r < R. Since on the closed intervl [ r, r] the series converges uniformly, nd the terms of the series re continuous functions, the sum S is continuous on [ r, r]. In prticulr, it is continuous t x 0. The sum S of the power series intervl contined in ( R, R). n x n n=0 is uniformly continuous on ny compct 32 Arithmetics of power series Let n x n nd b n x n be power series with rdii of convergence R 1 nd R 2, n=0 n=0 respectively, where 0 R 1 R 2. Then: - the sum ( n + b n ) x n n=0 - the sclr product k n x n n=0 - the Cuchy product c n x n, c n = n=0 n b n k k=0 ll hve rdius of convergence t lest R 1. Moreover if n x n hs the sum f(x) nd n=0 b n x n hs the sum g(x), then: n=0 ( n + b n ) x n = f(x) + g(x) n=0 (k n ) x n = k f(x) n=0 60

61 c n x n = f(x) g(x). n=0 Proof. These clims concerning the sum nd sclr product follow from the sum nd sclr product rules for series. To estblish the Cuchy product result, note tht n x n nd b n x n re bsolutely convergent for x < R 1. Since c n x n = n=0 the series n=0 n ( k x k )(b n k x n k ), c n x n is bsolutely convergent for x < R 1, nd hs the sum stted. n=0 Much of the preceding discussion cn be modified to pply to series of the form n (x ) n. n=0 k=0 33 Differentible functions Intuitively, function f : A R 1 R 1 is differentible t c A if tngent cn be drwn to the curve t the point P (c, f(c)). Figure 33.1: The slope of the chord P Q in Figure 33.1 is f(x) f(c) x c nd s Q moves closer to P it is required tht the slope P Q pproches the slope of the tngent line t P. This geometric ide motivtes the following forml definition: Definition A function f : A R 1 R 1 is differentible t c A if f(x) f(c) lim x c x c exists. f (c) is written for the vlue of this limit, clled the derivtive of f t c. An lterntive form of the limit is obtined by setting x = c + h. Then f(x) f(c) lim x c x c f(c + h) f(c) = lim. h 0 h 61

62 Exmple The function f(x) = x 2 is differentible for ll x. Solution: Consider f(x) f(c) x c f(x) f(c) lim x c x c for ny x c, where c is fixed. Now x 2 c 2 = lim x c x c = lim (x + c) = 2 c. x c Hence, f is differentible t c nd f (c) = 2 c. Since c ws rbitrry, the derivtive function f cn be defined s f (x) = 2 x. Exmple The function f(x) = x is not differentible t c = 0. Solution: Consider Hence f(x) f(0) x 0 = x { 1 for x > 0 x = 1 for x < 0. f(x) f(0) f(x) f(0) lim = 1 nd lim = 1. x 0 + x 0 x 0 x 0 Since these right nd left limits differ, f is not differentible t 0. It is esy to show tht f is differentible for ll x 0 nd f (x) = 1 for x > 0 nd f (x) = 1 for x < 0. In generl, points where f is not differentible cn be often detected by exmining the f(x) f(c) left nd the right limits of s x c. x c f(x) f(c) The left limit lim is clled the left derivtive of f t c nd is denoted by x c x c f (c). f(x) f(c) Similrly, the right limit lim is clled the right derivtive of f t c nd is x c + x c denoted by f +(c). Clerly, f (c) exists if nd only if f (c) nd f +(c) both exist nd re equl. The next result estblishes tht only continuous functions cn be differentible. Theorem If f is differentible t c, then f is continuous t c. Proof. Define the function F c (x) = f(x) f(c) x c if x c f (c) if x = c. Since f is differentible t c, lim x c F c (x) = F c (c) nd hence, F c is continuous t c. Now f(x) = f(c) + F c (x) (x c) for ll x. Since F c nd the identity nd constnt functions re ll continuous t c, f is continuous t c. 62

63 Note tht Exmple 33.2 shows tht there re continuous functions which re not differentible. The following tble gives certin elementry functions nd their derivtives. Function f derivtive f f(x) = k constnt f (x) = 0 f(x) = x n, n N f (x) = n x n 1 f(x) = x f (x) = 1 2 x f(x) = sin x f (x) = cos x f(x) = cos x f (x) = sin x f(x) = tn x f (x) = 1 cos 2 x f(x) = cot x f (x) = 1 sin 2 x f(x) = e x f (x) = e x f(x) = ln x f (x) = 1 x 34 Rules of differentibility Sum rule Let f nd g be functions differentible t c. Then, their sum f + g is differentible t c nd (f + g) (c) = f (c) + g (c). Product rule Let f nd g be functions differentible t c. Then, their product f g is differentible t c nd (f g) (c) = f (c) g(c) + f(c) g (c). Reciprocl rule 1 Let f be function which is non-zero nd is differentible t c. Then f is differentible t c nd ( ) 1 (c) = f (c) f f 2 (c). 63

64 Proof of product rule. For x c As x c, (f g)(x) (f g)(c) x c f(x) f(c) x c f(x) g(x) f(c) g(c) = = x c f(x) g(x) f(x) g(c) + f(x) g(c) f(c) g(c) = x c f(x) f(c) = g(c) + f(x) x c f (c) nd g(x) g(c) x c g(x) g(c). x c g (c). = Therefore (f g)(x) (f g)(c) x c x c f (c) g(c) + f(c) g (c). The product nd the reciprocl rules cn be combined s follows, to give the quotient rule. Quotient rule t c nd If f nd g re differentible t c nd g(x) 0, then f is differentible g ( ) f (c) = f (c) g(c) f(c) g (c). g [g(c)] 2 Exmple Use the bove rules to prove tht ech of the following functions is differentible t the points indicted: i) f(x) = x 2 + sin x, x R 1 ; ii) f(x) = x 2 sin x, x R 1 ; iii) f(x) = tn x, x R 1 nd x (2n + 1) π 2, n integer; iv) f(x) = x n, n Z, n 0; v) f(x) = sec x, x (2n + 1) π 2 ; vi) f(x) = csc x, x nπ; vii) f(x) = tn x, x (2n + 1) π 2 ; viii) f(x) = cot x, x nπ. 64

65 Squeeze rule Let f, g nd h be three functions such tht g(x) f(x) h(x) for ll x in some neighborhood of c nd such tht g(c) = f(c) = h(c). If g nd h re differentible t c, then so is f nd f (c) = g (c) = h (c). Proof. The given inequlities imply g(x) g(c) x c f(x) f(c) x c h(x) h(c) x c for ll x > c nd the inequlity signs re reversed for x < c. The result follows by squeeze rule for limits of functions provided tht g (c) = h (c) cn be estblished. To this end let g(x) g(c) if x c G c (x) = x c g (c) if x = c nd h(x) h(c) if x c H c (x) = x c h (c) if x = c. Since g nd h re differentible t c, G c nd H c re continuous t c. Let k(x) = G c (x) H c (x). Thus k is continuous t c. The erlier inequlities imply tht if x > c, then k(x) 0 nd if x < c, then k(x) 0. Hence, k(c) = 0 nd so G c (c) = H c (c). In other words, g (c) = h (c). Exmple The function x 2 sin 1 if x 0 f(x) = x 0 if x = 0. cn be squeezed between h(x) = x 2 nd g(x) = x 2 t x = 0. Since g nd h re differentible t 0 with common derivtive of vlue 0, the squeeze rule gives tht f is differentible t x = 0. By other rules, f is lso differentible for x 0. Moreover, 2x sin 1 f (x) = x cos 1 if x 0 x 0 if x = 0. Note tht lim x 0 f (x) does not exist so tht f (0) exists but f is not continuous t 0. Composite rule Let f be differentible t c, nd g be differentible t b = f(c). Then g f is differentible t c nd (g f) (c) = g (f(c)) f (c). 65

66 Proof. Let nd f(x) f(c) F c (x) = x c f (c) if x c if x = c g(y) g(b) if y b G b (y) = y b g (b) if y = b. Then F c is continuous t x = c nd, for ll x, G b is continuous t y = b nd, for ll y Now f(x) = f(c) + (x c) F c (x). g(y) = g(c) + (y b) G b (y). (g f)(x) =g(f(x)) = g(y) = g(b) + (y b) G b (y) = =g(f(c)) + (f(x) f(c)) G b (f(x)) = =g(f(c)) + (x c) F c (x) G b (f(x)). So (g f)(x) (g f)(c) = F c (x) G b (f(x)). x c The function on the right-hnd side of the bove equlity is continuous t x = c. Hence s required. (g f)(x) (g f)(c) lim x c x c = F c (c) G b (f(c)) = f (c) g (f(c)), The composite rule is often clled the chin rule nd the formul given for the derivtive of composite is more suggestive in Leibnitz nottion. Let x = h nd y = f(x+h) f(x). Then f (x) = lim h 0 f(x + h) f(x) h y = lim x 0 x. The Leibnitz nottion for this limit is dy. Write y = g(u) where u = f(x). Then dx f (x) = du dx nd g (f(x)) = dy du nd (g f) (x) = dy. The chin rule cn now be written dx s dy dx = dy du du dx. Exmple Show tht h(x) = sin x 2 is differentible. Solution: Let g(x) = sin x nd f(x) = x 2, then h = g f. Since f nd g re everywhere differentible, the composite rule gives tht h is differentible. Moreover, h (x) = g (f(x)) f (x) = 2x cos x. 66

67 Inverse rule Suppose tht f : A B is continuous bijection where A nd B re intervls. If f is differentible t A nd f () 0, then f 1 is differentible t b = f() nd (f 1 ) (b) = 1 f (). Proof. For A, let F is continuous t x = nd for ll x A f(x) f() if x F (x) = x f () if x =. f(x) = f() + (x ) F (x). Given f() = b nd letting f(x) = y for x A we hve F (x) = y b x for x. Consider so G b (y) = G b (y) = x y b for y b x f(x) f() = 1 F (x) = 1 (F f 1 )(y) for y b. Since f bijective nd continuous, so f 1 is continuous too. Also f 1 (b) = nd F is continuous t x =. Hence F f 1 is continuous t y = b nd (F f 1 )(b) = F (f 1 (b)) = F () = f () = (f f 1 )(b) 0 So In other words x y b = G b(y) = 1 (F f 1 )(y) 1 (F f 1 )(b) = 1 f () f 1 (y) f 1 (b) lim y b y b ( = 1 f f 1 ) (b) = 1 f (). s y b. If the function f : A f(a) is differentible on the intervl A nd f () 0 for ny A then f 1 is differentible on f(a) nd (f 1 ) (f()) = 1 f (). Exmple The function f : (0, ) (0, ) given by f(x) = x 2 is bijection. Its inverse function is given by f 1 (x) = x. Now f is differentible for x > 0 nd f (x) = 2x 0. Hence, by the inverse rule, f 1 (x) = x is differentible nd (f 1 ) (x) = 1 2 x. 67

68 ( Exmple The function g : π 2, π ) ( 1, 1) given by g(x) = sin x is bijection 2 with inverse given by g 1 (x) = rcsin x. Now g is differentible nd (g g 1 )(x) = cos(rcsin x) for x < 1. Hence, by the inverse rule g 1 is differentible nd (g 1 ) (x) = 1 1 x 2 for x < Locl extremum In this section we present result which helps to locte the locl mxim nd minim of differentible function. Definition A function f hs locl mximum vlue t c if c is contined in some open intervl I for which f(x) f(c) for ech x I. If f(x) f(c) for ech x I, then f hs locl minimum vlue t c. Theorem 35.1 (Locl extremum theorem, Fermt). If f is differentible t c nd possesses locl mximum or locl minimum t c, then f (c) = 0. Proof. Consider the cse of locl minimum t x = c. There is n open intervl I such f(x) f(c) tht f(x) f(c) 0 for ll x I. If x > c, then 0 nd if x < c, then x c f(x) f(c) 0. Thus f x c +(c) 0 nd f (c) 0. But f (c) exists nd so f +(c) = f (c). Thus f (c) = 0. Note tht lthough f must vnish t locl extremum this is not sufficient for such point. For exmple, consider the behvior of f(x) = x 3 t x = 0. Here f (0) = 0, but 0 is neither locl mximum, nor locl minimum. Exmple Loclize the locl mximum nd locl minimum of the function f(x) = x (x 1) (x 2) 36 Theorems concerning bsic properties of differentible functions This section estblishes some bsic properties of differentible functions. Theorem 36.1 (Rolle s theorem). Let f be differentible on (, b) nd continuous on [, b]. If f() = f(b) then there exists c (, b), such tht f (c) = 0. 68

69 Proof. Since f is continuous on [, b], it ttins mximum vlue f(c 1 ) nd minimum vlue f(c 2 ) on [, b] by the boundedness property. If f(c 1 ) = f(c 2 ), then f is constnt for ll x [, b], hence f (x) = 0 for ll x [, b] nd the result follows. If f(c 1 ) f(c 2 ), then t lest one of c 1 nd c 2 is not or b. Hence f hs locl mximum or minimum (or both) inside the intervl [, b]. By the locl extremum theorem f is zero t t lest one point inside [, b]. Theorem 36.2 (Men vlue theorem, Lgrnge). Let f be differentible on (, b) nd continuous on [, b]. Then there exists c (, b), such tht f (c) = f(b) f(). b f(b) f() Proof. Let g(x) = f(x) λx, where λ =. g is differentible on (, b) nd b continuous on [, b]. The choice of λ mens tht g() = g(b). Applying Rolle s theorem there is c (, b), such tht g (c) = 0. Hence, f (c) λ = 0, so f f(b) f() (c) =. b Theorem 36.3 (the incresing-decresing theorem). If f is differentible on (, b) nd continuous on [, b] then (1) f (x) > 0 for ll x (, b) implies f is strictly incresing on [, b]; (2) f (x) < 0 for ll x (, b) implies f is strictly decresing on [, b]; (3) f (x) = 0 for ll x (, b) implies f is constnt on [, b]. Proof. Let x 1, x 2 [, b] with x 1 < x 2. Since f stisfies the hypothesis of the men vlue theorem on the intervl [x 1, x 2 ] we hve f(x 2 ) f(x 1 ) x 2 x 1 = f (c) for some c, x 1 < c < x 2. But f (c) > 0 nd so f(x 2 ) > f(x 1 ). In other words, f is strictly incresing on [, b]. The proof of (2) nd (3) is similr. Comment The incresing-decresing theorem is useful for finding nd clssifying locl extrem nd estblishing inequlities between functions. Exmple Find nd describe the locl extrem of f(x) = x 2 e x. Solution: f is everywhere differentible nd f (x) = e x (2 x) x. Locl extrem occur only when f (x) = 0 nd so x = 0 or x = 2. Since e x > 0, if x < 0 then f (x) < 0, if x (0, 2) then f (x) > 0 nd if x > 2 then f (x) < 0. Thus f is decresing on (, 0), incresing on (0, 2) nd decresing gin on (2, + ). This mens tht x = 0 gives locl minimum nd x = 2 gives locl mximum of f. 69

70 Exmple Prove tht e x 1 + x for ll x. Solution: Let f(x) = e x 1 x. f is differentible nd f (x) = e x 1. Hence f (x) > 0 for x > 0 nd so f(x) > f(0) = 0 for x > 0. Since f (x) < 0 for x < 0 it follows tht f(x) > f(0) = 0 for x < 0. Finlly we obtin tht f(x) 0 for ll x R 1. Tht mens e x 1 + x. The next result is difficult to interpret geometriclly but it will be needed to prove l Hôspitl s rule. This is rule which is well suited to the evlution of limits of form f(x) lim x x 0 g(x), where f(x 0) = g(x 0 ) = 0. Theorem 36.4 (Cuchy s men vlue theorem). Let f nd g be differentible on (, b) nd continuous on [, b]. Then there exists c (, b), such tht f (c) g (c) provided tht g (x) 0 for ll x (, b). = f(b) f() g(b) g() Proof. First note tht g() g(b), otherwise Rolle s theorem pplied to g on [, b] would men tht g vnished somewhere on (, b). Let h(x) = f(x) λg(x) where f(b) f() λ = g(b) g(). By the sum nd product rules for continuity nd differentibility nd our choice of λ, h stisfies ll the hypotheses of Rolle s theorem. Hence there is c (, b), such tht h (c) = 0. This gives f (c) = λ g (c) nd the result now follows. Theorem 36.5 (l Hôspitl s rule, version A). Let f nd g stisfy the hypotheses of Cuchy s men vlue theorem nd let x 0 stisfy x 0 (, b). If f(x 0 ) = g(x 0 ) = 0, then f(x) lim x x 0 g(x) = lim f (x) x x 0 g (x) provided tht the ltter limit exists. Proof. Apply Cuchy s men vlue theorem to f nd g on the intervl [x 0, x] where x 0 < x b. Hence there exists c, x 0 < c < x such tht Now lim x x + 0 f (c) g (c) = f(x) f(x 0) g(x) g(x 0 ) = f(x) g(x). f(x) g(x) = lim c x + 0 f (c) g (c) = lim x x + 0 f (x) g (x) provided tht the ltter exists. A similr rgument pplied on the intervl [x, x 0 ] where x < x 0 gives tht: lim x x 0 gin provided tht the ltter exists. The rule now follows. f(x) g(x) = lim x x 0 70 f (x) g (x)

71 Exmple Show tht sin x lim x 0 x = 1. Solution: The functions f(x) = sin x nd g(x) = x stisfy the hypotheses of l Hôspitl rule. Moreover sin x lim x 0 x = lim cos x = 1. x 0 1 Exmple Show tht lim (1 + x) 1 x = e. x 0 Solution: Vi the composite rule for limits of functions we hve ( ) ( ) ln lim (1 + x) 1 x = lim ln(1 + x) 1 ln(1 + x) x = lim. x 0 x 0 x 0 x By l Hôspitl s rule Hence ln(1 + x) lim x 0 x 1 = lim x 0 x + 1 = 1. lim x 0 (1 + x) 1 x = e. L Hôspitl s rule cn be used to evlute mny indeterminte limits once they hve been expressed s the limit of quotient of differentible functions, provided of course tht f(x) lim cn be evluted. x x 0 g(x) Often this finl limit is itself indeterminte (in other words f (x 0 ) = g (x 0 ) = 0) nd it my be tempting to pply l Hôspitl s rule gin. But this requires tht f nd g re themselves differentible. 37 Higher-order derivtives nd differentils Definition If the derived function f of given differentible function f is itself differentible it is sid tht f is twice differentible nd f or f (2) denotes (f ) its second derivtive. Definition In generl, f is n times differentible if f is (n 1) times differentible nd its (n 1)-th derivtive is differentible. The n-th derivtive (f (n 1) ) is denoted by f (n). If moreover, f (n) is continuous function, then f is sid to be n times continuously differentible. Exmple If f(x) = x m, m N then f (n) (x) = m! (m n)! xm n for n m 0 for n > m. 71

72 ( 2. If f(x) = sin x then f (n) (x) = sin x + nπ 2 ) for n If f(x) = ln x then f (n) (x) = ( 1)n 1 (n 1)! x n for n 1. The successive differentition of products of simple functions re: (f g) =f g + f g (37.1) (f g) (2) =f (2) g + 2 f g + f g (2) (37.2) (f g) (3) =f (3) g + 3 f (2) g + 3 f g (2) + f g (3). (37.3) Generlly the following formul pplies. This cn be proved by induction on n. Theorem 37.1 (Leibnitz formul). Let f nd g be n times continuously differentible. Then h = f g is lso n times continuously differentible nd: h (n) = n Cn k f (n k) g (k). k=0 Theorem 37.2 (l Hôspitl s rule, version B). Let f nd g be n times continuously differentible on the intervl (, b) nd x 0 stisfies < x 0 < b. If f (k) (x 0 ) = g (k) (x 0 ) = 0 for 0 k n 1 nd then Exmple Prove tht nd g (n) (x 0 ) 0 f(x) lim x x 0 g(x) = f (n) (x 0 ) g (n) (x 0 ). 1 cos x lim = 1 x 0 x 2 2 ( ) 1 lim x 0 x cot x = Tylor polynomils Suppose tht n x n is power series with rdius of convergence R > 0. Let f(x) be n=0 the sum of series for x < R. It cn be proved tht f is differentible nd tht f (x) = n n x n 1 for x < R. 72

73 Continully differentiting in this mnner leds to f (k) (x) = n n (n 1)... (n k + 1) x n k for x < R. n=k If x = 0, then k = f (k) (0) k! for k = 1, 2,... Thus the coefficient of x n in ny power series is f (n) (0) where f(x) is the sum of the given n! power series. Hence f (n) (0) f(x) = x n for x < R. n! n=0 For smll vlues of x, the sum f(x) cn be pproximted by the polynomil for ny vlue of N. f(0) + f (0) 1! x + f (2) (0) 2! x f (N) (0) x N N! This section investigtes how good n pproximtion this polynomil is when f(x) is not necessrily the sum of given power series. Definition Let f n n times continuously differentible function t 0. The Tylor polynomil of degree n for f t 0 is defined by: T n f(x) = f(0) + f (0) 1! x + f (2) (0) 2! x f (n) (0) x n. n! Exmple Let f(x) = e x. For k = 1, 2,... f (k) (0) = 1. Thus nd so on. T 0 f(x) =1 T 1 f(x) =1 + x T 2 f(x) =1 + x x2 The first result provides n estimte for the difference between f(b), the vlue of given function t x = b, nd T n f(b), the vlue of its Tylor polynomil of degree n t x = b. Theorem 38.1 (The first reminder theorem). Let f be (n + 1) times continuously differentible on n open intervl contining the points 0 nd b. Then the difference between f nd T n f t x = b is given by for some c between 0 nd b. f(b) T n f(b) = bn+1 (n + 1)! f (n+1) (c) 73

74 Proof. For simplicity ssume tht b > 0. Let h n (x) = f(b) n k=0 f (k) (x) (b x) k x [0, b]. k! Then h n (b) = 0 nd h n (0) = f(b) T n f(b). Let ( ) n+1 b x g(x) = h n (x) h n (0) x [0, b]. b The function g is continuous on [0, b] nd differentible on (0, b) nd g(0) = g(b) = 0. Hence by Rolle s theorem g (c) = 0 for some c between 0 nd b. Now fter strightforwrd clcultion. Thus h n(x) = f (n+1) (x) (b x) n n! nd so leding to g (x) = f (n+1) (x) (b x) n (n + 1)(b x)n + h n! b n+1 n (0) 0 = g (c) = f (n+1) (c) (b c) n (n + 1)(b c)n + h n! b n+1 n (0) h n (0) = bn+1 (n + 1)! f (n+1) (c). Denote f(b) T n f(b) by R n f(b) nd cll it the reminder term t x = b. Thus f(b) = T n f(b) + R n f(b) nd so the error in pproximting f(b) by T n f(b) is given by the reminder term R n f(b). Since f (n+1) is continuous on closed intervl contining 0 nd b, it is bounded on tht intervl. So there exists number M such tht f (n+1) (c) M nd so R n f(b) b n+1 (n + 1)! M. Thus, for fixed n, the reminder term will be smll for b close to zero. In other words Tylor polynomils provide good pproximtions of the function ner x = 0. The next exmple illustrtes this. Exmple Let f(x) = sin x. Then T 7 f(x) = x x3 3! + x5 5! x7 7! for some c between 0 nd x. By the first reminder theorem R 7 f(x) = x8 ( sin c) 8! R 7 f(0.1) ! =

75 It cn now be shown how Tylor polynomils cn be used to generte power series expnsions for functions f which re infinitely differentible on n open intervl contining 0 nd x. For x we hve: f(x) = T n f(x) + R n f(x). Now lim T nf(x) = n k=0 f (k) (0) k! x k for x < R where R is the rdius of convergence of the resulting power series. If it cn be shown tht lim n R n f(x) = 0 for x < R < R for some R, then f(x) = n=0 f (n) (0) n! x n for x < R. This power series is clled the McLurin series for f(x). Exmple Derive the McLurin series for f(x) = e x. Solution: Firstly xn+1 T n f(x) = 1 + x 1! + x2 2! xn n! nd R n f(x) = (n + 1)! ec for some c (0, x). Now the series convergent for ll x by the rtio test so for ny fixed rel number x n=0 x n n! is bsolutely lim T nf(x) = n n=0 x n n!. By vnishing condition xn n! 0. Thus n R n f(x) = x n+1 (n + 1)! ec 0 s n for fixed x. Hence, for ny x, lim n R n f(x) = 0. Hence e x = n=0 x n n! = 1 + x 1! + x2 2! xn n! +... In similr mnner, power series cn be generted for ll the stndrd functions. In ech cse the sum of the power series is just lim T n f(x) nd the rnge of vlidity is precisely n those x for which: ) the resulting power series converges nd b) the reminder R n f(x) 0 s n. 75

76 In deriving the following list, the trickiest prt is estblishing b): e x = n=0 sin x = x n n! n=0 x R 1 (38.1) ( 1) n x 2n+1 (2n + 1)! x R 1 (38.2) ( 1) n x 2n cos x = x R 1 (38.3) (2n)! n=0 (1 + x) t t (t 1)... (t n + 1) = 1 + x n t / N, x ( 1, 1) (38.4) n! ( 1) n 1 x n ln(1 + x) = 1 < x 1. (38.5) n The form of the reminder found in the first reminder theorem is clled the Lgrnge form. The first few terms of the McLurin series for given function f provide good pproximtion of f(x) close to 0. But wht hppens when pproximtions for x close to some other rel number re required? Polynomils must be considered not in powers of x but in powers of (x ). Definition Let f be n times continuously differentible on n open intervl contining fixed rel number. Define the Tylor polynomil of degree n for f t by T n, f(x) = f() + f () 1! (x ) + f (2) () 2! The first reminder theorem cn now be generlized. (x ) f (n) () (x ) n. n! Theorem 38.2 (Tylor s theorem). Let f be (n + 1) times continuously differentible on n open intervl contining the points nd b. Then the difference between f nd T n, f t b is given by (b )n+1 f(b) T n, f(b) = f n+1 (c) (n + 1)! for some c between nd b. Proof. For ech t between nd b where f(b) = f(t) + f (t) 1! (b t) f (n) (t) (b t) n + F (t) n! F (t) = R n,t f(b) = f(b) T n,t f(b). 76

77 Differentiting with respect to t gives: ( 0 =f (t) + f (t) + f ) ( (2) (t) (b t) + 1!... + ( f (n) (t) Cnceltion now gives tht f (2) (t) 1! (n 1)! (b t)n 1 + f ) (n+1) (t) (b t) n + F (t). n! (b t) + f ) (3) (t) (b t) ! F (t) = f (n+1) (t) (b t) n. n! Apply Cuchy s men vlue theorem to the functions F nd G on the intervl with endpoints nd b, where G(t) = (b t) n+1. Thus, there is number c between nd b such tht Hence or f (n+1) (c) F (b) F () G(b) G() = F (c) (b c) n G (c) = n! (n + 1)(b c). n (f(b) T n, f(b)) (b ) n+1 = f(b) T n, f(b) = f (n+1) (c) (b c) n n! (n + 1)(b c) n (b )n+1 f n+1 (c). (n + 1)! The error in pproximting f(b) by the polynomil T n, f(b) is just the reminder term: R n, f(b) = (b )n+1 f n+1 (c) (n + 1)! where c lies between nd b. The pproximtion is good for b close to. Just s before, power series cn be generted in powers of (x ), clled Tylor series, for suitble functions of x. The rnge of vlidity is gin those x for which ) the resulting power series converges, nd b) R n, f(x) 0 s n. 39 Clssifiction theorem for locl extrem A form of Tylor s theorem much used in numericl nlysis is derived below nd used to round off the investigtion of locl extrem. From Tylor s theorem, it follows tht f(x) = T n, f(x) + R n, f(x) = = f() + f () 1! (x ) + f (2) () 2! (x ) f (n) () (x ) n + n! 77 (b )n+1 f n+1 (c) (n + 1)!

78 for some c between nd x. Let x = h. Then c lies between nd + h. Thus c = + θ h for some θ (0, 1). Hence the following result holds: for some θ (0, 1). f( + h) = f() + h 1! f () hn n! f (n) () + hn+1 (n + 1)! f n+1 ( + θ h) This expression emphsizes tht the vlue of f t + h is determined by the vlues of f nd its derivtives t with θ mesuring the degree of indetermincy. When the extrem of function f which hs sttionry point t x = (i.e. f () = 0) is investigted, it is required tht the sign of f(+h) f() for ll smll h be determined. The bove expression reltes f( + h) f() to the derivtives of f t, thus enbling the following to be proved. Theorem 39.1 (Clssifiction theorem for locl extrem). If f is (n + 1) times continuously differentible on neighborhood of nd f (k) () = 0 for k = 1, 2,..., n (in prticulr f () = 0 nd so x = is sttionry point of f) nd f (n+1) () 0 then: (1) n + 1 even nd f (n+1) () > 0 implies tht f hs locl minimum t x =. (2) n + 1 even nd f (n+1) () < 0 implies tht f hs locl mximum t x =. (3) n+1 odd implies tht f hs neither locl mximum nor locl minimum t x =. Proof. Since f (k) () = 0 for k = 1, 2,..., n f( + h) f() = hn+1 (n + 1)! f (n+1) ( + θh) where 0 < θ < 1. Since f (n+1) () 0 nd f (n+1) is continuous, there is δ > 0 such tht f (n+1) (x) 0 for x < δ. Thus for ll h stisfying h < δ, f (n+1) ( + θh) hs the sme sign s f (n+1) (), so f( + h) f() hs the sme sign s h n+1 f (n+1) () for ll h, h < δ. (1) If n + 1 is even nd f (n+1) () > 0, then f( + h) f() > 0 on the open intervl ( δ, + δ). Hence, x = gives locl minimum of f. (2) If n + 1 is even nd f (n+1) () < 0, then f( + h) f() < 0 on the open intervl ( δ, + δ). Hence, x = gives locl mximum of f. (3) If n + 1 is odd, the sign of f( + h) f() chnges with the sign of h. It is sid tht x = gives horizontl point of inflection. Exmple Determine the nture of the sttionry points of f(x) = x 6 4x 4. 78

79 40 The Riemnn-Drboux integrl It is intended to give one form of the definition of the Riemnn integrl f(x) dx. The definition involves the res of the rectngles nd pplies to wider clss of functions thn continuous ones. Definition Let [, b] be given finite intervl. A prtition P on [, b] is finite set of points {x 0, x 1,..., x n } stisfying = x 0 < x 1 <... < x n = b. Suppose now tht f is function defined nd bounded on [, b] (if f were continuous on [, b] this would certinly be the cse). Then f is bounded on ech of the subintervls [x i 1, x i ]. Hence f hs lest upper bound M i, nd gretest lower bound m i in [x i 1, x i ]. Definition The upper Drboux sum of f relted to P is defined by U f (P ) = n M i (x i x i 1 ) i=1 where M i = sup{f(x) x i 1 x x i }. The lower Drboux sum of f relted to P is defined by L f (P ) = n m i (x i x i 1 ) i=1 where m i = sup{f(x) x i 1 x x i }. Now f is bounded bove nd below on the whole [, b]. So there exist numbers m nd M with m f(x) M for ll x [, b]. Thus for ny prtition of [, b]: Hence the set is bounded bove nd the set is bounded below. So L f = sup L f nd U f = inf U f exist. m(b ) L f (P ) U f (P ) M(b ) L f = {L f (P ) P is prtition of [, b]} U f = {U f (P ) P is prtition of [, b]} The first result estblishes the intuitively obvious fct tht L f U f. Proposition If f is defined nd bounded on [, b], then L f U f. 79

80 Proof. Let P be prtition of [, b] nd P be the prtition P {y} where x i 1 < y < x i for one prticulr i, 1 i n. In other words, P is obtined by dding one more point to P. It is now shown tht L f (P ) L f (P ) nd U f (P ) U f (P ). Let M i = sup{f(x) x i 1 x y} nd M i = sup{f(x) y x x i }. Clerly, M i M i nd M i j=1 M i. Hence: i 1 U f (P ) = M j (x j x j 1 ) + M i(y x i 1 ) + M i (x i y) + i 1 M j (x j x j 1 ) + M i (y x i 1 ) + M i (x i y) + = j=1 n M j (x j x j 1 ) = U f (P ). j=1 n j=i+1 n j=i+1 M j (x j x j 1 ) M j (x j x j 1 ) = In similr wy, it cn be shown tht L f (P ) L f (P ). It now follows tht if P = P {y 1, y 2,..., y m }, where y i re distinct numbers in [, b], then L f (P ) L f (P ) nd U f (P ) U f (P ). Now suppose tht P 1 nd P 2 re two prtitions of [, b] nd let P 3 = P 1 P 2. Thus, L f (P 1 ) L f (P 3 ) nd U f (P 2 ) U f (P 3 ). Since L f (P 3 ) U f (P 3 ) it cn be deduced tht L f (P 1 ) U f (P 2 ). In other words, the lower sum relted to given prtition of [, b] does not exceed the upper sum relted to ny prtition of [, b]. Hence every lower sum is lower bound for the set of upper sums. So L f (P ) U f for ll possible prtitions P. But then U f is n upper bound for the set of lower sums. Thus L f U f. Definition A function defined nd bounded on [, b] is Riemnn-Drboux integrble on [, b] if L f = U f. This common vlue is denoted by f(x) dx = L f = U f. Exmple Prove tht f(x) = x is Riemnn-Drboux integrble on [0, 1]. Solution: For n N let P n = { 0, 1, 2,..., 1}. Hence U n n f (P n ) = n + 1 2n nd L f(p n ) = n 1 2n. So n 1 2n L f U f n + 1 2n. Letting n it cn be deduced tht L f = U f = 1 2. Exmple Show tht the function { 1 if x is rtionl f(x) = 0 if x is irrtionl is not Riemnn-Drboux integrble on ny intervl [, b]. 80

81 Solution: For ny prtition P it follows tht L f (P ) = 0 nd U f (P ) = b, since ny intervl of rel numbers contins infinitely mny rtionls nd irrtionls. Hence L f = 0 nd U f = b nd so L f U f. This definition of f(x) dx is only one of the mny wys of ssigning res to bounded regions. There re others, notbly the Lebesgue integrl; ll however, give the sme nswer for res under the grphs of continuous functions. It will be proved tht ll continuous functions re Riemnn-Drboux integrble nd net method of evluting the integrl involved is derived. Firstly, some elementry properties of the Riemnn integrl must be estblished - ll of which re essentilly properties of res. 41 Properties of the Riemnn-Drboux integrl Proposition If f nd g re Riemnn-Drboux integrble on [, b] then ll the integrls below exist nd (1) (2) (α f(x) + β g(x)) dx = α f(x) dx = c f(x) dx + f(x) dx + β c f(x) dx c b. g(x) dx α, β R 1. (3) if f(x) g(x) on [, b] then (4) f(x) dx f(x) dx. f(x) dx g(x) dx. Property (1) is described s the linerity of the integrl nd (2) is clled the dditive property. Proof of (1). It is sufficient to prove tht the following equlities hold αf(x) dx = α f(x) dx nd (f(x) + g(x)) dx = f(x) dx + g(x) dx The equlity αf(x) dx = α f(x) dx is true for ny α 0, provided by the equlities L αf (P ) = αl f (P ) nd U αf (P ) = αu f (P ) for ny α 0 nd ny prtition P of [, b]. The equlity αf(x) dx = α f(x) dx holds provided by U f (P ) = L f (P ), for ny 81

82 prtition P of [, b]. The equlity (f(x) + g(x)) dx = f(x) dx + g(x) dx is obtined by observing tht for ny prtition P of [, b] the followings hold: L f (P ) + L g (P ) L f+g (P ) U f+g (P ) U f (P ) + U g (P ) from where: L f + L g L f+g U f+g U f + U g These inequlities together with: L f = U f = f(x) dx nd L g = U g = g(x) dx prove tht: L f+g = U f+g = (f(x) + g(x)) dx = f(x) dx + g(x) dx Proof of (2). Let P 1 nd P 2 be prtitions of [, c] nd [c, b] respectively. Then P = P 1 P 2 is prtition of [, b]. Clerly, L f (P ) = L f (P 1 ) + L f (P 2 ). Let nd Since L f (P ) Hence L f (P 1 ) Hence L f (P 2 ) b L 1 = sup{l f (P 1 ) P 1 is prtition of [, c]} L 2 = sup{l f (P 2 ) P 2 is prtition of [c, b]}. f(x) dx (by definition) we hve L f (P 1 ) + L f (P 2 ) f(x) dx L f (P 2 ) nd so L 1 f(x) dx L 1 nd so L 2 f(x) dx. f(x) dx L f (P 2 ). f(x) dx L 1 or L 1 + L 2 Now consider upper sums nd observe tht U f (P ) = U f (P 1 ) + U f (P 2 ). Let nd Since U f (P ) U 1 = inf{u f (P 1 ) P 1 is prtition of [, c]} U 2 = inf{u f (P 2 ) P 2 is prtition of [c, b]}. f(x) dx (by definition) we hve U f (P 1 ) + U f (P 2 ) 82 f(x) dx. f(x) dx.

83 Hence U f (P 1 ) Hence U f (P 2 ) Thus b f(x) dx U f (P 2 ) nd so U 1 f(x) dx U 1 nd so U 2 L 1 + L 2 f(x) dx U f (P 2 ). f(x) dx U 1 or U 1 + U 2 f(x) dx U 1 + U 2. f(x) dx. Since f is Riemnn integrble on [, b], for ny ε > 0, P cn be chosen such tht U f (P ) L f (P ) < ε. Then: Hence U f (P 1 ) L f (P 1 ) + U f (P 2 ) L f (P 2 ) = [U f (P 1 ) + U f (P 2 )] [L f (P 1 ) + L f (P 2 )] = = U f (P ) L f (P ) < ε. 0 U f (P 1 ) L f (P 1 ) < ε nd U f (P 2 ) L f (P 2 ) < ε. Hence L 1 = U 1 nd L 2 = U 2. In other words, f is Riemnn-Drboux integrble on both [, c] nd [c, b]. Hence the dditive property is estblished. Proof of (3). It is sufficient to prove tht if f(x) 0 on [, b] then: f(x) dx 0 The bove inequlity follows from the inequlities which re vlid for ny prtition P of [, b]. 0 L f (P ) U f (P ) Proof of (4). We first prove tht f is Riemnn-Drboux integrble on [, b]. We consider the functions f +, f : [, b] R defined by: { f(x) if f(x) 0 f + (x) = 0 if f(x) 0 nd nd we remrk tht: { f (x) = 0 if f(x) 0 f(x) if f(x) 0 f(x) = f + (x) f (x) nd f(x) = f + (x) + f (x) We will show now tht f +, f : [, b] R re Riemnn-Drboux integrble on [, b]. The boundedness of f + nd f is obvious. Consider prtition P of [, b] nd denote: m + i = inf{f + (x) x [x i 1, x i ]} nd M + i = sup{f + (x) x [x i 1, x i ]} Remrk tht: M + i m + i M i m i i = 1, 2,..., n 83

84 where: m i = inf{f(x) x [x i 1, x i ]} nd M i = sup{f(x) x [x i 1, x i ]} Hence, we obtin the inequlities 0 U f +(P ) L f +(P ) U f (P ) L f (P ) for ny prtition P. It follows tht f + is Riemnn-Drboux integrble on [, b]. In similr wy, we obtin tht f is Riemnn-Drboux integrble on [, b]. Using (1) nd the equlity f = f + +f we obtin tht f is Riemnn-Drboux integrble on [, b]. In order to obtin the inequlity (4) we use: nd we deduce tht: f(x) f(x) f(x) f(x) dx f(x) dx f(x) dx 42 Clsses of Riemnn-Drboux integrble functions Theorem If f is continuous on [, b], then f is Riemnn-Drboux integrble on [, b]. Proof. If ε > 0 then either or else f(x) f() < ε for ll x [, b] S = {x x [, b] nd f(x) f() = ε } is non empty nd by the intermedite vlue property, inf S exists. A prtition P of [, b] is constructed s follows: if f(x) f() < ε for ll x [, b] let x 0 = nd x 1 = b, otherwise let x 0 = nd x 1 = inf S. Hence x = x 1 is the first element of [, b] with f(x) f() = ε. If x 1 < b let x 2 be the first element of [x 1, b] with f(x) f(x 1 ) = ε, otherwise let x 2 = b. Define x 3, x 4,... nd so on in similr mnner. If this process continues indefinitely, sequence (x n ) hs been produced in which f(x n ) f(x n 1 ) = ε for ll n N. Now x n is n incresing sequence which is bounded bove nd so (x n ) tends to some limit x. Since f is continuous, this implies tht f(x n ) f(x). Hence f(x n 1 ) f(x) lso. But this contrdicts the condition f(x n ) f(x n 1 ) = ε. So there is n integer N such tht P = {x 0, x 1,..., x N } is 84

85 prtition of [, b] for which f(x n ) f(x n 1 ) = ε, i = 1, 2,..., N. For this prtition P, we hve Hence M i m i < 2ε for ll i = 1, 2,..., N. U f (P ) L f (P ) = n (M i m i )(x i x i 1 ) 2ε (b ). i=1 Now for ny ε > 0 consider ε ε = > 0 nd deduce tht there exists prtition P 2(b ) with U f (P ) L f (P ) < ε. Now L f L f (P ) > U f (P ) ε U f ε. Since ε is rbitrry, L f U f. Since L f U f, by n erlier result, L f = U f. Thus, f is Riemnn-Drboux integrble on [, b]. Definition A function f is clled piecewise continuous on [, b] if there exists prtition P = {x 0, x 1,..., x n } of [, b] nd continuous functions f i defined on [x i 1, x i ], such tht f(x) = f i (x) for x (x i 1, x i ), i = 1, 2,..., n. A prtition Q of [, b] cn be chosen to contin the points x 0, x 1,..., x n. Then Q = P 1 P 2... P n where P i is prtition for [x i 1, x i ]. Hence L f (Q) = n L fi (P i ) i=1 where L f (Q) is the lower sum of f relted to Q. For ech i, L fi (P i ) is the lower sum of n f i relted to P i. Thus L fi (P i ) L f, where L f is the supremum of ll the lower sums i=1 for f on [, b]. Since f i is continuous on [x i 1, x i ], f i is Riemnn integrble on [x i 1, x i ]. Hence: n ( xi ) f i (x) dx L f. x i 1 In similr fshion i=1 U f n i=1 ) f i (x) dx x i 1 ( xi where U f is the supremum of ll upper sums of f on [, b]. Hence, using L f U f, we obtin L f = U f. In other words, piecewise continuous function is Riemnn-Drboux integrble nd f(x) dx = n i=1 xi x i 1 f i (x) dx. Exmple The function given by f(x) = x [x] is piecewise continuous on [0, 3]. Compute 2 0 f(x) dx. 85

86 43 Men vlue theorem Theorem 43.1 (the integrl men vlue theorem). If f nd g re continuous on [, b] nd g(x) 0 for x [, b], then there exists c between nd b such tht f(x) g(x) dx = f(c) g(x) dx. Proof. By the intervl theorem pplied to f on [, b], m f(x) M for ll x [, b] where m is the infimum nd M is the supremum of f on [, b]. Since g(x) 0 we hve m g(x) f(x) g(x) M g(x) for x [, b]. Hence nd m g(x) dx f(x) g(x) dx M k = f(x) g(x) dx g(x) dx [m, M]. g(x) dx By the intermedite vlue property, there exists c [, b] with f(c) = k. Hence f(x) g(x) dx = f(c) g(x) dx. Corollry If f is continuous on [, b], then there exists c [, b] such tht f(x) dx = f(c)(b ). Appliction 43.1 (Integrl test for series). Let f : R + R + be continuous decresing n function nd let n = f(n) for ech n N. Let j n = f(x) dx. The series converges if nd only if (j n ) converges. n 1 Proof. Since f(n + 1) f(x) f(n) for ll x [n, n + 1], n N we hve Therefore nd so n+1 f(n + 1) dx n+1 f(x) dx n n n f(n + 1) n+1 n+1 n f(k) = f(k + 1) n n+1 f(x) dx f(n) n+1 f(x) dx k=2 k=1 1 k=1 f(n) dx. n f(k). 86

87 Now let n = f(n) nd j n = n f(x) dx. Then 1 n n 1 k j n k. k=2 If (j n ) converges, then the n-th prtil sums of nd so n is convergent series. Conversely, if k=1 n re incresing nd bounded bove n is convergent series, then (j n ) is incresing nd bounded bove nd hence, it is convergent sequence. 44 The fundmentl theorem of clculus Theorem If f is Riemnn-Drboux integrble on [, b] nd F (x) = x f(t) dt, then F is continuous on [, b]. Furthermore, if f is continuous on [, b], then F is differentible on [, b] nd F = f. Proof. Since f is integrble, it is bounded on [, b]. So there exists some number M with f(t) M for ll t [, b]. For fixed c, we hve x c x F (x) F (c) = f(t) dt f(t) dt = x x f(t) dt f(t) dt M dt. Since the constnt function x M hs U(P ) = M x c for the trivil prtition P of the intervl with endpoints x nd c, Now given ε > 0, choose δ = ε M. Hence F (x) F (c) M x c. x c < δ F (x) F (c) < ε. In other words F is continuous t c nd, since c ws rbitrry, F is continuous on [, b]. c c c 87

88 Let c [, b] nd consider x > c. Then x c F (x) F (c) f(c) x c = f(t)dt f(t)dt f(c) x c x x f(t)dt (f(t) f(c))dt c f(c) x c c x c x f(t) f(c) dt c x c (since f(c) is constnt) Given ε > 0, there exists δ > 0 such tht f(t) f(c) < ε for t c < δ. Hence, for 0 < x c < δ, x x F (x) F (c) f(c) x c f(t) f(c) dt ε dt c c x c x c < ε. In other words, F +(c) = f(c). Similrly F (c) = f(c). Hence F is differentible nd F = f. Remrk Suppose tht it is required to evlute x2 x 1 f(t) dt where x 1, x 2 [, b] nd f is continuous on [, b]. Using the dditive property we hve x2 x 1 f(t) dt = nd by the fundmentl theorem x2 where F (x) = x f(t) dt. x2 f(t) dt x1 x 1 f(t) dt = F (x 2 ) F (x 1 ) f(t) dt Remrk If f is continuous on [, b] nd Φ = f on [, b], then there exists constnt c such tht Φ(x) = F (x) + c for ny x [, b] where F (x) = (Φ F ) = 0 nd by the men vlue theorem, Φ F = c. x f(t) dt. Tht is becuse Hence if f is continuous on [, b] nd Φ is continuously differentible on [, b] such tht Φ = f, then for ny x 1, x 2 [, b] we hve: x2 x 1 f(t) dt = Φ(x 2 ) Φ(x 1 ). 88

89 Notice tht this method of evluting integrls determine Φ such tht Φ = f on [, b]. f(t) dt hinges on the bility to Definition Any function Φ such tht Φ = f is clled primitive for f. Unfortuntely, most functions do not posses primitives expressible in terms of the elementry functions lone. In such cses it is necessry to settle for numericl estimtes. Exmple ) b) (x 3 + 2) dx = 1 4 x4 + 2x (x 2 x)dx = x3 3 x = 9 4. = 5 6. c) Find the primitives for the following functions: 1. f(x) = x 2 + 3x 2; 2. f(x) = 1 + cos 3x; 3. f(x) = e x cosh 2x; 1 4. f(x) = ; 9 x 2 5. f(x) = x. 45 Techniques to find primitives In the pplictions of integrl clculus, it is necessry to find primitives (when primitives exist nd re expressible in terms of simple functions). Vrious techniques exist for determining primitives nd this short section looks t two of the most importnt: integrtion by prts nd chnge of vribles. Integrtion by prts Proposition If the functions f nd g re continuously differentible on [, b], then f(x) g (x) dx = f(x) g(x) f (x) g(x) dx. where f(x)g (x)dx represents the set of primitives of fg nd the set of primitives of f g. f (x)g(x)dx represents 89

90 Proof. The function h = f g is differentible nd its derivtive h is continuous on [, b]. By the product rule of differentition, we hve h (x) = f (x) g(x) + f(x) g (x). Let now ϕ f(x)g (x)dx nd ψ = ϕ fg. It is esy to see tht ψ = ϕ f g fg = f g nd therefore ψ f (x)g(x)dx. We obtin tht ϕ = fg + ψ nd ψ f (x)g(x)dx. In other words, ϕ fg f (x)g(x)dx. It cn be shown in the sme mnner tht for every ψ f (x)g(x)dx, the function ϕ = fg + ψ f(x)g (x)dx. The vlue of this formul lies in the hope tht the primitive on the right-hnd side is esier to evlute thn the originl one. Corollry If the functions f nd g re continuously differentible on [, b] then b f(x) g (x) dx = f(x) g(x) f (x) g(x) dx. Exmple Mny so-clled reduction formule cn be estblished by repeted integrtion by prts. For exmple, let I n = cos n x dx. Then Hence: I n = cos n 1 x sin x + (n 1) I n 2 (n 1) I n. n I n = cos n 1 x sin x + (n 1) I n 2 n 2. This formul, together with the fct tht I 0 = x nd I 1 = sin x, leds to the evlution of cos n x dx for n N. Exmple Show tht 2 1 x 2 e 2x dx = 1 4 e2 (3e 2 1). Chnge of vribles Proposition If the function g : [α, β] [, b] is continuously differentible bijection hving the property g(α) =, g(β) = b nd f : [, b] R 1 is continuous, then ( ) f(x) dx g = (f g)(t) g (t) dt. where f(x)dx represents the set of primitives of f nd (f g)(t) g (t) dt represents the set of primitives of (f g) g. 90

91 Proof. Let F f(x) dx nd G(t) = (F g)(t). By the composite rule of differentition Hence Let now G G (t) = F (g(t)) g (t) = f(g(t)) g (t) = (f g)(t) g (t). G (f g)(t) g (t) dt (f g)(t) g (t) dt nd consider g 1 : [, b] [α, β]. We hve (g 1 ) (x) = 1 g (t) where x = g(t) nd the function F = F g 1 verifies F (x) = G (g 1 (x))(g 1 ) (x) = (f g)(g 1 (x))[(g 1 ) (x)] 1 (g 1 ) (x) = f(x) ( ) Therefore, F f(x) dx nd G f(x) dx g. Exmple Evlute 1 x ln x dx. Solution: Let f(x) = 1 x ln x nd g(t) = et. Then g (t) = e t nd so 1 x ln x dx = 1 1 e t t et dt = dt = ln t = ln(ln x). t Corollry If the function g : [α, β] [, b] is continuously differentible bijection hving the property g(α) =, g(β) = b nd f : [, b] R 1 is continuous, then Exmple Evlute f(x) dx = 2 1 β α (f g)(t) g (t) dt. t 2 t 3 1 dt. Solution: Let f(x) = x nd g(t) = t 3 1. Then g (t) = 3t 2 nd therefore 2 g(2) t3 1 3t 2 dt = x dx. 1 g(1) Hence 2 1 t 2 t 3 1 dt = x dx = x The chnge of vribles formul is often clled integrtion by substitution where x = g(t) gives the substitution to be used. It is extensively used in elementry clculus books, where tril substitutions, which depend on the form of the integrnd, re suggested. 91

92 Remrk If f : [, ] R is piecewise continuous nd symmetric (f( x) = f(x)) then: f(x) dx = 2 f(x) dx nd if f is ntisymmetric (f( x) = f(x)) then: 0 f(x) dx = 0 In order to obtin the bove equlities, the integrl f(x) dx = 0 f(x) dx + 0 f(x) dx is written s: f(x) dx nd then, in the first integrl, the substitution x = t is mde. Remrk If f : R R is periodic of period T nd piecewise continuous, then for ny R, the following equlity holds: +T f(x) dx = T f(x) dx R 0 In order to prove this equlity, we write: +T f(x) dx = 0 f(x) dx + T f(x) dx + +T f(x) dx In the integrl +T f(x) dx the substitution x = t + T is mde, obtining: 0 T T +T 0 f(x) dx = f(x) dx T 46 Improper integrls The definition of the Riemnn-Drboux integrls pplies only to bounded functions defined on bounded intervls. This section relxes these conditions nd defines improper integrls. The integrl of function defined nd bounded on n intervl which is not bounded is defined below. This is clled n improper integrl of the first kind. 92

93 Definition Let f be function bounded on [, + ) nd Riemnn-Drboux integrble on [, b] for every b >. If lim Otherwise f(x) diverges. b f(x) dx exists, it is sid tht A completely nlogous definition holds for integrls of the form f(x). f(x) converges. Since it is necessry to preserve the dditivity of the integrl for improper integrls the following is defined: + f(x) = 0 f(x) f(x) provided both improper integrls on the right hnd side converge. Exmple ) The integrl b) The integrl c) The integrl d) The integrl x 2 converges to π 2. 1 converges to 1. x2 1 x diverges. sin x diverges. The integrl of function over bounded intervl where the function is not bounded will now be defined. This is clled improper integrl of second kind. Definition Let f be function defined on (, b] nd Riemnn-Drboux integrble on [ + ε, b], for ε (0, b ). If lim f(x) dx ε 0 + +ε exists, then it is sid tht Exmple f(x) dx converges. ) The integrl b) The integrl c) The integrl x dx converges to x 2 dx converges to π 2. 1 dx diverges. x 93

94 + 1 ) The integrl dx diverges. 0 x Theorem 46.1 (Comprison test for integrls). Let f nd g be defined on [, + ) nd Riemnn-Drboux integrble on [, b] for every b >. Suppose tht ) 0 f(x) g(x) for ll x ; b) Then + + g(x) dx converges. f(x) dx converges too. Proof. Now Since 0 f(x) g(x), 0 f(x) dx f(x) dx is incresing nd bounded bove. integrl. g(x) dx. g(x) dx increses to its limiting vlue s b +. Hence Therefore + f(x) dx is convergent A comprison test for improper integrls of the second kind is esily formulted nd proved in similr mnner. Exmple The integrl + 0 e x dx converges. 1 + x2 Solution: Consider the functions f(x) = comprison test. e x 1 + x nd g(x) = 1 nd pply the x2 47 Fourier series The ide tht function my be represented by its Tylor series hs lredy been discussed. We sw tht in order to be ble to write the Tylor series of function f t point x, it needs to be infinitely differentible. This is sever restriction tht most functions do not stisfy. Even when Tylor s theorem with reminder is employed, the function still needs to be differentible finite number of times nd this, like infinite differentibility, certinly implies tht the function must be continuous. Nevertheless, mny functions used to describe importnt physicl phenomen re discontinuous nd cnnot be represented by Tylor series. For exmple, the function used to describe the voltge behvior in time in circuit in which switch is suddenly operted is discontinuous, just like the functionl behvior of the gs pressure cross shock front. 94

95 In principle, t lest, Fourier series offer the possibility of representtion of continuous nd piecewise continuous functions, becuse wheres for Tylor series expnsion, function needs to be differentible, for Fourier series expnsion it would pper tht it only needs to be integrble; the Fourier coefficients cn be computed when f(x) is piecewise continuous. Definition The Fourier series of piecewise continuous function f(x) defined on the intervl [ π, π] is the series f(x) ( n cos nx + b n sin nx) in which the Fourier coefficients n, b n re given by Exmple n = 1 π b n = 1 π π π π π f(x) cos nx dx for n = 0, 1, 2,... f(x) sin nx dx for n = 1, 2,... ) Determine the Fourier series of the function f(x) = π 2 x for x [ π, π]. Solution: 0 = 4π2 3, n = ( 1) n+1 4 n for n = 1, 2,..., b 2 n = 0 for ll n. b) Determine the Fourier series of the function f(x) = x for x [ π, π]. Solution: 0 = π, n = ( 1) n+1 4 π(2n + 1) for n = 1, 2,..., b 2 n = 0 for ll n. c) Determine the Fourier series of the function { for π x < 0 f(x) = b for 0 x < π Solution: 0 = + b, n = 0 for n = 1, 2,..., b 2n = 0, b 2n+1 = n = 1, 2,... 2(b ) (2n + 1)π for Usully, until the convergence problem hs been resolved, it is customry to denote the reltionship between f(x) nd its Fourier series by the sign insted of n equlity. The min result of this section will be estblishing fundmentl theorem on the convergence of Fourier series of piecewise continuous function f(x). However, s this will require severl subsidiry results, which re importnt in their own wy, we now estblish them in the form of two lemms. Lemm 47.1 (Integrl representtion of S n (x)). The n-th prtil sum of the Fourier series of the function f(x) defined on the fundmentl intervl [ π, π], nd prolonged by periodic extension outside it, my be represented in the form: S n (x) = 1 π f(x u) sin ( ) n u π 2 sin 1u du 2 π 95

96 Proof. First, using the summtion formul for geometric progression it follows immeditely tht n e ikx = exp [ i ( ) ] [ n x exp 1 i x] 2 2 i sin 1x. 2 k=1 Hence, equting the rel prts of the two sides of this eqution, we deduce tht 1 n 2 + cos kx = sin ( ) n x 2 sin 1x. 2 k=1 Integrtion of this expression over the intervls [ π, 0] nd [0, π] shows tht 0 π sin ( n + 1 2) u 2 sin 1 2 u du = π 0 sin ( n + 1 2) u 2 sin 1 2 u du = 1 2 π since the only contribution from the left-hnd side comes from the constnt term. Now consider the n-th prtil sum S n (x) of the Fourier series of f(x): S n (x) = n ( k cos kx + b k sin kx). By the definition of the Fourier coefficients k nd b k we my write S n (x) = 1 π f(t) dt + 1 n [ π cos kx f(t) cos kt dt + sin kx 2π π π π k=1 k=1 π π ] f(t) sin kt dt. Tking the functions cos kx, sin kx under the integrl signs, nd employing the trigonometric identity cos k(x t) = cos kx cos kt + sin kx sin kt llows us to write Applying the identity S n (x) = 1 π π π [ ] 1 n f(t) 2 + cos k(x t) dt. k=1 1 n 2 + cos k(x t) = sin ( ) n (x t) 2 sin 1 (x t) 2 k=1 nd writing x t = u, this becomes S n (x) = 1 π x+π x π f(x u) sin ( ) n u 2 sin 1u du. 2 The trigonometric fctor in this integrnd hs period 2π so tht if, for the purpose of the study of its Fourier series, f(x) itself is lso regrded s periodic with period 2π, then the entire integrnd is periodic with period 2π. Consequently, definite integrl of this function tken over ny intervl of length 2π will be the sme, showing tht we my replce the limits x π nd x + π by π nd π, respectively. This ssumption of the periodicity of the function f(x) outside [ π, π] in fct plces no restriction of f(x), becuse the Fourier series cn only represent f(x) in the fundmentl intervl [ π, π], so tht the mnner in which f(x) is defined outside it is immteril. 96

97 Lemm For piecewise continuous function f(x) defined on [ π, π] the following equlities hold: π π ) lim f(x) cos nx dx = 0 nd lim f(x) sin nx dx = 0, nd n π n π ( b) lim f(x) sin n + 1 ) x dx = 0 if π < b π. n 2 Proof. Consider the identity: π π [f(x) S n (x)] 2 dx = π π π [f(x)] 2 dx 2 f(x) S n (x) dx + π π π [S n (x)] 2. From the definition of the Fourier coefficients, the orthogonlity property of the trigonometric system i.e. π π π π π π sin mx cos nx dx = 0 for ll m, n; { 0 for m n sin mx sin nx dx = π for m = n ; cos mx cos nx dx = 0 for m n π for m = n 0 2π for m = n = 0 ; nd from the form of the n-th prtil sum S n (x), it follows tht π [ ] π [S n (x)] 2 2 n 0 dx = f(x) S n (x) dx = π 2 + ( 2 k + b 2 k). π π π Combining the lst two equlities we hve: π [ ] π [f(x) S n (x)] 2 dx = [f(x)] 2 2 n 0 dx π 2 + ( 2 k + b 2 k). π s the integrnd of the left-hnd side integrl involves squre, it is either positive or zero, so we my conclude 2 0 n 2 + ( 2 k + b 2 k) 1 π k=1 π π k=1 f 2 (x) dx. This is known s the Bessel inequlity nd it is true for ll n. The fct tht the right-hnd side is finite by hypothesis, implies tht the sum of squres of the Fourier series coefficients must lwys be convergent. This result implies tht n 0 nd b n 0 for n. In terms of the definition of Fourier coefficients the limits n 0 nd b n 0 re seen to be equivlent to π lim n π k=1 π f(x) cos nx dx = 0 nd lim f(x) sin nx dx = 0. n π 97

98 Observe tht when then we hve 2 0 π lim n π [f(x) S n (x)] 2 dx = 0, 2 + ( 2 k + b 2 k) = 1 π k=1 This lst result is known s Prsevl s equlity. The convergence π π π π [f(x)] 2 dx. [f(x) S n (x)] 2 dx n 0 is usully known s the convergence in the men. We will show now tht if f is piecewise continuous function on [ π, π], then for, b such tht π < b π we hve ( f(x) sin n + 1 ) x dx = 0. 2 lim n For tht, we first remrk tht for ny α < β β ( sin n + 1 ) x dx α 2 = cos ( ( n + 1 2) α cos n + 1 2) β n n Now let = x 0 < x 1 <... x p = b be prtition of the closed intervl [, b] nd the corresponding decomposition of the integrl: ( f(x) sin n + 1 ) p 1 x dx = 2 i=0 xi+1 Consider m i = inf{f(x) x [x i, x i+1 ]} nd represent following form ( f(x) sin n + 1 ) x dx = 2 x i p 1 xi+1 i=0 x i p 1 xi+1 + m i i=0 ( f(x) sin n + 1 ) x dx. 2 ( f(x) sin n + 1 ) x dx in the 2 ( [f(x) m i ] sin n + 1 ) x dx+ 2 x i ( sin n + 1 ) x dx. 2 For ω i = M i m i, where M i = sup{f(x) x [x i, x i+1 ]} we hve f(x) m i M i m i = ω i. Now ( f(x) sin n + 1 ) p 1 p 1 x dx 2 2 ω i x i + m i n + 1/2 i=0 i=0 98

99 p 1 For ε > 0 we choose the prtition such tht ω i x i < ε. This is possible becuse the 2 piecewise continuous function f is integrble. i=0 Now we cn tke n > 4 M(b ), where M = sup{f(x) x [, b]} nd we obtin tht ε for such vlues of n we hve ( f(x) sin n + 1 ) x dx 2 < ε. Collecting the results together we rrive t Lemm Now we re redy to prove the fundmentl Fourier theorem on convergence. Theorem 47.1 (Fourier theorem). Let f be piecewise continuous function defined on the intervl [ π, π] nd extended by periodicity outside it. If f(x) hs finite left-hnd nd right-hnd side derivtives t its points of discontinuity, then: ) when x = x 0 is point of continuity of f, then lim S n(x 0 ) = f(x 0 ). n b) when x = x 0 is point of discontinuity of f, then lim S n(x 0 ) = 1 [ f(x + n 2 0 ) + f(x 0 ) ]. Proof. Consider function f(x) defined on [ π, π] being defined by periodic extension outside [ π, π]. Assume tht f is piecewise continuous on [ π, π] nd hs finite discontinuity t x 0. Denote f(x 0 ) = lim x x 0 f(x) nd f(x + 0 ) = lim f(x). x x + 0 Then from the Lemm 47.1 we my write: S n (x 0 ) = 1 π f(x 0 u) sin ( ) n u π 2 sin 1u du. 2 We hve lso: nd π 1 2 f(x+ 0 ) = 1 0 π π 1 2 f(x 0 ) = 1 π π 0 From the bove results we deduce: S n (x 0 ) 1 2 [ f(x + 0 ) + f(x 0 ) ] = 1 π + 1 π f(x + 0 ) sin ( ) n u 2 sin 1u du 2 f(x 0 ) sin ( ) n u 2 sin 1u du. 2 0 π π [f(x 0 u) f(x + 0 )] sin ( ) n u 2 sin 1u du + 2 [f(x 0 u) f(x 0 )] sin ( ) n u 2 sin 1u du

100 The integrnds on the right-hnd side re well defined everywhere except, mybe t u = 0, where they require exmintion. The first integrnd cn be written in the form ( F 1 (u) sin n + 1 ) u 2 where F 1 (u) = f(x 0 u) f(x + 0 ) u 1 u 2 sin 1u. 2 As u 0, the second fctor tends to 1 nd when the right-hnd side derivtive of f exists t x = x 0, the first fctor tends to f (x + 0 ). So, F 1 (0) = f (x + 0 ) nd the integrnd is well defined t u = 0. Similrly, if F 2 (u) = f(x 0 u) f(x 0 ) u 1 u 2 sin 1u 2 nd if the right-hnd derivtive of f exists t x = x 0, then F 2 (0) = f (x 0 ) nd the second integrnd is lso well defined t u = 0. We my thus write S n (x 0 ) 1 [ f(x ) + f(x 0 ) ] = 1 0 ( F 1 (u) sin n + 1 ) u du+ 1 π ( F 2 (u) sin n + 1 ) u du π π 2 π 0 2 Applying Lemm 47.2 we conclude tht If f is continuous t x 0, then lim S n(x 0 ) = 1 [ f(x + n 2 0 ) + f(x 0 ) ]. lim S n(x 0 ) = f(x 0 ). n We hve thus proved one form of the Fourier theorem on convergence of Fourier series. Exmple ) Deduce the Fourier series expnsion of f(x) = π 2 x 2 in the intervl [ π, π]. b) Deduce the Fourier series expnsion of the function f(x) = x in the intervl [ π, π]. c) Deduce the Fourier series expnsion of the function { if x [ π, 0] f(x) = b if x (0, π] in the intervl [ π, π]. 100

101 48 Different forms of Fourier series Theorem 48.1 (chnge of the origin of the fundmentl intervl). If f(x) is piecewise continuous function defined in the fundmentl intervl [ π, π] nd by periodic extension outside it, then for ny α, the Fourier coefficients n, b n re given by n = 1 π b n = 1 π α+π α π α+π α π f(x) cos nx dx for n = 0, 1, 2,... f(x) sin nx dx for n = 1, 2,... The Fourier series of f(x) converges t every point of continuity nd: f(x) = ( n cos nx + b n sin nx) for x [α π, α + π]. Theorem 48.2 (chnge of the intervl length). The Fourier expnsion of the piecewise continuous function f(x) defined on [ L, L] is the series f(x) = ( n cos nπx L + b n sin nπx L ) with nd Exmple n = 1 L b n = 1 L L L L L f(x) cos nπx L f(x) sin nπx L dx for n = 0, 1, 2,... dx for n = 1, 2,... ) Deduce the Fourier series expnsion of the function x for π 2 x < 0 f(x) = x for 0 x < π 2π x for π x 3π 2. b) Deduce the Fourier series expnsion of f(x) = x 3 for 1 x 1. When f(x) is n even function defined on the intervl [ π, π], then f( x) = f(x). Thus, it follows directly tht f(x) cos nx is n even function, becuse cos nx is even, nd f(x) sin nx is n odd function, becuse sin nx is odd. Consider the Fourier coefficients n of n even function f(x), tht we choose to write in the form n = 1 π 0 π f(x) cos nx dx + 1 π 101 π 0 f(x) cos nx dx.

102 Then, chnging the vrible in the first integrnd by writing u = x, employing the even nture of the integrnd to replce f( u) cos n( u) by f(u) cos nu nd chnging the sign of the integrl by reversing the limits, we find n = 2 π π 0 f(x) cos nx dx for n = 0, 1, 2,... The sme rgument pplied to the coefficients b n shows tht b n = 0 for n = 1, 2,... Consequently, if f(x) is n even function on [ π, π], its Fourier series contins only cosine functions nd is of the form f(x) = n cos nx for x [ π, π]. This is clled the Fourier cosine expnsion of the even function f(x) in [ π, π]. Exmple ) Deduce the Fourier series expnsion of the even function f(x) = x 2 in [ π, π]. b) Deduce the Fourier series expnsion of the even function f(x) = x in [ π, π]. When f(x) is n odd function defined on the intervl [ π, π], then f( x) = f(x). A similr rgument s in the cse of even functions leds us to nd b n = 2 π π 0 n = 0 for n = 0, 1,... f(x) sin nx dx for n = 1, 2,... from which it follows tht the Fourier series of n odd function defined on [ π, π] contins only sine functions nd is of the form f(x) = b n sin nx for x [ π, π]. This is clled the Fourier sine expnsion of the odd function f(x) in [ π, π]. This results cn be usefully interpreted in terms of ny rbitrry function f(x) which is to be expnded in the hlf intervl [0, π]. Defining new function g(x), by the rule { f( x) for π x < 0 g(x) = f(x) for 0 x π we see tht g(x) is n even function which is equl to f(x) in the required intervl [0, π]. Thus, s Fourier cosine expnsion of g(x) only requires the knowledge of g(x) in the hlf intervl [0, π] in which g(x) = f(x), it follows tht f(x) = n cos nx for x [0, π] 102

103 is the desired cosine expnsion of f(x) in [0, π]. Alterntively, we my expnd the sme function f(x) in the hlf intervl [0, π] in Fourier sine series s follows: define new function h(x) by the rule { f( x) for π x < 0 h(x) = f(x) for 0 x π. Then, h(x) is n odd function which is equl to f(x) in the required intervl [0, π]. The Fourier sine expnsion of h(x) only requires the knowledge of h(x) in the hlf intervl [0, π] where h(x) = f(x). So f(x) = b n sin nx for 0 x π provides the desired sine expnsion of f(x) for x [0, π]. These expnsions re often clled the hlf-rnge expnsions of f(x). We hve proved the following theorem: Theorem 48.3 (Fourier sine nd cosine series). If f(x) is n rbitrry function defined nd piecewise continuous on [0, π], then it my either be expnded s Fourier cosine series f(x) = n cos nx 0 x π in which n = 2 π or s Fourier sine series π 0 f(x) = f(x) cos nx dx for n = 0, 1, 2,... b n sin nx 0 x π in which b n = 2 π π 0 f(x) sin nx dx for n = 1, 2,

104 Prt III Functions of severl vribles 49 Topology in R n Definition The set R n is the collection of ll the finite sequences x = (x 1, x 2,..., x n ) of n rel numbers: R n = {(x 1, x 2,..., x n ) x i R 1, i = 1, 2,..., n} Definition A rel vlued function of n vribles ssocites to every finite sequence of n rel numbers of set A R n n unique rel number. Formlly, f : A R n R 1 is given by (x 1, x 2,..., x n ) = x f(x) = f(x 1, x 2,..., x n ) where x = (x 1, x 2,..., x n ) is n element of A R n. Exmple The function f : R 2 R 1 given by f(x 1, x 2 ) = x x 2 2 is rel vlued function of two vribles. Definition A vector vlued function f of n vribles ssocites to every finite sequence x = (x 1, x 2,..., x n ) of n rel numbers of the set A R n unique vector f(x) from R m. Formlly, f : A R n R m is given by x f(x 1, x 2,..., x n ) = (f 1 (x 1, x 2,..., x n ),..., f m (x 1, x 2,..., x n )) where x = (x 1, x 2,..., x n ) A nd (f 1 (x 1, x 2,..., x n ),..., f m (x 1, x 2,..., x n )) R m. Exmple The function f : R 3 R 2 given by f(x 1, x 2, x 3 ) = (x 1 + x 2 + x 3, x 1 x 2 x 3 ) is vector function of three vribles. Here f 1 (x 1, x 2, x 3 ) = x 1 +x 2 +x 3 nd f 2 (x 1, x 2, x 3 ) = x 1 x 2 x 3. If f(x 1, x 2,..., x n ) = (f 1 (x 1, x 2,..., x n ),..., f m (x 1, x 2,..., x n )), then f i re rel vlued functions of n vribles for i = 1, m nd re clled the sclr components of the vector function f. When functions of one rel vrible were discussed, it ws necessry to investigte rel numbers which were close to fixed rel number. This led to n interest in the quntity x. Anlogies of length nd distnce in R n cn be obtined s follows: 104

105 R n is orgnized s n n-dimensionl vector spce using the sum nd the sclr product defined by: (x 1, x 2,..., x n ) + (y 1, y 2,..., y n ) = (x 1 + y 1, x 2 + y 2,..., x n + y n ) k(x 1, x 2,..., x n ) = (kx 1, kx 2,..., kx n ) For x R n the norm (or length) of x is defined by x = n x 2 i x = x x 2 n. i=1 The distnce between x nd = ( 1, 2,..., n ) is tken to be x. So x = n (x i i ) 2 = (x 1 1 ) 2 + (x 2 2 ) (x n n ) 2. i=1 A neighborhood of R n is set V R n which contins hypersphere S r () centered in, S r () = {x R n x < r} r > 0 S r () V. Deleting from neighborhood V of, deleted neighborhood of will be obtined: V 0 = V \ {}. Exmple For ny ( 1, 2 ) R 2 = ( 1, 2 ). For exmple, the set hypersphere S r () is neighborhood of is neighborhood of = (0, 1). S 1 (0, 1) = {(x 1, x 2 ) x (x 2 + 1) 2 < 1} Now the limit of sequence (x k ) of points of R n cn be defined. A sequence (x k ) of points of R n is function whose domin is the set of nturl numbers nd whose vlues belong to R n. The vlue of the function corresponding to rgument k is denoted by x k. The sequence x 1, x 2,..., x k,... is denoted by (x k ). Definition A vector x R n is sid to be the limit of the sequence (x k ) if for ny ε > 0 there exists N = N(ε) > 0 such tht for ny k > N we hve x k x < ε. In this cse we write lim k x k = x. The limit of the sequence (x k ), if it exists, is unique. If sequence (x k ) converges to x, then the sequence is bounded, i.e. there exists M > 0 such tht x k < M for ny k N. If sequence (x k ) converges to x, then ny subsequence (x kl ) of the sequence (x k ) converges to x. 105

106 A sequence (x k ), x k = (x 1k, x 2k,..., x nk ) R n converges to x = (x 1, x 2,..., x n ) R n if nd only if the sequence (x ik ) converges to x i for ny i = 1, 2,..., n. According to Bolzno-Weierstrss theorem, ny bounded sequence (x k ) of points of R n contins convergent subsequence. The Cuchy s criterion for the convergence of sequence (x k ) of points x k R n sttes tht (x k ) converges if nd only if for ny ε > 0 there exists N ε such tht for p, q > N ε we hve x p x q < ε. Definition Let be A R n. A point x R n is clled n interior point of the set A if there exists hypersphere S r (x) such tht S r (x) A. For instnce, point y of hypersphere S r () is n interior point of the hypersphere. Definition The interior set of A R n is the set of ll interior points of the set A. Usully the interior of set A is denoted by Int(A). For instnce, if A is the hypersphere S r (), then Int(A) = S r () = A. Definition A set A R n is sid to be open if A = Int(A). For instnce, ny hypersphere S r (x) is n open set. A set A R n is open if nd only if it contins neighborhood of ech of its points. The union of ny fmily of open sets is open. The sets R n nd re open. The intersection of finite number of open sets is open. Definition A set A R n is sid to be closed if its complement is open. The intersection of ny fmily of closed sets is closed. The union of finite number of closed sets is closed. The sets R n nd re closed. A closed hypersphere S r () defined s: is closed. S r () = {x R n x r} Definition A point R n is limit point (or point of ccumultion) of the set A R n provided every deleted neighborhood of intersects A. Definition The closure A of set A R n is the intersection of ll closed sets contining A. The set of points in A nd not in the interior Int(A) of A is clled the boundry of A nd it is denoted by A. 106

107 The closure opertion hs the properties: A B = A B; A A; A = A; A = A A is closed; x A if nd only if every neighborhood V of x intersects A. Definition A set A R n is bounded if there exists r > 0 such tht A S r (0). Definition A set A R n is compct if it is both bounded nd closed. For instnce, closed hypersphere S (r) is compct set. Exmple The set D defined by D = {(x, y) x + y 1 nd x 0 nd y 0} is compct subset of R 2. Solution: D is bounded since it is contined in the hypersphere S 2 (0) = {(x, y) x 2 +y 2 < 4}. If is n element of the complement of D, then lies distnce r > 0 wy from t lest one of the lines x + y = 1, x = 0 or y = 0. Hence, the open hypersphere S r () lies in one of the regions x + y > 1, x < 0 or y < 0. So S r () lies in the complement of D. Thus, D is closed. Remrk If A R n is compct set, then every sequence (x k ) with x k A contins subsequence (x kl ) which converges to point x 0 A. Definition A set A R n is connected if there re no open sets G 1, G 2 such tht A G 1 G 2, A G 1, A G 2, nd (A G 1 ) (A G 2 ) =. 50 Limit of function t point Definition Let f : A R n R 1 be rel vlued n vrible function nd point of ccumultion of A (i.e. every deleted neighborhood of contins t lest one point A). The rel number L is clled the limit of f(x) s x tends to if for ny ε > 0, there exists δ = δ(ε) > 0 such tht 0 < x < δ f(x) L < ε. We write lim f(x) = L. x 107

108 Just like in the cse of functions of one vrible, this definition is techniclly difficult to implement except for the simplest functions. However, the obvious generliztion of the sum, product nd quotient rules cn be proved. Their use is illustrted in the following exmple. Exmple Evlute lim f(x, y) where f(x, y) = x2 y 2 (x,y) (2,1) x 2 + y. 2 Solution: As x 2 nd y 1, x 2 y 2 3 nd x 2 + y 2 5. Hence x 2 y 2 x 2 + y 2 (x,y) (2,1) Definition Let f : A R n R m be vector vlued function of n vribles nd point of ccumultion of A. L R m is clled the limit of f(x) s x tends to, if for ny ε > 0, there exists δ > 0 such tht < x < δ f(x) L < ε. We write L = lim x f(x). If f(x 1,..., x n ) = (f 1 (x 1,..., x n ),..., f m (x 1,..., x n )) nd L = (L 1,..., L m ), then the following sttement holds: lim f(x) = L lim f i(x) = L i i = 1, m. x x ( ) xy Exmple Evlute lim f(x, y) where f(x, y) = (x,y) (2,1) x 2 + y, x y. 2 Theorem 50.1 (Heine s criterion for the limit). The function f : A R n R m hs limit s x pproches if nd only if for ny sequence (x k ), x k A, x k, nd x k s k, the sequence (f(x k )) converges. Proof. As for the one vrible rel vlued functions. Theorem 50.2 (Cuchy-Bolzno s criterion for the limit). The function f : A R n R m hs limit s x if nd only if for ny ε > 0 there exists δ > 0 such tht 0 < x < δ nd 0 < x < δ f(x ) f(x ) < ε. Proof. As for the one vrible rel vlued functions. 51 Continuity Definition A rel vlued n vrible function f : A R n R 1 is continuous t A if lim x f(x) = f(). 108

109 This definition requests three things: firstly tht lim x f(x) exists, secondly tht f() is defined, nd finlly tht the previous two vlues re equl. In terms of ε, δ this is equivlent to the following: Definition A function f : A R n R 1 is continuous t A if for every ε > 0 there exists δ = δ(ε) > 0 such tht x < δ f(x) f() < ε. Definition A vector vlued n-vrible function f : A R n R m is continuous t A if for every ε > 0 there exists δ = δ(ε) > 0 such tht x < δ f(x) f() < ε. Exmple Use the ε, δ condition of continuity to prove tht the following functions re continuous t the mentioned points: ) f(x, y) = x 2 + y 2 t x = y = 0; b) f(x, y) = (x 2 y 2, x y) t x = 1, y = 1; c) f(x, y, z) = x + y + z t x = y = z = 0; d) f(x, y, z) = (x 2 + y 2 + z 2, x + y + z) t x = y = z = 1. The rules for continuous functions of one vrible cn be generlized to give corresponding rules for functions of severl vribles. These re stted in the next two theorems. Theorem Let f nd g be rel vlued functions of n vribles defined in neighborhood of. If f nd g re continuous t, then so re f + g, f g, nd, when f(x) 0, 1 f. Theorem Let f : A R n B R m be continuous t A nd g : B R m R p be continuous t f() = b R m. Then the composite function g f : A R p is continuous t. Discontinuities of functions of more thn one vrible re often difficult to spot. In the cse of function f of two vribles, ny discontinuities cn be visulized geometriclly by ppeling to the surfce in R 3 represented by the eqution z = f(x, y). Some, but by no mens ll, of the discontinuities of such surfce correspond to holes or ters. Exmple The function f : R 2 R 1 given by f(x, y) = x 2 + y 2 is continuous for ll (x, y). The surfce given by z = f(x, y) is prbolic bowl. To see this, notice tht horizontl section for z = k 0 gives the circle x 2 + y 2 = k nd verticl cross-section for fixed x or y gives prbol. Exmple Investigte the behvior of the function f given by xy if (x, y) (0, 0) x f(x, y) = 2 + y 2 0 if (x, y) = (0, 0) s (x, y) pproches (0, 0). 109

110 Solution: f is discontinuous t (0, 0) nd the discontinuity is much nstier thn the removble jump of finite discontinuities seen for functions of one vrible. Recll tht when estblishing the discontinuity of function of one vrible t some points often the right nd left-hnd limits existed, but were unequl. Clerly, if lim (x,y) (0,0) f(x, y) exists, its vlue must be independent on the wy in which (x, y) pproches (0, 0). Consider f(x, mx): lim lim x 0 mx 2 f(x, mx) = lim x 0 x 0 x 2 + m 2 x = m m. 2 But this quntity vries with m nd so f cnnot be continuous t (0, 0), no mtter wht vlue is specified for f(0, 0). Exmple Investigte the behvior of the function f given by xy 3 if (x, y) (0, 0) f(x, y) = x 2 + y 6 0 if (x, y) = (0, 0) s (x, y) pproches (0, 0). Solution: Firstly lim f(x, mx) = lim x 0 x 0 m 3 x 4 x 2 + m 6 x 6 = lim x 0 m 3 x m 6 x 4 = 0. However this is not sufficient evidence to suppose tht f(x, y) pproches 0 s (x, y) pproches (0, 0). In fct Hence f is discontinuous t (0, 0). lim f(x, 3 x) = lim x2 x 0 x 2 + x = Theorem Let f : A R n R m, f(x) = (f 1 (x),..., f m (x)) nd A. The function f is continuous t A if nd only if the functions f i, i = 1, 2,..., m re continuous t. Theorem 51.4 (Heine s criterion for continuity). The function f : A R n R m is continuous t A if nd only if for ny sequence (x k ), x k A, x k the sequence k (f(x k )) converges to f(). Theorem 51.5 (Cuchy-Bolzno s criterion for continuity). The function f : A R n R m is continuous t A if nd only if for ny ε > 0 there exists δ > 0 such tht x < δ nd x < δ f(x ) f(x ) < ε. Remrk If the function f : A R n R m is continuous t A, then the function f : A R n R 1 + defined by f (x) = f(x), is continuous t. Remrk Generliztions of some importnt theorems, proved for single vrible rel vlued continuous functions, requires higher dimensionl nlogues of the topology in R

111 52 Importnt properties of continuous functions Theorem 52.1 (The boundedness property). If f : A R n R m is continuous on the compct set A, then ) the set f(a) is bounded nd b) there exists A such tht f() = sup f(a). Proof. ) Assume tht f(a) is unbounded. Then for every k N there exists x k A such tht f(x k ) > k. The sequence (x k ) is bounded nd therefore there exists subsequence (x kl ) of the sequence (x k ) which converges towrds point x 0 A, x kl k l x 0. Hence, the sequence (f(x kl )) converges to f(x 0 ). Therefore, there exists N such tht for k l > N we hve f(x kl ) f(x 0 ) + 1. Absurd. b) Consider R = sup f(a) nd note tht for every k N there exists x k A such tht R 1 k < f(x k) < R. For the sequence (x k ) there exists subsequence x kl such tht x kl k l f(x 0). From the inequlity f(x kl ) k l f(x 0) nd f(x kl ) k l it follows tht f(x 0 ) = R. R 1 k l < f(x kl ) < R. x 0 A. Therefore Corollry If f : A R n R 1 is continuous on the compct set A, then: ) there exists m, M such tht m = inf{f(x) x A} M = sup{f(x) x A} b) there exist c nd d, c, d A such tht f(c) = m nd f(d) = M. Definition A function f : A R n R 1 is uniformly continuous (on A) if for every ε > 0 there exists δ = δ(ε) > 0 such tht for x, x A we hve x x < δ f(x ) f(x ) < ε. Theorem A vector vlued function f : A R n R m is uniformly continuous (on A) if nd only if its sclr components f 1, f 2,..., f m : A R 1 re uniformly continuous. Theorem If f : A R n R m is continuous on the compct set A, then f is uniformly continuous on A. Theorem Let A R n, A R m nd f : A A such tht f(a) = A. The function f is continuous on A if nd only if for every open set G R m, there exists n open set G R n such tht G A = f 1 (G A ). 111

112 Corollry The function f is continuous on A if nd only if for every closed set F R m, there exists closed set F R n such tht F A = f 1 (F A ). Theorem If the set A R n is connected nd the function f : A R n R m is continuous, then the set f(a) is connected. Corollry If the set A R n is compct nd connected nd the function f : A R n R 1 is continuous, then f(a) is closed intervl. 53 Differentition This section defines wht is ment by sying tht function of n vribles is differentible, but to begin, let s exmine the concept of prtil differentibility. Definition Let f : A R n R 1 be rel vlued function of n vribles nd n interior point of the set A. The function f is sid to be prtilly differentible with respect to x i t if f( 1,..., i 1, i + h, i+1,..., n ) f( 1,..., i,..., n ) lim h 0 h exists. The vlue of this limit is denoted by f x i () nd is clled the prtil derivtive of f with respect to x i t the point. Definition If f is prtilly differentible with respect to x i in neighborhood of, then the function x f x i (x) defined on tht neighborhood is clled the prtil derivtive of f with respect to x i. Remrk To clculte prtil derivtives, one hs to differentite (in the norml mnner) with respect to x i keeping ll the other vribles fixed. Hence, the obvious rules for prtilly differentiting sums, products nd quotients cn be used. Exmple Clculte the prtil derivtives of the function f given by f(x, y, z) = x 2 y + x sin y + y z. Solution: The prtil derivtives with respect to x is Similrly, f x f y = x2 + x cos y + 1 z = 2xy + sin y. nd f z = y z

113 Exmple Clculte the prtil derivtives of the function f given by n n f(x 1,..., x n ) = ij x i x j (x 1,..., x n ) R n. Solution: i=1 f x k = j=1 n ( kj + jk )x j. j=1 Definition Let f : A R n R m be n n vrible vector vlued function (i.e. f(x 1,..., x n ) = (f 1 (x 1,..., x n ),..., f m (x 1,..., x n ))) nd n interior point of the set A. The function f is sid to be prtilly differentible with respect to x i t if the sclr components f 1 (x 1 (,..., x n ),..., f m (x 1,..., x n ) re prtilly differentible with respect to x i f1 t. The vector (),..., f ) m () is clled the prtil derivtive of f with respect x i x i to x i t nd it is denoted by f x i (): f x i () = ( f1 (),..., f ) m (). x i x i If f = (f 1,..., f m ) is prtilly differentible with respect to x i in neighborhood of, function f defined on tht neighborhood, clled the prtil derivtive of f with respect x i to x i is obtined: x f x i (x) = ( f1 (x),..., f ) m (x). x i x i Exmple Clculte the prtil derivtives of the function f given by Solution: f x f(x, y, z) = (x + y + z, xy + xz + yz, xyz). = (1, y + z, yz) ; f y = (1, x + z, xz) ; f z = (1, x + y, xy). Definition Let f : A R n R 1 be rel vlued function of n vribles, n interior point of the set A nd u unit vector in R n (i.e. u = 1). If the limit f( + t u) f() lim t 0 t exists it is clled the directionl derivtive of f t the point nd it is denoted by f( + t u) f() u f() = lim. t 0 t Remrk Let e i = (0,..., 0, }{{} 1, 0,..., 0). The directionl derivtive of f t is i ei f() = f x i () i = 1, n. Hence, prtil derivtives re specil cses of directionl derivtives. 113

114 Exmple If u = (u x, u y ) nd f(x, y) = x y, then u f(x, y) = f x u x + f y u y = y u x + x u y. Definition Let f = (f 1,..., f m ) be vector vlued n vribles function f : A R n R m ; n interior point of A nd u unit vector in R n (i.e. u = 1). If the limit f( + t u) f() lim t 0 t exists, it is clled the directionl derivtive of f t the point nd it is denoted by u f(). It is esy to see tht u f() = ( u f 1 (),..., u f m ()). Remrk The directionl derivtive u f() exists if the directionl derivtives u f i (), i = 1, m exist. Exmple The directionl derivtive of the function f(x, y, z) = (xy +xz +yz, xyz) t the point (x, y, z) is u f(x, y, z) = ((y + z)u x + (x + z)u y + (x + y)u z, yz u x + xz u y + xy u z ). Theorem Let f be rel vlued function of n vribles, f : A R n R 1, nd n interior point of A. If the prtil derivtives f, i = 1, n exist in neighborhood of x i nd they re continuous t, then the following equlity holds: lim h 0 1 h [ f( + h) f() n i=1 f x i () h i Proof. Consider the vectors v j = ( 1, 2,..., j, j+1 + h j+1,..., n + h n ) for j = 0, n 1 nd v n = nd represent f( + h) f() in the form: n 1 f( + h) f() = [f(v j ) f(v j+1 )] = = j=0 n i=1 n 1 j=0 ] = 0. f x i (v i + θ i h i e i ) h i with 0 θ i 1 where e i = (0,..., 0, 1, 0,..., 0). Hence: [ ] 1 n f f( + h) f() () h i = 1 h x i h i=1 f x j+1 (v j+1 + θ j+1 h j+1 e j+1 )h j+1 n i=1 [ f (v i + θ i h i e i ) f ] () h i. x i x i Now remrk tht v i + θ i h i e i h, i = 1, n nd therefore, for ε > 0, there is δ > 0 such tht h < δ f (v i + θ i h i e i ) f () x i x i < ε, i = 1, n. n 114

115 So, h < δ 1 h n f( + h) f() f () h i x i < ε. i=1 The bove theorem shows tht in smll neighborhood of the function f cn be n f pproximted by the polynomil of first degree f() + () h i. x i Definition A rel vlued n vribles function f : A R n R 1 is sid to be differentible t if it is prtilly differentible t with respect to every vrible x i nd [ ] 1 n f lim f( + h) f() () h i = 0. h 0 h x i The function d f : R n R 1 defined by i=1 i=1 d f(h) = n i=1 f x i () h i, h R n is clled the Fréchet derivtive of f t. Remrk The Fréchet derivtive d f : R n R 1 of function f : A R n R 1 t is liner function on R n. It is polynomil of first degree in h 1, h 2,..., h n. Remrk For h = 1, we hve d (h) = h f(). Exmple Show tht the following functions re differentible nd compute their Fréchet derivtives. ) f(x, y) = x 2 + y 2, d (x,y) f(h x, h y ) = 2x h x + 2y h y ; b) f(x, y) = x y, d (x,y) f(h x, h y ) = y h x + x h y ; c) f(x, y, z) = x y+x z+y z, d (x,y,z) f(h x, h y, h z ) = (y+z) h x +(x+z) h y +(x+y) h z. Remrk If the rel vlued function f : R n R 1 is differentible t A, then f is continuous t. Definition Let f = (f 1,..., f m ) be vector vlued function of n vribles, f : A R n R m, nd n interior point of A. The function f is sid to be differentible t if every sclr component f j, j = 1, m of f is differentible t. The function d f : R n R m defined by ( m n ) f i d f(h) = () h i e j x i j=1 is clled the Fréchet derivtive of f t, where e j = (0,..., 0, 1, 0,..., 0) R m. i=1 115

116 The Fréchet derivtive is set of m polynomils of first degree in h 1, h 2,..., h n. Exmple Show tht f(x 1, x 2, x 3 ) = (x 1 x 2 x 3, x x x 2 3) is differentible t ny point nd compute its Fréchet derivtive. Solution: d f(h) = (x 2 x 3 h 1 + x 1 x 3 h 2 + x 1 x 2 h 3, 2x 1 h 1 + 2x 2 h 2 + 2x 3 h 3 ) Definition The mtrix of the liner function d f is clled the Jcobi mtrix of f t. This is m n mtrix nd is given by ( ) fi J (f) = () x i We hve d f(h) = J (f) h.. m n Exmple Consider the vector vlued function of n vribles f : R n R m defined by ( m n ) f(x 1,..., x n ) = ij x j e i. i=1 Show tht f is Fréchet differentible t ny point x R n nd the following reltions hold: ( m n ) d x f(h) = ij h j e i J (f) = i=1 j=1 j=1 ( ) fi () = ( ij ) m n. x i m n Remrk If the vector vlued function of n vribles f : A R n R m is differentible t A, then f is continuous t. Theorem Let f : A R n B R m nd g : B R m R p. If f is differentible t Int(A) nd g is differentible t f() = b Int(B), then h = g f is differentible t nd d h = d b g d f. Proof. f differentible t implies with ε 1 (x) 0 s x. g differentible t b = f() implies with ε 2 (y) 0 s y b. Hence f(x) f() = d f(x ) + ε 1 (x) x g(y) g(b) = d b g(y b) + ε 2 (y) y b h(x) h() =g(f(x)) g(f()) = d b g(f(x) f()) + ε 2 (f(x)) f(x) f() =d b g(d f(x ) + ε 1 (x) x ) + ε 2 (f(x)) d f(x ) + ε 1 (x) x =d b g d f(x ) + x d b g(ε 1 (x)) + d f(x ) + ε 1 (x) x ) ε 2 (f(x)). 116

117 Denote ε 3 (x) = h(x) h() d bg d f(x ) x =d b g(ε 1 (x)) + d f(x ) + ε 1 (x) x x ε 2 (f(x)) Hence nd ε 3 (x) 0 s x. ε 3 (x) d b g ε 1 (x) + ( d f + ε 1 (x) ) ε 2 (f(x)) Remrk The Jcobi mtrix of h t is the product of the Jcobi mtrix of g t b nd the Jcobi mtrix of f t : h i x j () = m k=1 g i y k (b) f k x j (), i = 1, p, j = 1, n. Exmple Consider f : R 2 R 2 defined by f(x 1, x 2 ) = (x 1 + x 2, x 1 x 2 ) nd g : R 2 R 2 defined by g(ρ, θ) = (ρ cos θ, ρ sin θ). Find h(ρ, θ) = (f g)(ρ, θ) nd h ρ, h θ. Corollry Let f : A R n B R n be bijection where A, B re open subsets of R n. If f is differentible t A nd f 1 is differentible t b = f(), then d f is bijection of R n on R n nd (d f) 1 = d f() f 1. The bove sttement follows from the equlity f 1 f = i A using the rule of differentible composite functions. 54 Bsic properties of differentible functions Men vlue theorem(lgrnge) Let x, h R n. Denote by [x, x + h] the set defined by: [x, x + h] = {x + th R n 0 t 1} This set is clled the closed segment which joins x nd x + h. Consider A R n, A n open subset, f : A R 1 nd x, x + h A. Theorem If the following conditions hold: ) [x, x + h] A b) f is differentible on [x, x + h] then there exists t 0 (0, 1) such tht: f(x + h) f(x) = d x+t0 hf(h) = n i=1 f x i (x + t 0 h) h i 117

118 Proof. Consider the function ϕ(t) = f(x+th) for t [0, 1]. The function ϕ is differentible on [0, 1] nd ϕ (t) = d x+th f(h). Since f(x + h) f(x) = ϕ(1) ϕ(0), pplying the men vlue theorem for the function ϕ on [0, 1], we obtin tht there exists t 0 (0, 1) such tht: n ϕ(1) ϕ(0) = ϕ f (t 0 ) = d x+t0 hf(h) = (x + t 0 h) h i x i i=1 Remrk If the function f : A R n R m with m 2, then the bove theorem is flse. In order to illustrte this, consider for exmple the function f : [0, 2π] R R 2, f(t) = (cos t, sin t). In the cse f : A R n R m (m 2) the men vlue theorem is the following: Theorem If the following conditions hold: 1) [x, x + h] A 2) f is differentible on [x, x + h] 3) d x+th f M t [0, 1] then f(x + h) f(x) M h. Proof. Consider gin the function ϕ(t) = f(x + th) for t [0, 1] nd n ψ(t) = (ϕ i (1) ϕ i (0)) ϕ i (t) for t [0, 1]. For ψ there exists t 0 [0, 1] such tht i=1 ψ(1) ψ(0) = ψ (t 0 ). Hence ϕ(1) ϕ(0) 2 = i=1 m [ϕ i (1) ϕ i (0)] 2 = i=1 i=1 m [ϕ i (1) ϕ i (0)] ϕ i(t 0 ) [ m ] 1 [ 2 m ] 1 2 [ϕ i (1) ϕ i (0)] 2 [ϕ i(t 0 )] 2 M h ϕ(1) ϕ(0) i=1 nd ϕ(1) ϕ(0) M h Locl extremum Definition A function f : A R n R 1 hs locl mximum vlue t c A if there exists some open neighborhood V A of c for which f(x) f(c), for ny x V. It f(x) f(c) for ny x V, then f hs locl minimum vlue t c. Theorem 54.3 (Fermt). If f is differentible t c nd possesses locl mximum or locl minimum t c, then f x i (c) = 0 for i = 1, n. 118

119 Proof. Consider h R n nd for t R 1 sufficiently close to 0, the function ϕ(t) = f(c+th). The function f possesses locl mximum or locl minimum t c if nd only if ϕ possesses locl mximum or locl minimum t t = 0. Since the derivtive of ϕ t locl extremum is equl to 0 it follows tht ϕ (0) = n i=1 f x i (c) h i = 0 h R n Therefore f x i (c) = 0 for i = 1, n Note tht lthough f x i (c) = 0 t locl extremum c, this is not sufficient condition for such point. Exmple If f(x, y) = xy, then f = y nd f = x nd hence (0, 0) is x y the only possible locl extremum of f. However, for ny δ > 0, f(δ, δ) = δ 2 nd f( δ, δ) = δ 2 < 0. Hence, f tkes both positive nd negtive vlues in ny neighborhood of (0, 0). Thus (0, 0) is neither locl mximum nor locl minimum. Definition A point c is sttionry point of f if f x i (c) = 0 for i = 1, n. Just like in the cse of functions of one vrible, sttionry points for functions of severl vribles cn be clssified with the id of Tylor pproximtions. Definition Let f : A R n R m be differentible function on the open set A. If the prtil derivtives A x f i x j re continuous, i = 1, m, j = 1, n, then f is sid to be continuously differentible. Theorem 54.4 (of locl inversion). If the function f : A R n R n is continuously differentible on the open set A nd the Jcobi mtrix of f, ( f i x j ()) n n, is not singulr (i.e. det( f i x j ()) 0), then there exists n open neighborhood U of nd n open neighborhood V of f() = b such tht f : U V is bijective. Moreover, the inverse f 1 : V U is differentible t b = f() nd the following equlity holds: d b f 1 = (d f) 1 Proof. The proof of this theorem is rther technicl nd will be skipped. Exmple Show tht if ρ 0, then the function f(ρ, θ) = (ρ cos θ, ρ sin θ) is loclly invertible. Exmple Show tht if ρ 0, then the function f(ρ, θ, ϕ) = (ρ sin θ cos ϕ, ρ sin θ sin ϕ, ρ cos θ) is loclly invertible. 119

120 Implicit functions Consider A R n nd B R m two open subsets nd let be the function f : A B R m. Denote by f 1, f 2,..., f m the sclr components of f, i.e. f = (f 1, f 2,..., f m ) nd consider the system of equtions: ( ) f 1 (x 1, x 2,..., x n, y 1, y 2,..., y m ) = 0 f 2 (x 1, x 2,..., x n, y 1, y 2,..., y m ) = 0 f m (x 1, x 2,..., x n, y 1, y 2,..., y m ) = 0 Definition If there exists function ϕ : A A B, ϕ = (ϕ 1, ϕ 2,..., ϕ m ) such tht the following equlities hold: f 1 (x 1, x 2,..., x n, ϕ 1 (x 1, x 2,..., x n ),..., ϕ m (x 1, x 2,..., x n )) = 0 f ( ) 2 (x 1, x 2,..., x n, ϕ 1 (x 1, x 2,..., x n ),..., ϕ m (x 1, x 2,..., x n )) = 0 f m (x 1, x 2,..., x n, ϕ 1 (x 1, x 2,..., x n ),..., ϕ m (x 1, x 2,..., x n )) = 0 for ny (x 1, x 2,..., x n ) A, then the function ϕ = (ϕ 1, ϕ 2,..., ϕ m ) is sid to be defined implicitly by the system of equtions ( ). It is cler tht the system of equtions ( ) cn be written shortly s f(x, y) = 0 where x = (x 1, x 2,..., x n ) nd y = (y 1, y 2,..., y n ). Theorem 54.5 (Implicit function theorem). If the function f hs the following properties: 1) there exists A nd b B such tht f(, b) = 0 2) f is continuously differentible on A B 3) the Fréchet derivtive d b f : R m R m is bijective where f : B R m is defined by f (y) = f(, y), then there exists n open neighborhood U of nd n open neighborhood V of b nd function ϕ : U V hving the following properties: i) ϕ() = b ii) f(x, ϕ(x)) = 0 x U iii) f is continuously differentible on U nd the following equlity holds: d x ϕ = (d y f x ) 1 (d x f y ) where f x (y) = f(x, y), f y (x) = f(x, y) nd y = ϕ(x). Note tht the function ϕ defined implicitly cnnot be lwys written s n explicit formul. Exmple ) Find the function defined implicitly by the eqution x 2 + y 2 = 1. b) Show tht the eqution y 5 + y x = 0 defines implicitly function y = y(x). 120

121 55 Higher order prtil differentibility Let f : A R n R m be function prtilly differentible with respect to every vrible x j, j = 1, n on the open set A. Definition If the prtil derivtives x f i x j re prtilly differentible t A with respect to every vrible x k, it is sid tht f is twice prtilly differentible t with respect to every vrible. The prtil derivtive with respect to the vrible x k of the prtil derivtive f i will be 2 f i denoted by x k x j (), i.e. second-order of f. x k ( f i x j )() = x j 2 f i x k x j () nd will be clled prtil derivtive of For sclr component f i there exists n 2 prtil derivtives of second-order. Exmple Consider f(x, y) = x 2 + y 2 + e x cos y. The first order prtil derivtives of f exist t every point (x, y) R 2 nd re given by f x = 2x + ex cos y f y = 2y ex sin y The first-order prtil derivtives of f themself re prtilly differentible t ny (x, y) R 2 with respect to x nd y nd we hve: x ( f x ) = 2 f x 2 = 2 + ex cos y y ( f x ) = 2 f y x = ex sin y x ( f y ) = 2 f x y = ex sin y y ( f y ) = 2 f y = 2 2 ex cos y The prtil derivtives re the second-order prtil derivtives of f. Definition In generl, f is sid k-times prtilly differentible t A with respect to every vrible if f is (k 1)-times prtilly differentible with respect to every vrible on n open neighborhood of nd every (k 1)-th order prtil derivtive is prtilly differentible with respect to every vrible x jk t. We denote k 1 f i k f i ( )() = () x jk x jk 1 x j1 x jk x jk 1 x j1 nd we cll it k-th order prtil derivtive of f. k 1 f i x jk 1 x j1 Exmple Find the k-th order prtil derivtive of the function f(x, y) = x 2 + y 2 + e x cos y. Definition If the prtil derivtives of first order x f i x j point A, it is sid tht f is twice differentible t. re differentible t Definition The second Fréchet derivtive of f t is denoted by d 2 f nd defined s function d 2 f : R n R n R m by the formul ( m n ) n d 2 2 f i f(u)(v) = () u j v k e i x j x k i=1 j=1 k=1 u, v R n, e i = (0,..., 0, 1, 0,..., 0), i = 1, n. 121

122 The second Fréchet derivtive d 2 f is system of m biliner forms in u 1, u 2,..., u n ; v 1, v 2,..., v n. The second Fréchet derivtive of f t stisfies lim u 0 1 u d +uf(v) d f(v) d 2 f(u)(v) = 0 for every v R n. In other words, the polynomil d +u f(v) cn be pproximted by the polynomil [d f + d 2 f(u)](v). Theorem 55.1 (mixed derivtive theorem of Schwrz). If the function f is twice differentible t, then the following reltions hold: 2 f i x j x k () = 2 f i x k x j () i = 1, m j, k = 1, n Proof. The proof of this theorem is rther technicl nd will be skipped. Exmple Consider f(x, y, z) = (xy + xz + yz, xyz) nd verify tht: 2 f x y = 2 f y x 2 f x z = 2 f z x 2 f y z = 2 f z y Theorem 55.2 (Criterion for second-order differentibility). If the prtil derivtives of second-order 2 f i x j x k exist in neighborhood of nd they re continuous t, then f is twice differentible t. Definition If the prtil derivtives of order k 1, A it is sid tht f is k-times differentible t. k 1 f i x jk 1 x j1 re differentible t The Fréchet derivtive of order k of f t is defined s the function d k f : R n R n R m given by ( m n ) n n d k f(u 1 )(u 2 ) (u k k f i ) = () u 1 j x jk x 1 u 2 j 2 u k j k e i j1 i=1 j 1 =1 j 2 =1 j k =1 The Fréchet derivtive of order k of f t stisfies: lim u k 0 1 u k dk 1 +u k f(u 1 )(u 2 ) (u k 1 ) d k 1 f(u 1 )(u 2 ) (u k 1 ) d k f(u 1 )(u 2 ) (u k ) = 0 If the function is k-times differentible t, then the following reltions hold: k f i x j1 x j2 x jk () = k f i () x σ(j1 ) x σ(j2 ) x σ(jk ) Theorem 55.3 (Criterion for k-times differentibility). If the prtil derivtives of k- th order exist in neighborhood of nd they re continuous t, then f is k-times differentible t. 122

123 56 Tylor s theorems Theorem 56.1 (Tylor s formule with integrl reminder). If the prtil derivtives of order m + 1 of the function f : A R n R p re continuous nd the closed segment [x, x + h] is included in the open set A, then: f(x + h) = f(x) + 1 1! d xf(h) + 1 2! d2 xf(h)(h) m! dm x f + 1 m! 1 0 (1 t) m d m+1 {}}{ (h) (h) dt x+th f m+1 m {}}{ (h) (h) + Proof. The function g(t) = f(x + th) is considered for t [0, 1]. For g the following reltions hold d k k g dt = {}}{ k g(k) (t) = d k x+thf (h) (h) k = 1, m + 1 On the other hnd d dt [g(t) + 1 t g (t) + + 1! (1 t)m g m (t)] = m! (1 t)m g m+1 (t) m! Hence g(1) [g(0) + 1 1! g (0) m! gm (0)] = 1 m! 1 f(x + h) = f(x) + 1 1! d xf(h) + 1 2! d2 xf(h)(h) m! dm x f 0 (1 t) m g m+1 (t) dt m {}}{ (h) (h) m! 1 0 (1 t) m d m+1 {}}{ (h) (h) dt x+th f m+1 Theorem 56.2 (Tylor s formul with the Lgrnge reminder). If the function f : A R n R p is m + 1 times differentible on A nd d m+1 y f M on the closed segment [x, x + h] which is included in A, then: f(x + h) f(x) 1 1! d xf(h) 1 m! dm x f m {}}{ (h) (h) M h m+1 (m + 1)! Proof. The proof of this theorem is rther technicl nd it will be skipped. Theorem 56.3 (Tylor s formule with O( h m ) reminder). If the function f is (m 1)- times differentible on A nd m-times differentible t x A, then: f(x + h) f(x) 1 1! d xf(h) 1 m! dm x f m {}}{ (h) (h) = O( h m ) Proof. By induction. 123

124 57 Clssifiction theorem for locl extrem Theorem If f : A R n R 1 hs continuous prtil derivtives of second-order on the open set A nd d f = 0 for n A, then: i) if f hs locl minimum t, then d 2 f(h)(h) 0 ii) if f hs locl mximum t, then d 2 f(h)(h) 0 Proof. Consider the Tylor formule f( + h) = f() + 1 1! d f(h) + 1 2! d2 f(h)(h) + O( h 2 ) nd hence f( + h) f() = 1 2! d2 f(h)(h) + O( h 2 ) i) if is locl minimum for f, then there exists r > 0 such tht for h < r we hve f( + h) f() = 1 2! d2 f(h)(h) + O( h 2 ) 0 Let be h R n, h 0 nd t R 1 such tht t < We hve r. h 1 2! d2 f(th)(th) + O( th 2 ) 0 t 2 [ 1 2! d2 f(h)(h) + h 2 O( th 2 ) th 2 ] 0 d 2 f(h)(h) + h 2 O( th 2 ) th 2 0 It follows tht d 2 f(h)(h) 0. ii) The second sttement is proved similrly. Theorem 57.2 (Sufficient condition for locl extrem). Assume tht f : A R n R 1 hs continuous prtil derivtives of second-order on A nd d f = 0 for A. i) If d 2 f(h)(h) 0 for h R n nd det( 2 f x i x j ()) 0, then f hs locl minimum t ; ii) If d 2 f(h)(h) 0 for h R n nd det( 2 f x i x j ()) 0, then f hs locl mximum t. Proof. The proof is rther technicl nd will be skipped. Exmple Find nd clssify the sttionry points of f(x, y) = x 4 y 4 2(x 2 y 2 ) Exmple Determine the vlues of k for which possesses minimum t (0, 0). f(x, y) = k(e y 1) sin x cos x cos 2y

125 58 Conditionl extrem Let us consider function f : A R n R 1, A open set nd Γ A, defined by: Γ = {x A g i (x) = 0 i = 1, p} where g i : A R 1, p < n. It will be ssumed tht f nd g i, i = 1, p hve continuous first order prtil derivtives on A. Definition If the restriction f / Γ hs n extremum t Γ, then we cll this extremum conditionl (by the equtions g i (x) = 0 i = 1, p). Theorem If d g 1,, d g p re linerly independent nd f hs conditionl extremum t, then there exist p constnts λ 1,, λ p such tht or f x k () = d f = p i=1 p λ i d g i i=1 λ i g i x k () k = 1, n Proof. The proof is rther technicl nd it will be skipped. Exmple Find the conditionl extrem of the following functions: ) f(x, y) = x 3 if x 2 + 6xy + y 2 = 1 b) f(x, y) = xy if 2x + 3y = 1 c) f(x, y, z) = x 2 + y 2 + z 2 if x + y + z = 1 59 Jordn mesurble subsets of R 2 Let be the set of one dimensionl bounded intervls of the form (, b), [, b), (, b], [, b], where, b R. The crtesin product = I 1 I 2 of two intervls of this type will be clled rectngle in R 2. The re of such rectngle is defined by re( ) = length(i 1 )length(i 2 ) Consider the set P of ll finite reunions of rectngles, i. e. P P if nd only if there exists 1, 2,, n such tht n P = Proposition If P 1, P 2 P, then P 1 P 2 P nd P 1 \ P 2 P. i=1 i 125

126 Proof. Direct verifiction. Proposition For ny P P there exists 1, 2,, n such tht P = i j = if i j. Proof. The proof is rther technicl nd will be skipped. Definition For P P the re is defined by n i nd i=1 re(p ) = n re( i ) i=1 where P = n i nd 1, 2,, n re disjoint. i=1 Proposition The re defined in this wy for P P stisfies: 1. re(p ) > 0 for P P 2. if P 1, P 2 P nd P 1 P 2 =, then re(p 1 P 2 ) = re(p 1 ) + re(p 2 ) 3. it is independent on the decomposition of P in finite union of disjoint intervls. Definition For A R 2, A bounded we define re i (A) = sup re(p ) re e (A) = inf re(p ) P A,P P P A,P P Definition A bounded set A R 2 is sid Jordn mesurble if re i (A) = re e (A) Definition If the bounded set A R 2 is Jordn mesurble, then the re of A is defined s re(a) = re i (A) = re e (A) Proposition A bounded set A R 2 is Jordn mesurble if nd only if for ny ε > 0 there exist P ε, Q ε P such tht P ε A Q ε nd re(q ε ) re(p ε ) < ε. Proposition A bounded set A R 2 is Jordn mesurble if nd only if there exist two sequences (P n ), (Q n ), P n, Q n P nd P n A Q n such tht In this cse we hve lim re(p n) = lim re(q n ) n n re(a) = lim n re(p n ) = lim n re(q n ) Proposition A bounded set A R 2 is Jordn mesurble if nd only if the re of its boundry is equl to zero. 126

127 Proposition If A 1 nd A 2 re Jordn mesurble sets, then A 1 A 2 nd A 1 \ A 2 re Jordn mesurble nd if A 1 A 2 =, then re(a 1 A 2 ) = re(a 1 ) + re(a 2 ). Proposition Let M R 2 be bounded set. If for ny ε > 0 there exist two Jordn mesurble sets A nd B such tht A M B nd re(b) re(a) < ε, then the set M is Jordn mesurble. Proposition If there exist two sequences (A n ), (B n ) of Jordn mesurble sets such tht A n M B n nd then M is Jordn mesurble nd lim re(a n) = lim re(b n ) n n re(m) = lim n re(a n ) = lim n re(b n ) Proof. The proof of the bove sttements is rther technicl nd it will be skipped. 60 The Riemnn-Drboux integrl of functions of two vribles Let A be given bounded nd Jordn mesurble subset of R 2. Definition A prtition P of A is finite set of subsets A i, i = 1, n of A stisfying: n A i = A, every A i is Jordn mesurble, if i j, then A i A j =. i=1 The dimeter of the set A i is the number d(a i ) defined by d(a i ) = The norm of the prtition P is the number mx (x,y ),(x,y ) A i (x x ) 2 + (y y ) 2 ν(p ) = mx{d(a 1 ), d(a 2 ),, d(a n )} Suppose now tht f is function defined nd bounded on A, f : A R 1. Then f is bounded on ech prt A i. Hence f hs lest upper bound M i nd gretest lower bound m i on A i. Definition The upper Drboux sum of f relted to P is defined by n U f (P ) = M i re(a i ) where M i = sup{f(x, y) (x, y) A i }. The lower Drboux sum of f relted to P is defined by n L f (P ) = m i re(a i ) where m i = inf{f(x, y) (x, y) A i }. i=1 i=1 127

128 Definition The Riemnn sum of f relted to P is defined by σ f (P ) = n f(ξ i, η i ) re(a i ) i=1 where (ξ i, η i ) A i. Remrk It is obvious tht we hve L f (P ) σ f (P ) U f (P ) Now f is bounded bove nd below on A. m f(x, y) M for ll (x, y) A. Thus for ny prtition P of A we hve m re(a) = m Hence the set n re(a i ) L f (P ) U f (P ) M i=1 is bounded bove nd the set So there exist numbers m nd M with n re(a i ) = M re(a) i=1 L f = {L f (P ) P is prtition of A} U f = {U f (P ) P is prtition of A} is bounded below. So L f = sup L f nd U f = inf U f exist. P P The firs result estblishes the intuitively obvious fct tht for bounded function L f U f. Theorem If f is defined nd bounded on A, then L f U f. Proof. Let P be prtition of A nd P the prtition P = P {A i, A i } where A i A i = A i for one prticulr i, 1 i n. In other words, P is obtined by decomposing A i in two mesurble subsets. It is now shown tht L f (P ) L f (P ) nd U f (P ) U f (P ). Let M i = sup{f(x, y) (x, y) A i} nd M i = sup{f(x, y) (x, y) A i }. Clerly M i M i nd M i M i. Hence U f (P ) = = i 1 M j re(a j ) + M i re(a i) + M i re(a i ) + j=1 i 1 M j re(a j ) + M i re(a i) + M i re(a i ) + j=1 n M j re(a j ) = U f (P ) j=1 n j=i+1 n j=i+1 M j re(a j ) M j re(a j ) = In similr fshion it cn be shown tht L f (P ) L f (P ). It now follows tht if P = P {A i 1, A i 1,, A i m, A i m }, then L f (P ) L f (P ) nd U f (P ) U f (P ). 128

129 Now suppose tht P 1 nd P 2 re two prtitions of A, P 1 = {A 1,, A m } nd P 2 = {B 1,, B n } nd let P 3 the prtition P 3 = {A i B j i = 1, m, j = 1, n}. Thus L f (P 1 ) L f (P 3 ) nd U f (P 2 ) U f (P 3 ). Since L f (P 3 ) U f (P 3 ) it cn be deduced tht L f (P 1 ) U f (P 2 ). In other words the lower sum relted to given prtition of A does not exceed the upper sum relted to ny prtition of A. Hence every lower sum is lower bound for the set of upper sums. So L f (P ) U f for ll possible prtitions P. But then U f is n upper bound for the set of lower sums. Thus L f U f. Definition A function f defined nd bounded on A is Riemnn-Drboux integrble on A if L f = U f. This common vlue is denoted by f(x, y) dx dy nd it is clled the double integrl of f. A Theorem The function f defined on A nd bounded on A is Riemnn-Drboux integrble on A if nd only if for ny ε > 0 there exists P such tht U f (P ) L f (P ) < ε. Proof. It follows from Theorem 60.1 Theorem The function f defined on A nd bounded on A is Riemnn integrble on A if nd only if there exists number I(= L f = U f ) such tht for ny ε > 0 there exists δ > 0 such tht for ν(p ) < δ we hve σ(p ) I < ε. Proof. It follows from Remrk 60.1 nd Theorem Remrk The constnt function f(x, y) = 1 is Riemnn integrble on A nd f(x, y) dx dy = re(a) Remrk This definition of the integrl A f(x, y) dx dy is only one of the mny wys to define the integrl of two vribles function. There re others, notbly Lebesque integrl; ll, however, give the sme thing in the cse of continuous functions. A 61 Integrble functions Theorem If f is continuous on A nd A is Jordn mesurble, then f is Riemnn- Drboux integrble on A. Proof. Since A is compct the function f is uniformly continuous. For ε > 0 there exists δ > 0 such tht [(x x ) 2 + (y y ) 2 ] 1 2 < δ f(x, y ) f(x, y ε ) < re(a) 129

130 Choose P such tht for i = 1, n. Hence (x, y ), (x, y ) A i (x x ) 2 + (y y ) 2 < δ U f (P ) L f (P ) = n (M i m i )re(a i ) < i=1 ε re(a) n re(a i ) i=1 Hence f is Riemnn-Drboux integrble on A. Definition A function f is clled piecewise-continuous on A if there exists prtition P = {A 1,, A n } of A nd continuous functions f i, i = 1, n defined on A i such tht f(x) = f i (x) for x Int(A i ). Theorem A piecewise-continuous function is Riemnn-Drboux integrble nd A f(x, y) dx dy = n i=1 A i f i (x, y) dx dy Proof. The proof is rther technicl nd it will be skipped. 62 Properties of the Riemnn-Drboux integrl Theorem If f nd g re Riemnn-Drboux integrble on A, then ll the integrls below exist nd the following reltions hold: (1) [αf(x, y) + βg(x, y)] dx dy = α f(x, y) dx dy + β g(x, y) dx dy, α, β R 1 A A f(x, y) dx dy + (2) f(x, y) dx dy = f(x, y) dx dy where A 1 A 2 = A nd A A 1 A 2 A 1 A 2 = (3) if f(x, y) g(x, y) on A, then f(x, y) dx dy g(x, y) dx dy A (4) f(x, y) dx dy A f(x, y) dx dy A A A Property (1) is clled the linerity of the integrl nd (2) is clled the dditive property. Proof. Is mde using the definition of the Riemnn-Drboux integrl. 130

131 Theorem 62.2 (The men vlue theorem). Let f : A R 1 be integrble on A nd stisfying m f(x, y) M for (x, y) A Then m re(a) f(x, y) dx dy M re(a) A Proof. Use property (3) from Theorem Riemnn-Drboux integrl clculus when A is rectngulr We intend to show tht in some conditions the clculus of the integrl of two vribles function reduces to the itertive clculus of the integrls of one vrible functions. Assume tht A is given by A = [, b] [c, d] nd f : A R 1. Theorem If the function f is integrble on A nd if for every x [, b] (x-fixed) the function f x (y) = f(x, y) is integrble on [c, d] i.e. the integrl I(x) = d f x (y) dy = d f(x, y) dy c c exists, then the itertive integrl d dx c f(x, y) dy exists too nd the following equlity holds: A f(x, y) dx dy = d dx c f(x, y) dy Proof. Consider the prtitions P x = { = x 0 < x 1 < < x i < < x n = b} of [, b] nd P y = {c = y 0 < y 1 < < y j < < y m = d} of [c, d]. Hence P = {A ij } i=0,n 1,j=0,m 1 is prtition of A, A ij = [x i, x i+1 ) [y j, y j+1 ). Denote by m ij = inf{f(x, y) (x, y) A ij } nd M ij = sup{f(x, y) (x, y) A ij } For (x, y) A ij we hve m ij f(x, y) M ij i = 0, n 1, j = 0, m 1 131

132 Hence m ij [y j+1 y j ] y j+1 y j f(x, y) dy M ij [y j+1 y j ] for i = 0, n 1, j = 0, m 1 nd x [x i, x i+1 ]. Therefore y j+1 m ij [y j+1 y j ] inf { f(x, y) dy x [x i, x i+1 ]} x nd hence sup x y j { y j+1 y j M ij [y j+1 y j ] f(x, y) dy x [x i, x i+1 ]} m ij [y j+1 y j ][x i+1 x i ] [x i+1 x i ] inf x { [x i+1 x i ] y j+1 y j [x i+1 x i ] sup x y j+1 y j y j+1 y j f(x, y) dy f(x, y) dy x [x i, x i+1 ]} f(x, y) dy { y j+1 y j M ij [y j+1 y j ][x i+1 x i ] f(x, y) dy x [x i, x i+1 ]} Hence where n 1 L f (P ) L I(x) (P x ) = [x i+1 x i ] L I(x) (P x ) = U I(x) (P x ) = i=0 n 1 L f (P ) = m 1 j=0 m 1 j=0 n 1 i=0 n 1 i=0 U f (P ) = i=0 m 1 j=0 y i+1 [x i+1 x i ] inf x { [x i+1 x i ] sup x m 1 j=0 y i f(x, y) dy U I(x) (P x ) U f (P ) m ij [x i+1 x i ][y j+1 y j ] { y j+1 y j f(x, y) dy x [x i, x i+1 ]} y j+1 y j f(x, y) dy x [x i, x i+1 ]} n 1 M ij [x i+1 x i ][y j+1 y j ] i=0 132

133 Since we hve sup L f (P ) = inf U f (P ) = sup L I(x) (P x ) = inf U I(x) (P x ) = A f(x, y) dx dy d dx f(x, y) dy nd f(x, y) dx dy = d dx c f(x, y) dy Chnging x with y we obtin lso A c A f(x, y) dx dy = d c dy f(x, y) dx Exmple Consider A = [1, 3] [1, 2] nd f(x, y) = 1. Evlute (x+y) 2 Exmple Evlute f(x, y) dx dy in the following cses: A f(x, y) dx dy. A ) A = [1, 3] [2, 5] nd f(x, y) = 5x 2 y 2y 3 b) A = [0, 1] [0, 1] nd f(x, y) = x2 1+y 2 c) A = [0, 1] [0, 1] nd f(x, y) = y (1+x 2 +y 2 ) 3 2 Theorem If A = [, b] [c, d] nd f : A R 1 is continuous, then the function F (x, y) = x y f(u, v) du dv = f(u, v) du dv [,x] [b,y] hs continuous prtil derivtives of first order: c F y x = c f(x, v)dv F x y = f(u, y)du The second order prtil derivtive 2 F x y 2 F x y exists nd = f(x, y) for (x, y) A 133

134 Proof. Represent F (x, y) in the following form x y x F (x, y) = f(u, v) du dv = y du f(u, v) dv = y x dv f(u, v) du c c c Theorem If there exists Φ : A R 1 such tht 2 Φ x y = f(x, y) then A f(x, y) dx dy = Φ(, c) Φ(b, c) + Φ(b, d) Φ(, d) Proof. Consider the prtitions P x = { = x 0 < x 1 < < x i < < x m = b} nd P y = {c = y 0 < y 1 < < y j < < y n = d} of [, b] nd [c, d] respectively. Now let be P = {A ij } i=0,m 1,j=0,n 1 prtition of A where A ij = [x i, x i+1 ] [y j, y j+1 ] nd pply twice the men vlue theorem for the expression obtining Φ(x i+1, y j+1 ) Φ(x i+1, y j ) Φ(x i, y j+1 ) + Φ(x i, y j ) Φ(x i+1, y j+1 ) Φ(x i+1, y j ) Φ(x i, y j+1 ) + Φ(x i, y j ) = = 2 Φ x y (ξ ij, η i,j )(x i+1 x i )(y j+1 y j ) = f(ξ ij, η i,j )(x i+1 x i )(y j+1 y j ) where x i ξ ij x i+1 nd y j η ij y j+1. Hence m 1 i=0 n 1 f(ξ ij, η i,j )(x i+1 x i )(y j+1 y j ) = Φ(b, d) Φ(b, c) Φ(, d) + Φ(, c) j=0 nd the equlity f(x, y) dx dy = Φ(, c) Φ(b, c) + Φ(b, d) Φ(, d) A 64 Riemnn-Drboux integrl clculus when A is not rectngle Let A the set defined by A = {((x, y) x [, b] nd y [g(x), h(x)]} where g, h re continuous functions stisfying g(x) h(x) for every x [, b]. 134

135 Theorem If the function f : A R 1 is integrble on A nd for every x [, b] the integrl h(x) I(x) = f(x, y) dy exists, then the itertive integrl g(x) dx h(x) f(x, y) dy g(x) exists too nd the following equlity holds: f(x, y) dx dy = dx h(x) f(x, y) dy A g(x) Proof. This cse cn be reduced to the cse when A is rectngle. Exmple Compute y 2 R 2 x 2 dx dy where A = {(x, y) x 2 + y 2 R 2 }. Exmple Compute A (x 2 + y) dx dy where A = {(x, y) y 2 x 0 nd x 2 y 0}. A Let A, B Jordn mesurble sets. Theorem If T : B A is bijection such tht T nd T 1 hve continuous prtil derivtives, then x x ξ η re(a) = dx dy = dξ dη where T (ξ, η) = (x(ξ, η), y(ξ, η)) for every (ξ, η) B. A Proof. Consider prtition P B = {B 1, B 2,, B n } of B nd the coresponding prtition P A = {A 1, A 2,, A n } of A with A i = T (B i ). If (P B ) is smll, then x x ξ η re(a i ) = re(b i) Hence n re(a i ) = i=1 y ξ n i=1 Considering sequence of prtitions PB n result. B y η x ξ y ξ x η y η y ξ y η re(b i) with ν(pn B ) n 135 0, we obtin the stted

136 Theorem If A, B R 2 re Jordn mesurble sets, T : B A is bijection such tht T nd T 1 hve continuous prtil derivtives nd f : A R 1 is n integrble function, then the following equlity holds: x x A f(x, y) dx dy = B f(x(ξ, η), y(ξ, η)) ξ y ξ η y η dξ dη Proof. Similrly s before. Exmple Evlute y 2 R 2 x 2 dx dy where A = {(x, y) x 2 + y 2 R 2 } by n pproprite chnge of vribles. A 65 Jordn mesurble subsets of R n Let be the set of one dimensionl bounded intervls of the form (, b), [, b), (, b], [, b], where, b R. The crtesin product = I 1 I n, where I i re intervls of this type, is clled hypercube in R n. The volume of such hypercube is defined by vol( ) = length(i 1 )length(i 2 ) length(i n ) Consider the set P of the finite unions of hypercubes, i. e. P P if nd only if exist 1, 2,, k such tht k P = It is esy to verify tht the following sttement hold: l=1 l P 1, P 2 P P 1 P 2 P nd P 1 \ P 2 P Proposition for ny P P there exist 1, 2,, k such tht P = p q = if p q. Definition For P P the volume is defined s k l nd l=1 vol(p ) = k vol( l ) l=1 k where P = l nd 1, 2,..., k re given by Proposition l=1 Proposition The volume defined bove for P P stisfies: 1. vol(p ) 0 for P P 136

137 2. P 1, P 2 P, P 1 P 2 = vol(p 1 P 2 ) = vol(p 1 ) + vol(p 2 ) 3. vol(p ) is independent on the decomposition of P. Definition For A R n, A bounded we define vol i (A) = sup vol(p ) vol e (A) = inf vol(p ) P A,P P P A,P P Definition A bounded set A R n is sid Jordn mesurble if vol i (A) = vol e (A) Definition If the bounded set A R n is Jordn mesurble, then the volume of A is defined s vol(a) = vol i (A) = vol e (A) Proposition A bounded set A R n is Jordn mesurble if nd only if for ny ε > 0 there exist P ε, Q ε P such tht P ε A Q ε nd vol(q ε ) vol(p ε ) < ε. Proposition A bounded set A R n is Jordn mesurble if nd only if for ny ε > 0 there exist two sequences (P k ), (Q k ), P k, Q k P nd P k A Q k such tht In this cse we hve lim vol(p k) = lim vol(q k ) k k vol(a) = lim k vol(p k ) = lim k vol(q k ) Proposition A bounded set A R n is Jordn mesurble if nd only if the volume of its boundry is equl to zero. Proposition If A 1 nd A 2 re Jordn mesurble sets, then A 1 A 2 nd A 1 \ A 2 re Jordn mesurble nd if A 1 A 2 =, then vol(a 1 A 2 ) = vol(a 1 ) + vol(a 2 ). Proposition Let M R n be bounded set. If for ny ε > 0 there exists two Jordn mesurble sets A nd B such tht A M B nd vol(b) vol(a) < ε, then the set M is Jordn mesurble. Proposition If there exist two sequences (A k ) nd (B k ) of Jordn mesurble sets such tht A k M B k nd then M is Jordn mesurble nd lim vol(a k) = lim vol(b k ) k k vol(m) = lim k vol(a k ) = lim k vol(b k ) The proof of the bove sttements is rther technicl nd it will be omitted. 137

138 66 The Riemnn-Drboux integrl of n vrible function Let A be bounded nd Jordn mesurble subset of R n. Definition A prtition of A is finite set of subsets A l, l = 1, k of A hving the following properties: 1) every A l is Jordn mesurble 2) k A l = A l=1 3) if p q then A p A q = The dimeter of the set A l is the number The norm of the prtition P is the number d(a l ) = sup x,y A l x y ν(p ) = mx{d(a 1 ),, d(a k )} Suppose now tht f is rel vlued n vribles function defined nd bounded on A, f : A R 1. Then f is bounded on ech prt A l, l = 1, k. Hence f hs lest upper bound M l nd gretest lower bound m l on A l (l = 1, k). Definition The upper sum of f relted to P is defined by U f (P ) = k M l vol(a l ) l=1 where M l = sup{f(x) x A l }. The lower sum of f relted to P is defined by L f (P ) = k m l vol(a l ) l=1 where m l = inf{f(x) x A l }. The Riemnn sum of f relted to P is defined by σ f (P ) = k f(ξ l ) vol(a l ) l=1 where ξ l A l. 138

139 Remrk It is cler tht the following inequlities hold: L f (P ) σ f (P ) U f (P ) Now f is bounded bove nd below on A. So there exist numbers m nd M such tht m f(x) M for x A Thus for ny prtition P of A we hve m vol(a) m k vol(a l ) U f (P ) σ f (P ) L f (P ) M l=1 k vol(a l ) = M vol(a) l=1 Hence, the set is bounded bove nd the set L f = {L f (P ) P is prtition of A} U f = {U f (P ) P is prtition of A} is bounded below. So L f = sup L f nd U f = inf U f exist. The first result estblishes the intuitively obvious fct tht for bounded function L f U f. Theorem If f is defined nd bounded on A, then L f U f. Proof. The sme s in two dimensions. Definition A function f defined nd bounded on A is Riemnn-Drboux integrble on A if L f = U f. This common vlue is denoted by f(x) dx or f(x 1,, x n ) dx 1 dx n A A nd is clled the Riemnn-Drboux integrl of f. Theorem The function f defined nd bounded on A is Riemnn-Drboux integrble on A if nd only if for ny ε > 0 there exists P such tht we hve Proof. The sme s in two dimensions. U f (P ) L f (P ) < ε Remrk The constnt function f(x) = 1 for x A is Riemnn integrble on A nd 1 dx 1 dx n = vol(a) A 139

140 67 Integrble functions of n vribles Theorem If f is continuous on A nd A is Jordn mesurble, then f is Riemnn- Drboux integrble on A. Proof. Since A is compct the function f is uniformly continuous. For ε > 0 there exists δ > 0 such tht x x < δ f(x ) f(x ε ) < vol(a) Choose P such tht x, x A i x x < δ for i = 1, n. Hence n ε U f (P ) L f (P ) = (M i m i ) vol(a i ) < vol(a) i=1 Hence f is Riemnn integrble on A. n vol(a i ) Definition A function f is clled piecewise-continuous on A if there exists prtition P = {A 1,, A k } of A nd continuous functions f i, i = 1, k defined on A i such tht f(x) = f i (x) for x Int(A i ). Theorem A piecewise-continuous function is Riemnn integrble nd A f(x) dx = Proof. Is technicl nd it will be omitted. k i=1 A i f i (x) dx i=1 68 Properties of the Riemnn-Drboux integrl of n- vrible functions Theorem If f nd g re Riemnn-Drboux integrble on A R n then ll the integrls below exist nd the reltions hold: (1) [αf(x) + βg(x)] dx = α f(x) dx + β g(x) dx, α, β R 1 A (2) f(x) dx = f(x) dx + A A 1 A 2 (3) if f(x) g(x) on A, then A A f(x) dx where A 1 A 2 = A nd A 1 A 2 = f(x) dx g(x) dx (4) f(x) dx f(x) dx A A A A 140

141 Property (1) is clled the linerity of the integrl nd (2) is clled the dditive property. Proof. It is proved using the definition of the Riemnn-Drboux integrl. Theorem 68.2 (The men vlue theorem). Let f : A R 1 stisfying m f(x) M for x A Then m vol(a) f(x) dx M vol(a) be integrble on A nd A Proof. Use property (3) from Theorem Riemnn-Drboux integrl clculus for n-vrible functions when A is hypercube We intend to show tht in some conditions the clculus of the integrl of n vribles function reduces to the itertive clculus of the integrls of one vrible functions. Assume tht A is hypercube A = [ 1, b 1 ] [ 2, b 2 ] [ n, b n ] nd f : A R 1. Theorem If the function f is integrble on A nd if for every fixed x 1 [ 1, b 1 ] the function f x1 (x 2,, x n ) = f(x 1, x 2,, x n ) is integrble on A 1 = [ 2, b 2 ] [ n, b n ] i.e. the integrl I(x 1 ) = A 1 f x1 (x 2,, x n ) dx 2 dx n = 2 2 n n f(x 1, x 2,, x n ) dx 2 dx n exists, then the itertive integrl n I(x 1 ) dx 1 = dx exists too nd the following equlity holds: n f(x 1, x 2,, x n ) dx 2 dx n A n f(x) dx = I(x 1 ) dx 1 = dx n f(x 1, x 2,, x n ) dx 2 dx n Proof. Similr s for the two vribles functions. dxdydz Exmple Evlute for the set A bounded by the plnes: (1 + x + y + z) 3 x = 0, y = 0, z = 0 nd x + y + z = 1. A 141

142 Exmple Evlute Exmple Evlute z 2 c 2 1. A A z dx dy dz for the set A defined by x2 + y2 + z b 2 c 2 ( x2 + y2 2 b + z2 x2 ) dx dy dz for the set A defined by 2 c2 + y2 + 2 b 2 Remrk The bove theorem reduces successively the evlution of the integrl to the evlution of integrls for functions of one vrible. 1 1 n n f(x 1,, x n ) dx 1 dx 2 dx n = 1 1 dx dx 2 n n f(x 1,, x n ) dx n Remrk Theorem 69.1 is vlid lso for more complex sets A s in 2-dimensionl cse. Theorem If A, B R n re Jordn mesurble nd T : B A is bijection, T nd T 1 hving continuous prtil derivtives, then vol(a) = 1 dx = dx 1 dx n = D(x 1,, x n ) D(ξ 1,, ξ n ) dξ 1 dξ n A A where T (ξ 1,, ξ n ) = (x 1 (ξ 1,, ξ n ),, x n (ξ 1,, ξ n )). Proof. As in 2-dimensionl cse. Theorem If A, B R n re Jordn mesurble sets, T : B A is bijection, T nd T 1 hving continuous prtil derivtives nd f : A R 1 is n integrble function, then the following equlity holds: where D(x 1,,x n ) D(ξ 1,,ξ n ) Proof. As in two dimensions. A f(x) dx = B B f(x(ξ)) D(x 1,, x n ) D(ξ 1,, ξ n ) is the determinnt of the Jcobi mtrix of T. dξ Exmple Compute the volume of the set A R 3 bounded by x 2 + y 2 + z 2 R 2. xyz Exmple Compute the integrl dx dy dz where A is bounded bove x 2 + y2 by the surfce (x 2 + y 2 + z 2 ) 2 = 2 xy nd below by the surfce z = 0. A 142

143 70 Elementry curves nd elementry closed curves The wy of defining line integrl is quite similr to the fmilir wy of defining definite integrl known from clculus. In order to do this, we must introduce the concepts of curve nd rc length. We will present these concepts in prticulr frmework which cn be extended in nturl wy. Definition An elementry curve (elementry rc) is set of points C R 3 for which there exists closed intervl [, b] R nd function ϕ : [, b] C hving the following properties: ) ϕ is bijective; b) ϕ is of clss C 1 nd ϕ (t) 0, t [, b]. The points A = ϕ() nd B = ϕ(b) re clled the end points of the curve. The function ϕ is clled prmetric representtion of the curve nd the vector ϕ (t) is tngent to the curve t the point ϕ(t). Figure 70.1: Definition An elementry closed curve is set of points C R 3 for which there exists closed intervl [, b] R nd function ϕ : [, b] C with the following properties: ) ϕ is bijective from [, b) to C nd ϕ() = ϕ(b) ; b) ϕ is of clss C 1 nd ϕ (t) 0, t [, b]. The function ϕ is clled prmetric representtion of the curve nd the vector ϕ (t) is tngent to the curve t the point ϕ(t). Exmple If x 0 = (x 0 1, x 0 2, x 0 3) nd h = (h 1, h 2, h 3 ) the closed segment [x 0, x 0 + h] which joins the points x 0, x 0 + h is n elementry curve. In this cse, we cn tke [, b] = [0, 1] nd ϕ : [, b] C is given by ϕ(t) = (x th 1, x th 2, x th 3 ). Exmple The circle C defined by x x 2 2 = 1 nd x 3 = 0 is n elementry closed curve C. In this cse we cn tke [, b] = [0, 2π] nd ϕ : [, b] C is given by ϕ(t) = (cos t, sin t, 0). 143

144 Figure 70.2: Remrk Any elementry or elementry closed curve possesses n infinity of prmetric representtions. Remrk The end points A, B of n elementry curve re independent of the prmetric representtion of the curve. This mens tht for every prmetric representtion ψ : [c, d] C of the curve we hve {ψ(c), ψ(d)} = {ϕ(), ϕ(b)} = {A, B}. Using prmetric representtion of n elementry curve or of n elementry closed curve we cn define the curve length. Definition The length of the elementry curve (elementry closed curve) C is given by: l = ϕ (t) dt = ϕ 2 1 (t) + ϕ 2 2 (t) + ϕ 2 3 (t) dt where ϕ(t) = (x 1 (t), x 2 (t), x 3 (t)) is prmetric representtion of C nd ϕ i (t) = dϕ i (t), dt i = 1, 2, 3. Remrk The curve length is independent of the prmetric representtion of the curve C. In other words, if ψ : [c, d] C is second prmetric representtion of C, then d c ψ (τ) dτ = ϕ (t) dt Exmple The length of the closed segment [x 0, x 0 +h] represented by ϕ(t) = x 0 +th, t [0, 1] is 1 l = h h h 2 3 dt = h 0 nd the length of the circle C = {(x 1, x 2, x 3 ) R 3 x x 2 2 = 1, x 3 = 0} represented by ϕ(t) = (cos t, sin t, 0), t [0, 2π] is l = 2π 0 sin 2 t + cos 2 t dt = 2π 144

145 Exmple 70.4 (Shows tht the curve length is independent of the prmetric representtion). For the circle C = {(x 1, x 2, x 3 ) R 3 x x 2 2 = 1, x 3 = 0} the prmetric representtion ϕ : [0, π] C, ϕ(t) = (cos 2t, sin 2t, 0) is chosen. Using this representtion we hve π l = 2 sin 2 2t + cos 2 2t dt = 2π 0 This is the sme vlue s the one obtined in the cse of the prmetric representtion ϕ(t) = (cos t, sin t, 0), t [0, 2π]. Remrk Since ϕ (t) 0, t [, b], n elementry curve C hs only two end points. In other words, if C is n elementry curve then there exist two points A, B C such tht for ny prmetric representtion ψ : [c, d] C of the curve C we hve {ψ(c), ψ(d)} = {A, B}. When ψ(c) = A nd ψ(d) = B, then if τ moves from c to d, then ψ(τ) moves from A to B. When ψ(c) = B nd ψ(d) = A, then if τ moves from c to d, then ψ(τ) moved from B to A. The two wys to cover the curve C, from A to B or from B to A, re clled orienttions of C nd the bove presented fcts show tht on n elementry curve there re two orienttions. Moreover, the covering given by n rbitrry representtion of C is one of the bove mentioned orienttions. In other words, the set of representtions is divided in two clsses: for ll the representtions which belong to one of the clsses we hve one orienttion (sy from A to B) nd for the other clss the opposite orienttion (from B to A). Figure 70.3: Consider now n elementry curve C with the prmetric representtion ϕ : [, b] C for which the orienttion of the curve is from A to B (ϕ() = A, ϕ(b) = B). If in the formul l = ϕ (τ) dτ we replce the fixed upper limit b with vrible upper limit t, the integrl becomes t s A (t) = ϕ (τ) dτ 145

146 The vlue s A (t) represents the rc length AA C, where A = ϕ(t). The function s A is defined on the closed intervl [, b] nd it is bijection from [, b] to [0, l], s A () = 0 nd s A (b) = l. More, s A nd s 1 A re continuously differentible. The function s A cn be used in order to define new prmetric representtion of the curve C, nmely: x A : [0, l] C, x A = ϕ s 1 A. In this representtion of C, s [0, l] serves s prmeter nd x A (0) = A, x A (l) = B. When s moves from 0 to l, then x A (s) moves from A to B. The prmetric representtion x A is cnonic when the orienttion of C is from A to B, i.e. the prmeter s is the rc length A x A (s) nd there is no other representtion with this property. Consider now for the sme elementry curve C with the end points A nd B prmetric representtion ψ : [c, d] C for which the orienttion is from B to A (ψ(c) = B nd ψ(d) = A). If in the formul l = d c ψ (τ) dτ we replce the fixed upper limit d with vrible upper limit t, the integrl becomes s B (t) = t c ψ (τ) dτ The vlue s B (t) represents the rc length BB C, where B = ψ(t). The function s B : [c, d] [0, l] is bijection nd s B (c) = 0, s B (d) = l. The function s 1 B cn be used in order to define new prmetric representtion x B = ψ s 1 B of C, such tht x B (0) = B, x B (l) = A. If s moves from 0 to l, then x B (s) moves from B to A. The representtion x B is cnonic if the orienttion of C is from B to A, i.e. the prmeter s is the rc length B x B (s) nd there is no other representtion with this property. Exmple In the cse of the closed intervl [x 0, x 0 +h] which joins the points A = x 0 nd B = x 0 + h, if the prmetric representtion ϕ(t) = x 0 + th, t [0, 1] is chosen, then ϕ(t) moves from A to B when t moves from 0 to 1. If the prmetric representtion ψ(τ) = x 0 + (2 τ)h, τ [1, 2] is chosen, then ψ(τ) moves from B = x 0 + h to A = x 0 when τ moves from 1 to 2. Using the representtion ϕ, the rc length AA is given by s A (t) = t 0 h h h 2 3 dτ = t h with s A (0) = 0 nd s A (1) = h. The function s 1 A : [0, h ] [0, 1] is s 1 A (s) = representtion x A : [0, h ] C is given by s, nd the cnonic prmetric h x A (s) = (ϕ s 1 A )(s) = x0 + 1 h s h. 146

147 Using the represention ψ, the rc length BB is given by s B (t) = t 1 ψ (τ) dτ = t 1 h h h 2 3 dτ = (t 1) h with s B (1) = 0 nd s B (2) = h. The function s 1 s B : [0, h ] [1, 2] is given by s 1 B (s) = 1+, nd the cnonic prmetric h representtion x B : [0, h ] C is x B (s) = (ψ s 1 B )(s) = x0 + (2 1 s h ) h = x0 + (1 s h ) h The representtion x A (s) = x s h is cnonic in the orienttion from A to B, nd h the representtion x B (s) = x 0 + (1 s ) h is cnonic in the orienttion from B to A. h In both cses, the prmeter s moves from 0 to l. Now consider n elementry closed curve C. In this cse, there ren t two end points A nd B, nd we cnnot spek bout the orienttion from A to B or from B to A. In the followings, we will show how to proceed in order to introduce two orienttions in the cse of n elementry closed curve C. For the elementry closed curve C let s consider prmetric representtion ϕ : [, b] C nd the point A = ϕ() C. Since ϕ (t) 0, if t moves from to b then ϕ(t) describes the curve, moving in unique wy. This movement of ϕ(t) is one orienttion of the curve. The opposite movement on C is the opposite orienttion. If ψ : [c, d] C is n rbitrry representtion of C then when t increses on [c, d], ψ(t) moves on C ccording to one of the orienttions: d ( if ϕ ψ 1 (τ) ) = dt > 0 then ϕ(t) nd ψ(τ) move in the sme wy when t moves dτ dτ from to b nd τ moves from c to d; d ( if ϕ ψ 1 (τ) ) = dt < 0 then ϕ(t) nd ψ(τ) move in opposite senses s t moves from dτ dτ to b nd τ moves from c to d. As in the cse of n elementry curve C we cn consider the function s A : [, b] [0, l] defined by s A (t) = t ϕ (τ) dτ If A = ϕ(t) then s A (t) is the rc length AA described by ϕ(τ) when τ moves from to t. The function s A : [, b] [0, l] is bijection nd cn be used in order to define new representtion of C: x : [0, l] C, x A ϕ s 1 A. In this representtion, s [0, l] is the prmeter nd x A (0) = x A (l) = A. When s moves from 0 to l then x A (s) moves on C nd describes the curve C moving in the sme sense s ϕ(t). The function x A : [0, l] C defined by x A (s) = x A (l s) is the representtion of C which corresponds to the opposite orienttion. For instnce, in the cse of the circle: C = {(x 1, x 2, x 3 ) x x 2 2 = 1, x 3 = 0} 147

148 Figure 70.4: considering the prmetric representtion ϕ(t) = (cos t, sin t, 0) with t [0, 2π] we hve: ϕ(0) = (1, 0, 0) = A, s A (t) = t 0 dτ = t; x A (s) = (cos s, sin s, 0), s [0, π]; x A (0) = (1, 0, 0) = A; x A (s) moves on the circle s in Fig. 70.4, nd x A (s) = (cos(2π s), sin(2π s), 0) moves on the circle s in Fig. 70.5: Figure 70.5: It follows tht n elementry curve nd n elementry closed curve cn be represented s: X(s) = (X 1 (s), X 2 (s), X 3 (s)), 0 s l where l is the curve length nd s [0, l] is the rc length X(0)X(s) C. For n elementry curve C, there exist two representtions of this kind corresponding to the two orienttions of C. For n elementry closed curve C if we fix point A on C, then we lso hve two representtions of this kind corresponding to the two orienttions of C. 71 Line integrl of first type Let now n elementry curve C nd one of its prmetric representtions x(s) = (x 1 (s), x 2 (s), x 3 (s)) 0 s l in function of the rc length s. In order to mke choice ssume tht x(s) moves from A nd B when s moves from 0 to l. Let now f(x 1, x 2, x 3 ) be given function which is defined (t lest) t ech point of C nd is continuous function of s, i.e. s f(x 1 (s), x 2 (s), x 3 (s)) is continuous. We subdivide C into n portions in n rbitrry mnner: 148

149 Figure 71.1: Let P 0 (= A), P 1, P 2,..., P n 1, P n (= B) the end points of these portions nd let s 0 (= 0) < s 1 < s 2 < < s n (= l) the lengths of rcs AQ i ; s i = length(aq i ). Then we choose n rbitrry point on ech portion, sy, point Q 1 between P 0 nd P 1, point Q 2 between P 1 nd P 2 etc. Tking the vlues of f t these points Q 1, Q 2,, Q n we form the sum I n = n f(q m )(s m s m 1 ) m=1 We do this for n = 2, 3, in completely independent mnner, so tht the gretest s m = s m s m 1 pproches zero s n pproches infinity. We obtin sequence of rel numbers I 2, I 3,. The limit of this sequence is clled the line integrl of first type of f long C from A to B nd is denoted by f ds C The curve C is clled the pth of integrtion. Since, by ssumption, f is continuous nd C is smooth, tht limit exists nd is independent of the orienttion nd of the choice of subdivisions nd points Q m. In fct, the position of point P on C is determined by the corresponding vlue of the rc length s; since A nd B correspond to s = 0 nd s = l respectively, we hve l f ds = f(x 1 (s), x 2 (s), x 3 (s)) ds C 0 It is esy to see tht if C is represented by the continuously differentible vector function ϕ : [, b] R 3, ϕ = ϕ(t) then C f ds = f(ϕ 1 (t), ϕ 2 (t), ϕ 3 (t)) ϕ 2 1(t) + ϕ 2 2(t) + ϕ 2 3(t) dt Hence the line integrl of first type is equl to the definite integrl nd fmilir properties of ordinry definite integrls re eqully vlid for line integrls. Proposition ) k f ds = k f ds k = constnt C C 149

150 b) (f + g) ds = f ds + g ds c) C C C f ds = f ds + f ds, where the pth C is subdivided into two disjoint rcs C 1 C C 1 C 2 nd C 2 Figure 71.2: Remrk If C is n elementry closed curve, the line integrl of first type is defined similrly. For line integrl over closed pth C, the symbol (insted of ) is sometimes used in the literture. C C Exmple Evlute the following line integrls: 1) 2) C xy ds, where c is the segment which joins the points A(0, 0) nd B(1, 1); (x + y) ds, where c is the closed curve given by the prmetric representtion C x = cos t, y = sin t, t [0, 2π]. 72 Line integrls of second type In mny pplictions the integrnds ppering in the line integrls of first type re of the form: C f dx 1 ds or g dx 2 ds or h dx 3 ds where dx 1, dx 2, dx 3 re the derivtives of the functions occurring in the prmetric ds ds ds representtion of the pth of integrtion. The integrls f dx 1 ds ds, g dx 2 ds ds, h dx 3 ds re clled line integrls of second type. ds C C Their vlues depend on the orienttion of C; chnging the orienttion of C the integrls re multiplied by

151 We simply denote these integrnds by: f dx 1 ds ds = C C C g dx 2 ds ds = h dx 3 ds ds = C C C f dx 1 g dx 2 h dx 3 In terms of the considered prmetric representtions, these line integrls of second type re equl to the following Riemnn integrls: C fdx 1 = l 0 f(x 1 (s), x 2 (s), x 3 (s)) dx l 1 ds ds = 0 f(x 1 (s), x 2 (s), x 3 (s)) cos α(s)ds C C gdx 2 = hdx 3 = l 0 l 0 g(x 1 (s), x 2 (s), x 3 (s)) dx l 2 ds ds = h(x 1 (s), x 2 (s), x 3 (s)) dx l 3 ds ds = 0 0 g(x 1 (s), x 2 (s), x 3 (s)) cos β(s)ds h(x 1 (s), x 2 (s), x 3 (s)) cos γ(s)ds where α(s), β(s), γ(s) re the ngles of the tngent to the curve nd the coordinte xis Ox 1, Ox 2, Ox 3, respectivelly. In terms of n rbitrry prmetric representtion ϕ : [, b] C which corresponds to the sme orienttion, these line integrls of second type re equl to the following Reimnn integrls: fdx 1 = f(ϕ(t)) ϕ 1 (t) ϕ 2 1 (t) + ϕ 2 2(t) + ϕ 2 3(t) dt C C gdx 2 = hdx 3 = g(ϕ(t)) h(ϕ(t)) ϕ 2 (t) ϕ 2 1 (t) + ϕ 2 2(t) + ϕ 2 3(t) dt ϕ 3 (t) ϕ 2 1 (t) + ϕ 2 2(t) + ϕ 2 3(t) dt C All these integrls depend on the orienttion of the curve C. If the orienttion chnges, the vlue of the integrl chnges its sign. For the sums of these types of integrls long the sme pth C we dopt the simplified nottion f dx 1 + g dx 2 + h dx 3 C 151

152 which is equl to the Riemnn integrl l [f(x 1 (s), x 2 (s), x 3 (s)) cos α(s) + g(x 1 (s), x 2 (s), x 3 (s)) cos β(s) + h(x 1 (s), x 2 (s), x 3 (s)) cos γ(s)] ds 0 Exmple Evlute the line integrl I = [x 2 1x 2 dx 1 + (x 1 x 3 ) dx 2 + x 1 x 2 x 3 dx 3 ] C where C is the rc of the prbol x 2 = x 2 1 in the plne x 3 = 2 from A(0, 0, 2) to B(1, 1, 2). Exmple Evlute the bove line integrl where C is the segment of the stright line x 2 = x 1, x 3 = 2 from A(0, 0, 2) to B(1, 1, 2). 73 Trnsformtion of double integrls into line integrls Double integrls over plne region my be trnsformed into line integrls over the boundry of the region nd conversely. This trnsformtion is of prcticl s well s theoreticl interest nd cn be done mens of the following bsic theorem. Theorem 73.1 (Green s theorem in the plne). Let R be closed bounded region in the x, y plne whose boundry C consists of finite mny elementry closed curves. Let f(x, y) nd g(x, y) be functions which re continuous nd hve continuous prtil derivtives of first order everywhere in some domin contining R. Then the following equlity holds: ( g x f ) dx dy = f dx + g dy = [f cos α + g cos β] ds y R C The integrtion being tken long the entire boundry C of R such tht R is on the left s one moves on C. C Figure 73.1: 152

153 Proof. We first prove the theorem for specil region R which be represented in both of the forms: R = {(x, y) x b, u(x) y v(x)} nd R = {(x, y) c y d, p(y) x q(y)} Figure 73.2: Figure 73.3: In this cse we hve R f dx dy = y ( v(x) u(x) = f dy) dx = y f(x, u(x)) dx [f(x, v(x)) f(x, u(x))] dx f(x, v(x)) dx = f(x, y) dy f(x, y) dx = f(x, y) dx b C C C since y = u(x) represents the oriented curve C nd y = v(x) represents the oriented curve C. If portions of C re segments prllel to the y-xis such s C nd C, 153

154 Figure 73.4: then the result is the sme s before, becuse the integrls over these portions re zero nd my be dded to the integrls over C nd C to obtin the integrl over the whole boundry C. Similrly we obtin R g d dx dy = x c q(y) p(y) dx dy = C g(x, y) dy Therefore [ g x f ] dx dy = f dx + g dy y R C nd the theorem is proved for specil regions. We now prove the theorem for region R which itself is not specil region but cn be subdivided into finitely mny specil regions. In this cse we pply the theorem to ech subregion nd then dd the results; the left-hnd members dd up to the integrl over R while the right-hnd members dd up to the line integrl over C plus integrls over the curves introduced for subdividing R. Ech of the ltter integrls occur twice, tken once in ech direction. Hence these two integrls cncel ech other, nd we re left with the line integrl over C. Exmple Using Green s theorem, evlute the following integrls: ) y dx + 2x dy, where C is the boundry of the squre 0leqx 1, 0leqx 1 C (counterclockwise); b) y 3 dx + (x 3 + 3y 2 x) dy, where C is the boundry of the region y = x 2 nd y = x, C where 0leqx 1 (counterclockwise); c) 2xy dx + (e x + x 2 ) dy, where C is the boundry of the tringle with vertices (0, 0), C (1, 0), (1, 1) (clockwise); d) C xy 2 dx + x 2 y dy, where c is the boundry of the region in the first qudrnt bounded by y = 1 x 2 (counterclockwise). 154

155 Now we will present second theorem of Green which concern the trnsformtion of double integrl of the Lplcin of function into line integrl of its norml derivtive. Let w(x, y) be function which hs continuous second order prtil derivtives in domin D of the x, y-plne. Definition The Lplcin of w is by definition the function w : D R 1 defined by w = 2 w x + 2 w 2 y 2 Assume now tht D contins region R (R D) of the type indicted in Green s theorem. Theorem The following equlity holds: w w dx dy = n ds = R C C n w ds = C grdw n ds where n is the outwrd unit norml vector to C. Proof. Consider f = w w g nd g = nd remrk tht w = y x theorem in the plne, we obtin w dx dy = w w dx + y x dy = R C C f x y ( w y dx ds + w x dy ds ) ds =. Applying Green s C grdw n ds The integrnd of the lst integrl my be written s the dot product of the vectors tht is grd w = ( w x, w ) nd n = (dy y ds, dx ds ) n grd w = w x dy ds w y dx ds The vector n is the outwrd unit norml vector to C. tht is becuse the vector τ = ( dx, dy ) is the unit tngent vector to C nd τ n = 0. ds ds The dot product n grd w is the directionl derivtive w = n nw. Therefore we hve w w dx dy = n ds = n w ds R C C Let v(x, y) be vector vlued function v(x, y) = (f(x, y), g(x, y)) which hve continuous first order prtil derivtives in domin D of the x, y-plne. Definition The divergence of v is by definition the rel vlued function div v : D R 1 defined by div v = f x + g y 155

156 Theorem If D contins region R (R D) of the type indicted in Green s theorem, then the following equlity holds: div v dx dy = v n ds where n is the outwrd unit norml vector to c. R C Proof. Applying 73.1 we obtin div v dx dy = ( f x + g ) dx dy = y R R C g dx + f dy = C v n ds Exmple Verify this formul when v = (x, y) nd C is the circle x 2 + y 2 = Elementry Surfces We shll consider surfce integrls. This considertions will require knowledge of some bsic fcts bout surfces, which we shll now explin nd illustrte by simple exmples. Definition An elementry surfce is set of points S R 3 for which there exists bounded, open nd connected set D R 2 nd function ϕ : D S hving the following properties: ) ϕ is bijective; b) ϕ is of clss C 1 nd the vector ϕ u ϕ v = N = ( ϕ 2 u ϕ 3 v ϕ 3 u ϕ 2 v, ϕ 3 u ϕ 1 v ϕ 1 u ϕ 3 v, ϕ 1 u ϕ 2 v ϕ 2 u ϕ 1 v ) is different from 0 for ny (u, v) D. The function ϕ is clled prmetric representtion of S. The vectors ϕ u, ϕ v to the surfce S t the point ϕ(u, v). re tngents The vector N ϕ is clled the norml vector to the surfce S t the point ϕ(u, v) nd the vector n ϕ = N ϕ is the unit norml vector to the surfce S t the point ϕ(u, v). N ϕ Exmple The set S = {(x 1, x 2, 1) R 3 0 < x 1 < 1, 0 < x 2 < 1} is n elementry surfce. A bounded, open nd connected set D R 2 nd function ϕ : D S hving the properties ) nd b) re: D = {(u, v) R 2 0 < u < 1, 0 < v < 1} nd ϕ(u, v) = (u, v, 1) A norml vector to S t the point ϕ(u, v) is N ϕ = (0, 0, 1) which is ctully unit norml vector. The vectors ϕ ϕ = (1, 0, 0) nd = (0, 1, 0) re tngents to S t the point u v ϕ(u, v). 156

157 Exmple The set S = {(x 1, x 2, x 3 ) R 3 x x 2 2 = 1, x 1 > 0, x 2 > 0, 0 < x 3 < 1} is n elementry surfce. A bounded, open nd connected set D R 2 nd function ϕ : D S hving the properties ) nd b) re: D = {(u, v) R 2 0 < u < π 2, 0 < v < 1} nd ϕ(u, v) = (cos u, sin u, v) A norml vector to S t the point ϕ(u, v) is N ϕ = (cos u, sin u, 0) which is ctully unit norml vector. The vectors ϕ ϕ = ( sin u, cos u, 0) nd = (0, 0, 1) re tngents to S u v t the point ϕ(u, v). Exmple The set S = {(x 1, x 2, x 3 ) R 3 x 2 1 +x 2 2 +x 2 3 = 1, x 1 > 0, x 2 > 0, x 3 > 0} is n elementry surfce. A bounded, open nd connected set D R 2 nd function ϕ : D S hving the properties ) nd b) re: D = {(u, v) R 2 0 < u < π 2, 0 < v < π } nd ϕ(u, v) = (cos u sin v, sin u sin v, cos v) 2 The vector N ϕ = sin v (cos u sin v, sin u sin v, cos v) is norml vector to S nd n ϕ = (cos u sin v, sin u sin v, cos v) is unit norml vector to S. The vectors ϕ ϕ = ( sin u sin v, cos u sin v, 0) nd = (cos u cos v, sin u cos v, sin v) re tngents u v to S t the point ϕ(u, v). Remrk An elementry surfce possesses n infinity of prmetric representtions. The direction of N ϕ is independent of the prmetric representtion, but the orienttion of the norml vector N ϕ depends on the prmetric representtion of S. If insted of the prmetric representtion x = ϕ(u, v), (u, v) D we consider the prmetric representtion x = ψ(u, v ), where x = ψ(u, v ) = ϕ( u, v ), (u, v ) D nd T : D D is defined by T (u, v) = ( u, v), then the orienttion of N ϕ chnges; N ϕ = N ψ. In generl, if x = ϕ(u, v), (u, v) D nd x = ψ(u, v ), (u, v ) D re two prmetric representtions of S, then: - if the determinnt of the Jcobi mtrix of T = ψ 1 ϕ, T : D D, T (u, v) = (u, v ) is positive, then the orienttion of the norml vector to S does not chnge; - if the determinnt of the Jcobi mtrix of T = ψ 1 ϕ, T : D D, T (u, v) = (u, v ) is negtive, then the orienttion of the norml vector to S chnges; Remrk If n elementry surfce S is given by the prmetric representtion ϕ : D R 2 S, x = ϕ(u, v), nd C is n elementry curve on S (C S) then C = ϕ 1 (C) is n elementry curve in D. If u = g(t) nd v = h(t), t [, b] is prmetric representtion of C then prmetric representtion of C is obtined s x = ϕ(g(t), h(t)), t [, b]. Exmple Consider the elementry surfce S = {(x 1, x 2, 1) R 3 0 < x 1 < 1, 0 < x 2 < 1} 157

158 with the prmetric representtion ϕ(u, v) = (u, v, 1), (u, v) (0, 1) (0, 1), nd the elementry curve C on S (C S) defined by C = {(x 1, x 2, x 3 ) R 3 x 1 = x 2, x 3 = 1, x 1 [ 1 3, 2 3 ]} The curve C = ϕ 1 (C), C D hs the prmetric representtion: u = t, v = t, t [ 1, 2 ]. The prmetric representtion x = ϕ(g(t), h(t)) of C in this cse is given by 3 3 x(t) = (t, t, 1). Exmple In the cse of the elementry surfce S = {(x 1, x 2, x 3 ) R 3 x x 2 2 = 1, x 1 > 0, x 2 > 0, 0 < x 3 < 1} with the prmetric representtion ϕ(u, v) = (cos u, sin u, v), (u, v) D = {(u, v) R 2 0 < u < π, 0 < v < 1}, nd the elementry curve on S is defined by 2 C = {(x 1, x 2, x 3 ) R 3 x x 2 2 = 1, x 3 = 1 2, 1 2 < x 1 < 3 2, 3 2 < x 2 < 1 2 } The curve C = ϕ 1 (C), C D hs the prmetric representtion: u = t, v = 1 [ 2, π t 3, 2π ]. The prmetric representtion x = ϕ(g(t), h(t)) of the curve C in this cse 3( is x(t) = cos t, sin t, 1 ). 2 Remrk Using the formul of prmetric representtion of the elementry curve C S, x = ϕ(u(t), v(t)), we obtin tht the tngent vector t to C t ϕ(u(t), v(t)) is given by: ( dx dt = ϕ1 u du dt + ϕ 1 v dv dt, ϕ 2 u du dt + ϕ 2 v dv dt, ϕ 3 u du dt + ϕ 3 v dv ) dt The vectors: ( ϕ u = ϕ1 u, ϕ 2 u, ϕ ) 3 u nd ϕ ( v = ϕ1 v, ϕ 2 v, ϕ ) 3 v re tngent vectors to S t ϕ(u, v) nd dx dt = ϕ u du dt + ϕ v dv dt. If the elementry surfce S is given by the prmetric representtion x = ϕ(u, v), (u, v) D nd the elementry curve C on S (C S) is given by the prmetric representtion u = u(t), v = v(t), t [, b], then the length l of the curve C is given by: l = E u2 + 2F u v + G v 2 dt where: E = ϕ (u(t), v(t)) 2 F u G = ϕ (u(t), v(t)) 2 u v = ϕ u 158 ϕ (u(t), v(t)) (u(t), v(t)) v = du dt v = dv dt

159 The expression E u 2 + 2F u v + G v 2 is clled first fundmentl form of S. It is of bsic importnce becuse it enbles us to mesure lengths, ngles between curves nd res on the corresponding surfce. In fct, we hve lredy seen how we compute the length of curve. Now we consider the mesurement of the ngle between the curves C 1 : x = ϕ(g(t), h(t)) C 2 : x = ϕ(p(t), q(t)) Let be P S, P = ϕ(g(t 0 ), h(t 0 )) = ϕ(p(t 0 ), q(t 0 )) point of intersection of the two curves. The tngent vector t the point P to the curve C 1 is T 1 nd the tngent vector t the point P to the curve C 2 is T 2 : T 1 = ( ϕ 1 u ġ + ϕ 1 v ḣ, ϕ 2 u ġ + ϕ 2 v ḣ, ϕ 3 u ġ + ϕ 3 v ḣ) T 2 = ( ϕ 1 u ṗ + ϕ 1 v q, ϕ 2 u ṗ + ϕ 2 v q, ϕ 3 u ṗ + ϕ 3 v q) The ngle t the point of intersection P between C 1 nd C 2 is defined s the ngle γ between T 1 nd T 2 t P nd Since: cos γ = T 1 T 2 T 1 T 2 T 1 T 2 = E ġ ṗ + F (ġ q + ḣ ṗ) + G ḣ q T 1 = E ġ 2 + 2F ġ ḣ + G ḣ2 T 2 = E ṗ 2 + 2F ṗ q + G q 2 we obtin tht the ngle between to intersecting curves on surfce cn be expressed in terms of E, F, G nd the derivtives of the functions representing the curves, evluted t the point of intersection. We will show in the followings how to compute the res on surfce S. The re A of prt S of the elementry surfce S, represented by x = ϕ(u, v), (u, v) D, is given by: A = EG F 2 du dv where D is the prt of D corresponding to S. This formul cn be mde plusible by noting tht: D A = EG F 2 u v is the re of smll prllelogrm whose sides re the vectors x x u nd u v v From the definition of the vector product it follows tht: A = x x u v = x u v u x v u v = EG F 2 u v 159

160 The integrnd is obtined by subdividing S into prts S 1, S 2,..., S n nd pproximting the re of ech prt S k by the re of the prllelogrm from the tngent plne to S t point from S k nd forming the sum of ll the pproximting res. This is done for ech k = 1, 2,... so tht the dimensions of the lrgest S k pproches zero s n. The limit of these sums is the integrl. In vrious pplictions, surfce integrls occur for which the concept of orienttion of surfce is essentil. In the cse of n elementry surfce, for the unit norml vector n there exist two orienttions (see Fig. 74.1) nd we cn ssocite to ech of them one orienttion of the elementry surfce S (s for the elementry curves, using two wys to cover the curve). The set of representtions of the elementry surfce S is decomposed (ccording to these orienttions) in two disjoint clsses. For ll representtions belonging to one of these clsses, we hve one of the two orienttions of n (of S) nd for ll representtions from the other clss we hve to opposite orienttion of n (of S). Figure 74.1: If smooth surfce S is orientble, then we my orient S by choosing of the two possible directions of the unit norml vector n. If the boundry of the elementry surfce S is simple closed curve C, then we my ssocite with ech of two possible orienttions of S n orienttion of C s it is shown in Figs Figure 74.2: The rules is: looking the curve C from the top of the unit norml vector n the sense on the curve is lwys counterclockwise. Using this ide we my extend the concept of orienttion to surfces which cn be decomposed in elementry surfces, s follows: A surfce S which cn be decomposed in elementry surfces is sid to be orientble if we cn orient ech elementry piece of 160

161 S in such mnner tht long ech curve C which is common boundry of two pieces S 1 nd S 2 the positive direction of C relted to S 1 is opposite to the positive direction of C relted to S 2. Figure 74.3: However this my not hold in the lrge. There re non-orientble surfces. An exmple of such surfce is the Möbius strip. A model of Möbius strip cn be mde by tking long rectngulr piece of pper nd sticking the shorter sides together so tht the two points A nd the two points B coincide (see Fig.74.4). Figure 74.4: Möbius strip Definition A surfce S is orientble if chosen norml orienttion given t n rbitrry point P 0 S cn be continued in unique nd continuous wy to the entire surfce S. Hence, the surfce S is orientble if there does not exist closed curve C S pssing through P 0 such tht the chosen norml orienttion reverses by moving continuously long the curve C from P 0 nd bck to P Surfce integrls of first type Surfce integrls occur in mny pplictions, for exmple, in connection with the center of grvity of curved lmin, the potentil due to chrges distributed on surfces. Let S be n elementry surfce of finite re nd let f rel vlued function which is defined nd continuous on S. We subdivide S into n prts S 1, S 2,, S n of res A 1, A 2,, A n. In ech prt S k we choose n rbitrry point P k nd form the sum: n I n = f(p k ) A k k=1 161