MATHEMATICAL ANALYSIS I

Transcription

1 VŠB Technicl University of Ostrv Fculty of Electricl Engineering nd Computer Science MATHEMATICAL ANALYSIS I Jiří Bouchl Mrie Sdowská Ostrv 007

2 INSTEAD OF PREFACE It ws ll very well to sy Drink me, but the wise little Alice ws not going to do tht in hurry. No, I ll look first, she sid, nd see whether it s mrked poison or not ; for she hd red severl nice little histories bout children who hd got burnt, nd eten up by wild bests nd other unplesnt things, ll becuse they would not remember the simple rules their friends hd tught them: such s, tht red-hot poker will burn you if you hold it too long; nd tht if you cut your finger very deeply with knife, it usully bleeds; nd she hd never forgotten tht, if you drink much from bottle mrked poison, it is lmost certin to disgree with you, sooner or lter. However, this bottle ws not mrked poison, so Alice ventured to tste it, nd finding it very nice, (it hd, in fct, sort of mied flvour of cherry-trt, custrd, pine-pple, rost turkey, toffee, nd hot buttered tost,) she very soon finished it off. (Lewis Crroll, Alice s Adventures In Wonderlnd )

3 Contents A The Rel Number System; the Supremum Theorem Number Systems, Their Nottion nd Some Propositions Mimum, Minimum; Supremum; Infimum B Rel Functions of Single Rel Vrible Definition of Function Some Specil Properties of Functions Monotonic Functions Even nd Odd Functions Periodic Functions Injective Functions Bounded Functions Opertions with Functions Sum, Difference, Product, Quotient nd Composition of Functions Inverse Function Restriction of Function C Elementry Functions Bsic Elementry Functions Elementry Functions D Sequnces of Rel Numbers Limit of Sequnce Clculting Limits E Limit nd Continuity of Function Limit of Function Continuity of Function F Differentil nd Derivtive of Function Motivtion Differentil nd Derivtive; Clculting Derivtives G Bsic Theorems of Differentil Clculus Theorems on Function Increment l Hospitl s Rule H Function Behviour Intervls of Strict Monotonicity Etremes of Functions Locl Etremes Globl Etremes Conve nd Concve Functions Asymptotes of ( Grph of) Function Emintion of Function Behviour

4 I Approimtion of Function by Polynomil Tylor s Polynomil (Approimtion of Function in Neighbourhood of Point) Tylor s Polynomil Numericl Derivtive Interpoltion Polynomils (Approimtion of Function in n Intervl) J Antiderivtive (Indefinite Integrl) Antiderivtive Techniques of Integrtion (Methods of Clculting Antiderivtives) Integrtion of Rtionl Functions Prtil Frctions Decomposition of Rtionl Functions Integrtion of Prtil Frctions Integrtion of Some Other Specil Functions Integrls of the Type sin n cos m d Integrls of the Type R(sin, cos ) d Integrls of the Type ) R (, s +b d c+d 6.4 Integrls of the Type R(, + b + c) d K Riemnn s (Definite) Integrl Motivtion nd Inspirtion Definition of Definite Integrl Integrl with Vrible Upper Bound Methods of Clculting Definite Integrls Numericl Clcultion of the Riemnn Integrl Applictions of Definite Integrl Are of Plne Region Length of Plne Curve Volume of Rottionl Solid Are of Rottionl Surfce Improper Integrl References Inde

5 A THE REAL NUMBER SYSTEM; THE SUPREMUM THEOREM NUMBER SYSTEMS, THEIR NOTATION AND SOME PROPOSITIONS. N = {,, 3, 4, 5,...}... the set of ll nturl numbers. The following theorem is useful not only when thinking bout structure of the nturl numbers, but it is lso good instrument for proving mny mthemticl propositions.. Theorem (Principle of Mthemticl Induction). Let M be subset of N such tht i) M, ii) n M : n + M. Then M = N..3 Emple. Prove tht n N : n = n (n + ). PROOF. Let us denote M := {k N : k = k (k + ) }. The tsk is to prove tht M = N. According to Theorem. it is sufficient to show tht the premises i) nd ii) hold for M. The premise i) holds clerly since = ( + ). In order to prove ii) let us ssume tht n M, i.e., n = n (n + ). We shll show tht then n + M, i.e., n + (n + ) = (n + ) (n + + ). This is esy since from our ssumption it follows tht ( n) + (n + ) = n (n + ) + (n + ) = (n + ) (n + + ). Thus the premise ii) holds too. There is lso nother wy how to prove this proposition. Given n N, ( n) = n + + n + (n ) + (n ) = n (n + ). 5

6 .4 Z = {..., 3,,, 0,,, 3,...} = {n, n, 0}... the set of ll integer numbers. n N.5 Q = { p : p, q Z q 0}... the set of ll rtionl numbers. q There is lot of ides showing certin incompleteness of the rtionl number system, despite the fct tht between ech two - rbitrrily ner - distinct rtionl numbers there still lies n infinite number of them. For emple, i) ( ε > 0)( p, p Q) : ε < p < < p < + ε, ii) there is no rtionl number p such tht p =. Let us prove t lest the second proposition. PROOF. Conversely, we ssume tht there is rtionl number p such tht p =. Since p Q, there re integer coprime nonzero numbers m, n such tht p = m n. Thus we get m n = nd so m = n. This implies tht m hs to be even (note tht the squre of n even number is even nd the squre of n odd number is odd). Therefore there is k Z such tht m = k. By inserting this reltion into m = n, we obtin 4k = n. Hence it follows tht n = k. This implies tht lso n hs to be even which contrdicts our ssumption tht m nd n re coprime..6 R... the set of ll rel numbers. Let us recll tht we hve number of opertions defined in R (nd lso in N, Z nd Q): +,,, :,. An order of the rel numbers is their other essentil chrcteristic: for every two rel numbers, y ectly one of the following possibilities holds i) < y, ii) = y, iii) > y; for every three rel numbers, y, z it holds tht ( < y y < z) ( < z). Let us lso introduce the nottion R + = { R : > 0}... the set of ll positive rel numbers, R = { R : < 0}... the set of ll negtive rel numbers, R \ Q... the set of ll irrtionl numbers. 6

7 .7 R = R {+, }... the etended rel number system. Let us etend the order from R to R : R : < < +, < +. Let us lso define the following opertions in R : > : + (+ ) = + + = +, < + : + ( ) = + =, R + {+ } : (+ ) = + = +, ( ) = =, R { } : (+ ) = + =, ( ) = = +, R : = = 0, + = + = +. Insted of + (+ ) we usully write +. Similrly, insted of + ( ) we write. Also, we denote + briefly by..8 Remrk. N Z Q R R.9 Cution. We do not define: +, +, 0 (± ), (± ) (± ), 0 ( R ). MAXIMUM, MINIMUM; SUPREMUM, INFIMUM Alredy since secondry school we hve been used to deling with vrious sets of numbers. For instnce, let us consider the intervls (, ) nd (, ] nd number which in both intervls plys role of right bound : ny number lying to the right from it belongs neither to (, ) nor to (, ]. Also, every number from the set [, + ) {+ } hs this qulity (in while: to be n upper bound). However, number is the best since it is the smllest one. The fct tht in the first cse number does not belong to the given intervl nd in the second cse it does is n essence of the difference between the concepts supremum nd mimum. These concepts (together with other concepts defined below) re used for chrcteriztion of even much more complicted subsets of R.. Definitions. Let M R. A number k R is sid to be n upper bound of the set M if M : k. 7

8 A number l R is sid to be lower bound of the set M if. Definitions. Let M R. M : l. If n upper bound of M eists nd belongs to M, we cll it mimum of M nd denote it by m M. If lower bound of M eists nd belongs to M, we cll it minimum of M nd denote it by min M..3 Definitions. Let M R. A number s R is clled supremum of the set M if i) M : s (i.e., s is the upper bound of M), ii) ( k R, k < s) ( M) : > k (i.e., ny number smller thn s is not the upper bound of M). We write s = sup M. A number i R is clled n infimum of the set M if i) M : i (i.e., i is the lower bound of M), ii) ( l R, l > i) ( M) : < l (i.e., ny number lrger thn i is not the lower bound of M). We write i = inf M..4 Observtion. sup M is the lest upper bound of M nd inf M is the gretest lower bound of M..5 Emples. ) M = (, ]... min M does not eist, inf M =, sup M = m M =. ) M = R +... neither min M nor m M eists, inf M = 0, sup M = +. 3) M =... neither min M nor m M eists, inf M = +, sup M =. 4) M = { n : n N}... min M = inf M =, m M does not eist, sup M = 0..6 Definitions. Let M R. If sup M < +, we cll M bounded bove. If inf M >, we cll M bounded below. If M is bounded bove nd bounded below, we cll it bounded. 8

9 If M is not bounded, we cll it unbounded..7 Theorem (Supremum Theorem). Every subset of R hs ectly one supremum..8 Corollry. Every subset of R hs ectly one infimum..9 Eercises. ) Think over the reltion between m M nd sup M (min M nd inf M). ) Find out wht is the reltion between sup M nd inf( M), where M := { : M}, nd prove tht the eistence of infimum is relly consequence of Theorem.7. 3) Determine sup M nd inf M (nd lso m M nd min M, in cse they eist), if ) M = {q Q : q < 3}, b) M = { R : sin = }, c) M = { R : }. 4) Prove the proposition: A set M R is bounded k R + : M [ k, k]. 5) Prove tht inf { + : n N} = nd min { + : n N} fils to eist. n n 9

10 B REAL FUNCTIONS OF A SINGLE REAL VARIABLE 3 DEFINITION OF A FUNCTION 3. Definitions. We cll every mpping of R into R function (more precisely: rel function of single rel vrible). In other words, function f is prescription tht ssocites ech element D(f) R with ectly one vlue f() H(f) R (D(f)... the domin of f; H(f)... the rnge of f). If f is rel function of single rel vrible, we write f : R R. From now on we shll del only with functions whose domins re not empty. 3. Emples. ) f() := ; D(f) = [, ]... see Fig.. ) g() := ; D(g) = [, )... see Fig.. Cution: f g (f = g [D(f) = D(g) D(f) : f() = g()]) ) η() := { 0, < 0,, 0; Fig. Fig. D(η) = R, η... the so-clled Heviside function... see Fig. 3., < 0, 4) sgn() := 0, = 0,, > 0; D(sgn) = R, sgn... the so-clled sign function... see Fig. 4. 5) h() := ; D(h) = R ( Z : < + ),... the so-clled lower integer prt of number... see Fig. 5. 6) Id() := ; D(Id) = R, Id... the so-clled identity... see Fig. 6. 0

11 Fig. 3 Fig Fig. 5 Fig. 6 {, < 0, 7) l() := =, 0; D(l) = R,... the so-clled bsolute vlue of number... see Fig Fig. 7 { 0, Q, 8) χ() :=, R \ Q; D(χ) = R, χ... the so-clled Dirichlet function. 3.3 Definition. A grph of function f is defined by Grph f := { (, y) R R =: R : D(f) y = f() }. 3.4 Remrk nd convention. Now we know tht function is determined by its domin nd its prescription which ssocites ech element of the domin with ectly one vlue. We often determine function only by its prescription; in this cse the domin is set of ll rel numbers for which the prescription is meningful.

12 3.5 Emple. Let us determine the domin nd the grph of the function k() :=. SOLUTION. D(k) = { R : is defined } = { R : 0} = (, ]. Grph k = { (, y) R : (, ] y = } = = {(, y) R : (, ] y 0 y = }... see Fig Fig. 8 4 SOME SPECIAL PROPERTIES OF FUNCTIONS 4. Monotonic Functions 4.. Definitions. Let M R. A function f is sid to be incresing on the set M if, M : < f( ) < f( ), non-decresing on the set M if, M : < f( ) f( ), decresing on the set M if, M : < f( ) > f( ), non-incresing on the set M if, M : < f( ) f( ). A function is sid to be incresing (non-decresing, decresing, non-incresing) if it is incresing (non-decresing, decresing, non-incresing) on its domin. Incresing nd decresing functions re clled strictly monotonic, non-incresing nd non-decresing functions re clled monotonic.

13 4.. Emples. Let us consider the bove-mentioned functions. Then ) Id is incresing, ) η, sgn, h, Id re non-decresing, 3) k is decresing, 4) k is non-incresing Remrk. It is obvious tht every strictly monotonic function is monotonic. 4. Even nd Odd Functions 4.. Definitions. A function f is clled even if odd if D(f) : f( ) = f(), D(f) : f( ) = f(). 4.. Remrk. Note tht if f is even or odd, then D(f) : D(f) Emples. Let us consider the bove-mentioned functions gin. Then ) f, l, χ re even (g is not even!), ) sgn, Id re odd Observtion. Grph of n even function is symmetric to the line = 0. Grph of n odd function is symmetric to the origin. 4.3 Periodic Functions 4.3. Definitions. A function f is sid to be periodic if there eists T R + such tht D(f) : f() = f( + T ). We cll such T period of f Observtion. For ny periodic function f it holds: D(f) : + T D(f) Eercise. Prove the proposition: T R + Q χ is periodic with the period T. 3

14 4.4 Injective Functions 4.4. Definition. A function is sid to be injective if, D(f) : f( ) f( ) Emple. Functions Id nd k re injective. 4.5 Bounded Functions 4.5. Definitions. Let M D(f). A function f is sid to be bounded bove on the set M if set f(m) := {f() : M} is bounded bove. A function f is sid to be bounded bove if it is bounded bove on D(f). Below-bounded functions nd bounded functions re defined nlogously Emples. ) f, g, η, sgn re bounded, ) l, k re bounded below. 5 OPERATIONS WITH FUNCTIONS 5. Sum, Difference, Product, Quotient nd Composition of Functions 5.. Definitions. Let f nd g be functions. Then the functions f + g, f g, f g, re defined by the following prescriptions: (f + g) () := f() + g(), f g nd g f (f g) () := f() g(), (f g) () := f() g(), ( ) f () := f(), g g() (g f) () := g(f()). 5.. If we tke close look t the previous definitions, we note certin incorrectness there. For emple, in the reltion (f + g) () := f() + g() we use symbol + in two different menings. On the left side of this equlity it mens the opertion between two functions: the pir f nd g is ssocited with the function f +g; on the right side of the equlity symbol + mens the sum of two rel numbers f() nd g(). Similr incorrectness lso ppers in the definitions of the other opertions. This inccurcy is usul in mthemticl literture, but with little ttention we cnnot mke mistke. 4

15 5..3 Emples. ) Id = Id Id, ) k = f f, where f () = nd f () =, 3) = sgn() =. 5. Inverse Function 5.. Definition. Let f be function. A function f is clled n inverse of f if i) D(f ) = H(f), ii), y R : f () = y = f(y). 5.. Theorem (Eistence of n Inverse Function). Let f be function. Then f eists if nd only if f is injective. PROOF.? i) f eists f is injective Let us consider rbitrry, D(f) such tht f( ) = f( ) nd denote the vlue f( ) = f( ) by. This gives H(f) = D(f ). Hence it follows tht f () = nd f () =. We thus get =. ii) f is injective? f eists Let H(f). As f is injective there eists unique y D(f) such tht f(y ) =. Now we define function g on H(f) by the prescription g() := y. It is cler tht g = f Emple. Find the inverse, in cse it eists, of the function SOLUTION. i), D(v) = [, 0]: v() :=, D(v) = [, 0]... see Fig. 9. v( ) = v( ) = = = = = =. Hence v is injective nd thus v eists! ii) D(v ) = H(v) = [0, ]: v () = y = v(y) = y + y = y = y =. Therefore y = v () = nd we get v () :=, D(v ) = [0, ]... see Fig. 9. 5

16 v v Fig Observtions. Let f be n injective function. Then D(f ) : (f f )() =, D(f) : (f f)() =, (f ) = f, (, y) Grph f (y, ) Grph f (the grphs of f nd f re symmetric to the line y = ). 5.3 Restriction of Function 5.3. Definition. We sy tht function h is restriction of function f to set M (we write h = f M ) if the following conditions hold: i) M = D(h) D(f), ii) M : f() = h() Emples. ) g = f [, ), ) sgn R + = η R +. 6

17 C ELEMENTARY FUNCTIONS 6 BASIC ELEMENTARY FUNCTIONS Fig An eponentil function e (we shll denote it lso by ep())... see Fig. 0. This is undoubtedly the most importnt function in mthemtics. This is ectly how the wonderful Wlter Rudin s book Rel nd Comple Anlysis strts. Then it continues with n ect definition of the eponentil function nd with proof of its bsic properties. However, this wy is too difficult for us, in this moment we know too little. For illustrtion, let us note tht the eponentil function cn be defined using sum of series: ep() := + +! + 3 3! +... = + Since we do not know now wht the given sum of the series mens, we hve to mke do with wht we know bout the eponentil function from secondry school. In wht follows, we shll define other bsic elementry functions (ecept the goniometric functions) ectly. 6. A logrithmic function is defined s the inverse of the eponentil function, i.e., log := ep... see Fig.. n=0 n n!. 7

18 y Fig. 6.3 A constnt function is defined by f() := c (c R). In the cse c = 0 we spek bout zero function. For instnce: f() :=... see Fig...5 y Fig. 6.4 The power functions: power function with nturl eponent n N is given by For emple: f() :=... see Fig. 6, f() :=... see Fig. 3, f() := 3... see Fig. 4. f() = n := } {{ }. n times power function with negtive integer eponent n (n N) is given by f() = n := n =. For emple: f() := =... see Fig. 5, f() := =... see Fig. 6. 8

19 Fig. 3 Fig y y Fig. 5 Fig. 6 function nth root (n N, n > ) is defined by i) f() := ( n [0, + ) for every even n, ) ii) f() := ( n R ) for every odd n. We write f() = n. For emple: f() := =:... see Fig. 7, f() := 3... see Fig. 8. power function with rel eponent r R \ Z is defined by f() = r := e r log. For emple: f() := 5... see Fig. 9, f() := 5... see Fig. 0. 9

20 Fig. 7 Fig Fig. 9 Fig. 0 moreover, we define R : 0 :=. 6.5 Remrk. It cn be proved tht ( p, q Z, q ) ( R +) : p q = e p q log = q p. We cn further sk why we do not define for these p nd q when, moreover, p is even the function p p q by the prescription q := q p lso for negtive. The nswer is obvious. Such definition would not be correct since we could get 6.6 The goniometric functions: sin (sine)... see Fig., cos (cosine)... see Fig., = ( ) = ( ) = ( ) = =. tn := sin cos (tngent)... see Fig. 3, cot := cos sin (cotngent)... see Fig. 4. 0

21 Fig. Fig. We gin find ourselves in sitution when we work with the functions whose definitions re beyond our comprehension in this moment. Similrly s in the cse of the eponentil function, let us mention tht the functions sine nd cosine re defined by the following sums of the infinite series: sin := 3 3! + 5 5!... = + n=0 cos :=! !... = n=0 ( ) n n+ (n + )!, ( ) n n (n)!. 6 6 y 4 y Fig. 3 Fig Cution. Note tht the domin of every goniometric function is subset of R; we do not use the degrees t ll. In this contet it is good to recll tht equlities of the type 90 = π, 30 = π 6, etc., re meningless. 6.8 The cyclometric functions: ( ) rcsin := sin π [, π ] (rcsine)... see Fig. 5, rccos := ( cos [0, π] (rccosine)... see Fig. 6, ) ( ) rctn := tn π (, π ) (rctngent)... see Fig. 7, rccot := ( cot (0, π) (rccotngent)... see Fig. 8. )

22 Fig. 5 Fig Fig. 7 Fig The hyperbolic functions: sinh := e e (hyperbolic sine)... see Fig. 9, cosh := e +e (hyperbolic cosine)... see Fig. 30, tnh := sinh cosh = e e e +e (hyperbolic tngent)... see Fig. 3, coth := cosh sinh = e +e e e (hyperbolic cotngent)... see Fig The hyperbolometric functions: rg sinh := (sinh) (rgument of hyperbolic sine)... see Fig. 33, rg cosh := ( cosh [0, + ) (rgument of hyperbolic cosine)... see Fig. 34, ) rg tnh := (tnh) (rgument of hyperbolic tngent)... see Fig. 35, rg coth := (coth) (rgument of hyperbolic cotngent)... see Fig. 36.

23 y Fig. 9 Fig y Fig. 3 Fig Fig. 33 Fig ELEMENTARY FUNCTIONS 7. Definition. A function is sid to be elementry if it is formed by the bsic elementry functions using finite number of lgebric opertions (+,,, :) nd composition of functions. 3

24 4 3 y y Fig. 35 Fig Emples. ) The function f() = := e log ( R + ) is elementry. ( ) The inverse of ( R + \ {}) is elementry 3) sgn is not elementry. 4) = is elementry. 5) Every rel polynomil p, i.e., function given by log = log log ). p() := n n + n n ( i R, i = 0,,..., n), is elementry. 7.3 Eercises. Prove the following propositions: ) [, ] : rcsin + rccos = π, ) R : rctn + rccot = π, 3) R : cosh sinh =, 4) R : rg sinh = log ( + + ), 5) [, + ) : rg cosh = log ( + ), 6) (, ) : rg tnh = 7) R \ [, ] : rg coth = log +, log +, 8) u, v R : cosh(u + v) = cosh(u) cosh(v) + sinh(u) sinh(v), 9) u, v R : sinh(u + v) = sinh(u) cosh(v) + cosh(u) sinh(v). 4

25 D SEQUENCES OF REAL NUMBERS 8 LIMIT OF A SEQUENCE 8. Definitions. By sequence (more precisely: sequence of the rel numbers), we men function f whose domin equls to N. A sequence which ssocites every n N with number n R ( n... the so-clled nth term of the sequence ( n )) shll be denoted by one of the following wys:,, 3,... ; ( n ); { n } n=. 8. Cution. { n } n= { n : n N}... the rnge of sequence. 8.3 Emples. ) 3, 3, 3,... ; n := 3... constnt sequence, n N : n+ = n. ),, 3, 4, 5,... ; n := n... n rithmetic sequence, ( δ R)( n N) : n+ = n + δ. 3),, 4, 8, 6,... ; n := n... geometric sequence, ( q R)( n N) : n+ = q n. 4),, 3, 4, 5,... ; n := n... hrmonic sequence. 5),,, 3, 5, 8, 3,... ; = =, n N : n+ := n+ + n... Fiboncci sequence (defined recurrently). 6) 0,,,,, 3, 3, 0, 0, 7, 7,... = f(0), f(), f( ),..., f(n), f( n),..., where f() := ( )... see Fig Definitions. We sy tht sequence ( n ) hs limit R (we write lim n = or n ) if ( ε R + )( n 0 N)( n N, n > n 0 ) : n < ε. If sequence hs (finite) limit, we cll it convergent. In the opposite cse we cll the sequence divergent. 8.5 Emples. ) n :=

26 Fig. 37 PROOF. The proposition is obvious since ( ε R + )( n 0 N)( n N, n > n 0 ) : n 3 = 0 < ε. ) n := n 0. PROOF. We hve to prove tht i.e., ( ε R + )( n 0 N)( n N, n > n 0 ) : n 0 < ε, ( ε R + )( n 0 N)( n N, n > n 0 ) : n < ε. First of ll, let us note tht for n, ε > 0 we hve < ε < n. Now we fi ε R+ n ε nd choose n n 0 N such tht ε < n 0. This is certinly possible. For instnce, we cn consider n 0 = ε +. Then 8.6 Eercises. Prove tht ( n N, n > n 0 ) : n > n 0 > ε. ) the sequence { n } n= := {( )n } n= hs no limit; ) the sequence { n } n= := { n} n= is not convergent. 6

27 8.7 Theorem. Every convergent sequence is bounded. PROOF. The tsk is to show tht lim n = R k R + : { n : n N} [ k, k]. n ( ε R + )( n 0 N)( n N, n > n 0 ) : n < ε Let us tke such n n 0 nd put ( n 0 N)( n N, n > n 0 ) : n <. k = m{,,..., n0, + }. Clerly k R +. It remins to prove tht n N : n [ k, k]: n {,,..., n 0 } : n { n, n } [ k, k], ( n N, n > n 0 ) : n (, + ) [ k, k]. 8.8 Cution. Theorem 8.7 cnnot be reversed. More precisely, not every bounded sequence is convergent. For instnce, the sequence defined by n := ( ) n is bounded nd lim ( ) n fils to eist. 8.9 Definition. Let ( n ) be sequence. Then sequence { kn } n= = k, k,..., kn,..., where (k n ) is n incresing sequence of the nturl numbers, i.e., is clled subsequence of ( n ). n N : k n < k n+ k n N, 8.0 Emple. ( n ) =, 3, 3, 8,,,,..., ( kn ) =, 3,,,..., (k n ) =, 3, 6, 7, Theorem. Every bounded sequence contins convergent subsequence. 8. Definitions. A sequence ( n ) is sid to be Cuchy sequence if it stisfies the so-clled Bolzno-Cuchy criterion: ( ε R + )( n 0 N)( n, m N; n, m > n 0 ) : n m < ε. 7

28 8.3 Theorem. A sequence is convergent if nd only if it is Cuchy sequence. 8.4 Definitions. Let ( n ) be sequence. Then ( n ) hs limit + (we write lim n = + or n + ) if ( k R)( n 0 N)( n N, n > n 0 ) : n > k, ( n ) hs limit (we write lim n = or n ) if 8.5 Emples. ) n := n 3 +, ) n := n. ( l R)( n 0 N)( n N, n > n 0 ) : n < l. 8.6 Theorem. Every sequence hs t most one limit. PROOF. Let ( n ) be sequence nd, b R. Conversely, we suppose tht n, n b, b. Let, for emple, < b nd let us choose c (, b). Then there eist n, n N such tht ( n N, n > n ) : n < c, Hence which is impossible. ( n N, n > n ) : n > c. ( n N, n > m {n, n }) : c < n < c 8.7 Theorem (Limit of Subsequence). Let lim n = R nd ( kn ) be subsequence of the sequence ( n ). Then lim kn =. The bove-mentioned theorem cn be very useful, for emple, when proving tht sequence does not hve ny limit. 8.8 Emple. The sequence { n } n= := {( )n } n= does not hve ny limit. PROOF. By Theorem 8.7, it is sufficient to find two convergent subsequences of ( n ) whose limits differ. And tht is quite esy: for subsequence contining only even terms of ( n ) we hve n = ( ) n =, for subsequence contining only odd terms of ( n ) we hve n = ( ) n =. 8

29 8.9 Theorem (Limit of Monotonic Sequence). Let ( n ) be sequence. If ( n ) is non-decresing, then lim n = sup { n : n N}. If ( n ) is non-incresing, then lim n = inf { n : n N}. PROOF. We ssume, for emple, tht ( n ) is non-decresing (if ( n ) is non-incresing, we proceed nlogously, or we cn employ the fct tht ( n ) is non-decresing). We put s = sup { n : n N} nd split the proof into two prts. i) Let us consider primrily sitution when s = + (i.e., ( n ) is not bounded bove). The tsk is to prove tht lim n = + which mens tht ( k R)( n 0 N)( n N, n > n 0 ) : n > k. Let k R be given. Then k < + = sup { n : n N}, nd therefore there eists n n 0 N such tht n0 > k. Hence nd from the ssumption of monotonicity of ( n ) the desired proposition follows since ( n N, n > n 0 ) : n n0 > k. ii) Now if s R (i.e., ( n ) is bove-bounded), we hve to prove tht ( ε R + )( n 0 N)( n N, n > n 0 ) : s ε < n < s + ε. Let ε R + be given. Since s ε < s = sup { n : n N}, there eists n n 0 N such tht n0 > s ε. From the monotonicity of ( n ) nd from the fct tht supremum is lso n upper bound we finlly get tht 8.0 Emples. ( n N, n > n 0 ) : s ε < n0 n s < s + ε. ) It cn be shown tht the sequence ( n ), where n := ( n, + n) is incresing nd bounded bove. Therefore it is lso convergent. Furthermore, it cn be proved tht ( lim + n) n = e ( ). 9

30 ) The sequence ( n ), where n := n k= k = n, is clerly incresing, nd therefore its limit eists. However, for every n N we hve n n = n + + n n n n =. Hence ( n ) fils to be the Cuchy sequence, nd therefore its finite limit does not eist, by Theorem 8.3. So 3) The sequence ( n ), where ( lim ) + =: n n = +. n= n := n k= k = n, is clerly incresing. Since holds for every k N \ {}, we obtin = + n = n k= ( ) k < k (k ) = k k k < n (n ) = + ( ) ( ) ( n ) = n n, which holds for every n N \ {}. Hence ( n ) is bounded bove, nd therefore it is convergent. Moreover, it cn be shown tht lim ( n ) + =: n = π n= 9 CALCULATING LIMITS 9. Theorem. Let lim n = R nd lim b n = b R. Then i) lim n =, 30

31 ii) lim ( n ± b n ) = ± b whenever the right side of the equlity is meningful, iii) lim ( n b n ) = b whenever the right side of the equlity is meningful, iv) lim n b n = b whenever the right side of the equlity is meningful nd b n 0 for ll n N, v) lim k n = k whenever k N \ {}, R nd n 0 for ll n N. 9. Remrks. Let us think over the lst theorem in more detil. Every of the mentioned propositions gives the informtion (of course, on the ssumption tht the right side is meningful): i) tht the relevnt limit eists, ii) how to clculte it using numbers nd b. Proposition i) cnnot be reversed for 0. In other words, the sttement lim n eists lim n eists fils to be true. As contrry emple we cn consider the sequence {( ) n } n=. However, directly from the definition of the limit it follows tht lim n = 0 lim n = 0. Cution! If the right side in equlities ii) - iv) is meningless, it does not imply tht the relevnt limit does not eist. Let us hve look t the following emples: } i) n := n + b n := n + n b n = n +, } ii) n := n + b n := n + n b n = n, } iii) n := n + b n := n + n b n = ( R cn be chosen rbitrrily), } iv) n := n + b n := n ( ) n + n b n = ( ) n... this sequence hs no limit. The emples bove lso show why it is not resonble to define (+ ) (+ ). We cn lso find similr emples for other opertions. 9.3 Emples. ) lim n + 6n + 7 3n = lim n n 3 = lim( ) n n lim(3 ) n n = (+ ) 3 + (+ ) = = = 3. 3

32 ) ( lim + ) n ( = lim 3 + ) 3n ( = 3 lim + ) 3n = 3 e 3n 3n 3n (here we use the fct tht { ( + is subsequence of the convergent sequence n= { ( + )n}, nd therefore it hs the sme limit e). n n= 3n )3n} 9.4 Convention. To sy tht S(n) holds for ll lrge enough n N mens tht 9.5 Observtions. ( n 0 N)( n N, n > n 0 ) : S(n). n R ε R + : n < ε for ll lrge enough n N. Let limit of sequence ( n ) eist nd let (b n ) be sequence such tht n = b n for ll lrge enough n N. Then lim b n = lim n. 9.6 Definition. From now on by sequence we shll now lso men function defined (only) on set N \ K, where K N is some finite set. The bove-mentioned definitions of the limit remin (without ny chnge!) vlid lthough we hve generlized the concept of sequence. 9.7 Emples. ) lim = 0 (lthough is not defined for n = 3), n 3 n 3 ) lim +n+n3 (n 3)(n 007) of the sequence). = + (despite the numbers 3 nd 007 do not belong to the domin 9.8 Theorem (Pssing Limit in Inequlities). Let ( n ), (b n ) nd (c n ) be sequences nd let lim n = R nd lim b n = b R. i) If < b, then n < b n for ll lrge enough n N. ii) If n b n for ll lrge enough n N, then b. iii) If n c n b n for ll lrge enough n N nd = b, then lim c n eists nd lim c n = = b. iv) If n c n for ll lrge enough n N nd = +, then lim c n = +. v) If c n b n for ll lrge enough n N nd b =, then lim c n =. 9.9 Cution. The following proposition n < b n for ll lrge enough n N, n, b n b, < b fils to be true (it is enough to consider, for emple, n = 0 0, b n = n 0). 3

33 9.0 Emples. ) n := sin(007n3 log n+e 3n ) n 0. PROOF. We first observe tht n sin(007n3 log n + e 3n ) n n holds for ll n N. Now, since ± n 0, we obtin, by Theorem 9.8 iii), n 0. ) n := n n. PROOF. Let the sequence (h n ) be defined by n n = + hn, n N \ {}. By Theorem 9. ii), it suffices now to show tht h n 0. First of ll, let us note tht h n 0 for ll n N \ {}. Since n = ( + h n ) n = n k=0 ( ) n h k n k holds for ll n N \ {}, it follows tht lso n h n 0 ( ) n h n = n(n ) holds for ll n N \ {}. As n 0, we hve, by Theorem 9.8 iii), h n 0 nd hence, by Theorem 9. v), h n = h n = h n Eercises. Let ( n ) be sequence defined by where q R. Prove tht ) lim n does not eist if q, ) lim n = 0 if q <, 3) lim n = if q =, 4) lim n = + if q >. 9. Theorem. Let lim n = 0. n := q n, i) If n > 0 for ll lrge enough n N, then lim n ii) If n < 0 for ll lrge enough n N, then lim n = +. =. h n 33

34 E LIMIT AND CONTINUITY OF A FUNCTION 0 LIMIT OF A FUNCTION 0. Convention. By writing 0 n 0 we men tht n 0 nd n 0 for ll lrge enough n N. We understnd the reltions 0 < n 0 nd 0 > n 0 in the similr wy. 0. Definitions. We sy tht function f hs t 0 R limit R (we write lim 0 f() = ) if 0 n 0 f ( n ) (i.e., f ( n ) for ll sequences ( n ) stisfying 0 n 0 ), limit from the right R (we write limit from the left R (we write lim f() = ) if < n 0 f ( n ), lim f() = ) if 0 0 > n 0 f ( n ). 0.3 Emples. Let f() := (see Fig. 5). Then ) lim f() =, ) lim f() = 0, + 3) lim f() = 0, 4) lim f() = +, 0+ 5) lim f() =, 0 6) lim f() does not eist since, for emple, 0 n := ( )n 0 n does not hve ny limit. 0 nd f( n ) = ( ) n n 0.4 Definitions. Let 0 R nd δ R +. We define the following sets: U( 0, δ) := ( 0 δ, 0 + δ)... neighbourhood of 0 (with rdius δ), U( 0, δ) := [ 0, 0 + δ)... right neighbourhood of 0 (with rdius δ), 34

35 U ( 0, δ) := ( 0 δ, 0 ]... left neighbourhood of 0 (with rdius δ), U(+, δ) := { R : > δ } = ( δ, + ) {+ }... neighbourhood of + (with rdius δ), U(, δ) := { R : < δ } = (, δ ) { }... neighbourhood of (with rdius δ), P ( 0, δ) := U( 0, δ) \ { 0 }... n nnulr neighbourhood of 0 (with rdius δ) Anlogously we define P + ( 0, δ), P ( 0, δ), P (+, δ) nd P (, δ). If we do not cre bout the size δ of neighbourhood, we write briefly U( 0 ), P ( 0 ), Theorem. Let R. For ny 0 R, For ny 0 R, nd lim f() = ( U())( P ( 0 ))( P ( 0 )) : f() U(). 0 lim f() = ( U())( P + ( 0 ))( P + ( 0 )) : f() U() 0 + lim f() = ( U())( P ( 0 ))( P ( 0 )) : f() U(). 0 PROOF. We shll prove only the first equivlence for 0, R. To check the remining cses, it is enough to modify slightly the following steps. It is left it to the reder. i)? Conversely, we suppose tht ( ε R + ) ( δ R +) ( P ( 0, δ)) : [ / D(f) f() ε]. Hence it follows tht ( ε R + ) ( n N) ( n P ( 0, n)) : [n / D(f) f( n ) ε]. In this wy we obtin the sequence ( n ) stisfying clerly 0 n 0, but not f ( n ). This contrdicts our ssumption tht lim f() =. 0 ii)? We consider sequence ( n ) stisfying 0 n 0. Our tsk is to prove tht f( n ), i.e., ε R + : f ( n ) < ε for ll lrge enough n N. Given ε R +, there eists δ > 0 such tht P ( 0, δ) : f () < ε. Since n P ( 0, δ) (nd therefore f ( n ) < ε) for ll lrge enough n N, the proof is ctully completed. 35

36 0.6 Remrk. Neither eistence nor vlue of lim 0 f() depends on the eistence or vlue of f( 0 ). However, if lim 0 f() eists, then the function f hs to be defined on P ( 0, δ). 0.7 Emple. 9 lim = lim ( 3)( + 3) = lim ( 3) = 6. 3 The following three theorems re consequences of the definition of the limit of function nd corresponding theorems concerning the limit of sequence. 0.8 Theorem. A function f hs t most one limit t 0 R. 0.9 Theorem (Limit of Sum, Difference, Product nd Quotient of Functions). Let f, g : R R nd 0 R. Then i) lim (f() ± g()) = lim f() ± lim g() whenever the right side of the equlity is meningful, ii) iii) lim (f()g()) = lim f() lim g() whenever the right side of the equlity is meningful, f() lim = lim f() 0 0 g() lim g() 0 whenever the right side of the equlity is meningful. 0.0 Emples. ) lim + (3 ) = ) lim = lim lim + ( ( )) = +, ( + + ) = lim ( )( = lim +) 3) lim sin(nπ) = 0, but lim sin(nπ) does not eist. n + 0. Theorem. Let f, g, h : R R, 0, R, lim 0 f() = lim 0 g() =, ( P ( 0 ))( P ( 0 )) : f() h() g(). + =, Then lim 0 h() =. 36

37 0. Emple. PROOF. lim 0 ( sin ) = 0. The equlity follows directly from Theorem 0. since R \ {0} : sin, lim 0 ( ) = lim 0 = 0. (Note the fct tht lim sin does not eist.) Theorem. Let 0 R nd R. Then lim 0 f() = if nd only if lim f() = lim f() = CONTINUITY OF A FUNCTION. Definitions. Let 0 R. A function f is sid to be continuous t 0 if lim 0 f() = f( 0 ), continuous from the right t 0 if lim f() = f( 0), 0 + continuous from the left t 0 if lim f() = f( 0). 0 Note tht the continuity of f t 0 implies the eistence of U( 0 ) belonging to D(f), from the right t 0 implies the eistence of U + ( 0 ) belonging to D(f), from the left t 0 implies the eistence of U ( 0 ) belonging to D(f).. Theorem. Let f : R R nd 0 R. Then the following propositions re equivlent: i) f is continuous t 0, ii) 0 D(f) ( U(f( 0 )))( U( 0 ))( U( 0 )) : f() U(f( 0 )), iii) ( ε R + )( δ R + )( R) : 0 < δ f() f( 0 ) < ε, 37

38 iv) n 0 f( n ) f( 0 )..3 Emples. ) A constnt function is continuous t every 0 R. ) The function Id is continuous t every 0 R. 3) A function f defined by f() := is continuous t every 0 R. 4) The function sgn is continuous t every 0 R \ {0}, but not t 0 = 0. 5) The Dirichlet function χ is not continuous t ny point..4 Theorem (Continuity of Sum, Difference, Product nd Quotient of Functions). Let functions f nd g be continuous t 0 R. Then lso functions f + g, f g nd f g re continuous t 0. If, moreover, g( 0 ) 0, then the function f g is continuous t 0. PROOF. From the ssumptions lim f() = f( 0 ), 0 lim g() = g( 0 ), 0 the definition of opertions with functions nd Theorem 0.9 it follows tht lim (f + g)() = lim (f() + g()) = lim f() + lim g() = f( 0 ) + g( 0 ) = (f + g)( 0 ) Hence the function f + g is continuous t 0. We proceed similrly in the cse of the functions f g, f g nd f g..5 Theorem (Continuity of Composition of Functions). Let function f be continuous t 0 R nd let function g be continuous t f( 0 ). Then the function g f is continuous t 0. PROOF. We hve tht n 0 f( n ) f( 0 ) g(f( n )) g(f( 0 )), nd therefore, ccording to Theorem., the function g f is continuous t 0..6 Definition. A function f is continuous on n intervl I R if the following conditions hold: f is continuous t every interior point of the intervl I; if the bsepoint of I belongs to I, then f is continuous from the right t it; if the endpoint of I belongs to I, then f is continuous from the left t it. 38

39 .7 Theorem (Continuity of Bsic Elementry Functions). Let f be bsic elementry function nd let I D(f) be n intervl. Then f is continuous on I..8 Emple. sin lim 0 =. PROOF. By compring the res of the tringle OAC, sector of the circle OBC nd the tringle OBD (see Fig. 38), we obtin the following inequlities: ( 0, π ) : cos sin tn. D C tn 0.5 sin cos 0.5 A B 0.5 Fig. 38 Hence it follows tht ( 0, π ) : cos sin cos. Furthermore, since cosine nd sine re even nd odd function, respectively, it holds tht ( π, π ) \ {0} : cos sin cos. Finlly, the continuity of the functions cos nd t 0, i.e., cos implies, ccording to Theorem 0., tht lim cos = = lim 0 0 cos, sin lim 0 =..9 Theorem (Limit of Composition of Functions). Let 0,, b R nd ssume lim 0 f() =, 39

40 lim y g(y) = b, ( P ( 0 ))( P ( 0 )) : f() or g is continuous t. Then lim g(f()) = b. 0.0 Emples. Let us show tht the third ssumption of the previous theorem cnnot be omitted. ) Let f() := 0, g() :=. Then lim f() = 0, lim but lim g(f()) does not eist since D(g f) =. ) Let f() := 0, g() := g(y) = +, y 0 { 0, 98 = 0. Then but lim f() = 0, lim g(y) = +, y 0 lim g(f()) = lim g(0) = lim 98 = Eercises. ) Prove Theorem.9. ) Modify (nd prove) Theorem.9 for the cse of one-sided limits.. Emples. ) since lim 0 ( 5 ) = 0, lim y 0 sin y y =, R \ {0} : 5 0. sin ( 5 ) lim =

41 ) since ( lim cos sin ) = 0 lim sin = 0 (see Emple 0.), 0 the function cosine is continuous t 0 (i.e., lim cos y = ). y 0 (Note tht ( P (0))( P (0)) : sin = 0.) 4

42 F DIFFERENTIAL AND DERIVATIVE OF A FUNCTION MOTIVATION. It is often useful to substitute function f (t lest loclly, i.e., on neighbourhood) by simpler function, preferbly liner. However, this simplifiction (for non-liner f) cuses certin error. Let us try to find liner function pproimting function f on neighbourhood of point c so tht the pproimtion error is smll. More precisely, we try to find k R such tht f(c + h) f(c) + kh for ny smll h. Let us define function ω(h) (n error) by ω(h) := f(c + h) f(c) kh. Thus we wnt w(h) to be smll for ny smll h. We could require lim ω(h) = 0. h 0 However, this is not very resonble, since for continuous function f ny choice of k R complies with this ccurcy rte. It is more resonble to sk for i.e., f(c + h) f(c) kh lim h 0 h ω(h) lim h 0 h = 0, It follows tht f(c + h) f(c) k = lim h 0 h If f (c) R, then we cll function df c defined by ( ) f(c + h) f(c) = lim k = 0. h 0 h df c (h) := kh = f (c)h differentil of f t c. Note, by the wy, tht the line y = f(c) + f (c)( c) =: f (c). is sid to be tngent of grph of f t (c, f(c)). Number f (c) is the slope of the tngent.. Now let us consider mss point moving long line nd let us denote its position in time t by s(t). An verge velocity of the point on time intervl [c, c + h] cn be epressed by s(c + h) s(c). h If h tends to zero, then the verge velocity clerly tends to n immedite velocity of given mss point in the time c, i.e., s(c + h) s(c) v(c) = lim =: s (c). h 0 h 4

43 3 DIFFERENTIAL AND DERIVATIVE; CALCULATING DERIVATIVES 3. Definitions. Let f : R R nd R. If f( + h) f() lim h 0 h eists, we denote it by f () nd we cll it derivtive of the function f t the point. If f( + h) f() lim h 0+ h eists, we denote it by f +() nd we cll it derivtive from the right of the function f t the point. If f( + h) f() lim h 0 h eists, we denote it by f () nd we cll it derivtive from the left of the function f t the point. 3. Convention. Unless otherwise stted, we shll use the concept of derivtion in the mening of the proper (i.e., finite) derivtion. 3.3 Observtions. If f () eists (proper or improper), then there is U() such tht U() D(f). f( lim 0 +h) f( 0 ) f() f( = lim 0 ) h 0 h 0 0 whenever one side of the equlity is meningful. 3.4 Emples. ) If f is constnt, then f () = 0 for ll R. PROOF. we hve Let us ssume tht ( c R)( R) : f() = c, nd therefore for ll R f () = lim h 0 f( + h) f() h = lim h 0 c c h 0 = lim h 0 h = 0. ) (Id) = in R. PROOF. For ll R, we hve (Id) () = lim h 0 ( + h) () h h = lim h 0 h = lim =. h 0 43

44 3.5 Definitions. If there is k R such tht for function ω defined by ω(h) := f(c + h) f(c) kh it holds tht ω(h) lim h 0 h = 0, then the function f is sid to be differentible t the point c. A liner function df c defined by df c (h) := kh is clled differentil of the function f t the point c. 3.6 Theorem (Eistence of Differentil). A function f is differentible t point c R if nd only if the (finite!) derivtive of the function f t the point c eists. Moreover, in such cse h R : df c (h) = f (c) h. 3.7 Theorem (Continuity of Differentible Function). If function f is differentible t point 0, then it is continuous t 0. PROOF. The tsk is to prove tht lim (f() f( 0 )) = 0. 0 First, we note tht for ll, 0 D(f) such tht 0 we hve Hence it follows tht lim (f() f( 0 )) = lim 0 0 f() f( 0 ) = f() f( 0) 0 ( 0 ). [ ] f() f(0 ) ( 0 ) = f ( 0 ) 0 = 0. 0 (From the ssumption nd Theorem 3.6 it follows tht f f() f( ( 0 ) = lim 0 ) 0 0 R.) 3.8 Emples. ) The function sgn hs n (improper) derivtive t the point 0, but it is not continuous t 0. PROOF. sgn (0) = lim 0 sgn() sgn(0) 0 = lim 0 sgn() = lim 0 = +. 44

45 ) The function f() := 3 hs n (improper) derivtive t the point 0 nd it is continuous t 0. PROOF. f (0) = lim = lim 0 3 = lim 0 3 = +. 3) The function f() := is continuous t the point 0, but f (0) does not eist. PROOF. We cn esily clculte tht f +(0) = nd f (0) =, nd therefore, by Theorem 0.3, f (0) does not eist. 3.9 Theorem (Derivtive of Sum, Difference, Product nd Quotient of functions). Let R. Then i) (f ± g) () = f () ± g () whenever the right side of the equlity is meningful, ii) (fg) () = f ()g() + f()g () whenever f () nd g () eist finite, ( ) f () iii) g = f ()g() f()g () whenever f () nd g () eist finite nd g() 0. g () PROOF. i) (f ± g) (f ± g)( + h) (f ± g)() () = lim = h 0 h f( + h) f() g( + h) g() = lim ± = f () ± g () h 0 h h whenever f () ± g () is meningful (see Theorem 0.9). ii) ( f( + h)g( + h) f()g() (fg) () = lim h 0 h ( f( + h) f() = lim g( + h) + f() h 0 h where the lst equlity follows from g( + h) g() h ) f()g( + h) ± = h ) = f ()g() + f()g (), g () R g is continuous t lim h 0 g( + h) = g() nd the fct tht f () R. 45

46 iii) ( f g [ = lim h 0 ) () = lim h 0 g( + h)g() f(+h) f() g(+h) g() f( + h)g() f()g( + h) = lim = h h 0 hg( + h)g() ( )] f( + h) f() g( + h) g() g() f() = h h = g () (f ()g() f()g ()). 3.0 Remrk. Anlogous propositions hold lso for the one-sided derivtives. 3. Theorem (Derivtive of Composition of Functions). Let R nd let f () nd g (f()) eist finite. Then (g f) () = g (f())f (). 3. Remrk. For the ske of lucidity, we shll write, not very correctly, (f()) insted of f (). 3.3 Emples. ) R : sin = cos. PROOF. First, we recll tht Therefore for ll R we get R : sin = cos() sin sin( + h) sin = lim h 0 h ( = lim cos sin h h 0 h ( = lim h 0 cos h sin h cos sin h h sin sin h h = lim h 0 ) nd sin lim 0 =. sin cos h + cos sin h sin = lim = h 0 h ) ( cos sin h h sin ) sin h = h sin h = cos sin 0 = cos. ) R : cos = sin. PROOF. hve (cos ) = From Theorem 3. nd the previous emple it follows tht for ll R we ( ( π )) ( sin = sin π ) ( π ) ( π ) = cos (0 ) = sin. 46

47 3) D(tn) : tn = cos. PROOF. From Theorem 3.9 nd the previous emples, it follows tht for ll D(tn) we hve (tn ) = ( ) sin = (sin ) cos sin (cos ) cos cos = cos + sin cos = cos. 4) D(cot) : cot = sin. PROOF. D(cot) : (cot ) = ( cos ) sin cos = sin sin = sin. 5) R : (e ) = e. (Let us leve this proposition without proof... ) 3.4 Theorem (Derivtive of n Inverse Function). Let function f be continuous nd strictly monotonic on n intervl I R, let be n interior point of I, nd let f (f ()) eist. Then (f ) () eists nd is defined by (f ) () = f (f ()) if f (f ()) 0, + if f (f ()) = 0 nd f is incresing on I, if f (f ()) = 0 nd f is decresing on I. 3.5 Emples. ) R + : log =. PROOF. R + : log = ep (log ) = ep(log ) =. ) Let n N. Then R : ( n ) = n n. PROOF. We shll use the mthemticl induction. i) R : () = Id () =. ii) The tsk is to prove the impliction ( n N R : ( n ) = n n ) R : ( n+ ) = (n + ) n. ( n+ ) = ( n ) = ( n ) + n = n n + n = (n + ) n. 47

48 3) Let n N. Then R \ {0} : ( n ) = n n. PROOF. R \ {0} : ( n ) = ( ) = () n ( n ) = n ( n ) = ( n) n n = n n ( n) = n n. 4) Let r R. Then R + : ( r ) = r r. PROOF. R + : ( r ) = (e r log ) = e r log (r log ) = r r = rr. 5) Let R nd let f nd g be functions stisfying f() > 0 nd f (), g () R. Then (f() g() ) = ( e g() log f()) = e g() log f() (g() log f()) = ( = f() g() g () log f() + g() f ) (). f() 6) (, ) : rcsin =. PROOF. First, let us recll tht (, ) : rcsin ( π, π ) nd ( π, π ) : cos > 0. Therefore (, ) : (rcsin ) = = 7) (, ) : rccos =. sin (rcsin ) = cos(rcsin ) = cos(rcsin ) = sin (rcsin ) =. PROOF. Since (, ) : rccos (0, π) nd (0, π) : sin > 0, we get (, ) : (rccos ) = cos (rccos ) = sin(rccos ) = = sin(rccos ) = cos (rccos ) =. 48

49 7) R : rctn = +. PROOF. First, let us observe tht R : rctn ( π, π) nd ( π, π) : = + tn. Therefore = cos +sin cos cos R : rctn = 8) R : rccot = +. tn (rctn ) = cos (rctn ) = + tn (rctn ) = +. PROOF. First, let us observe tht R : rccot (0, π) nd (0, π) : = + cot. Therefore = sin +cos sin sin R : rccot = cot (rccot ) = sin (rccot ) 3.6 Definition. Let f be function. A function f defined by f () := f () = + cot (rccot ) = +. is sid to be derivtive of the function f. Anlogously we define functions f + nd f. 3.7 Definitions. Let I R be n intervl with end points, b R, < b. A function f is sid to be differentible on the intervl I if the following three conditions hold: i) (, b) : f () R, ii) if I, then f +() R, iii) if b I, then f (b) R; continuously differentible on the intervl I if the following three conditions hold: i) the function f is continuous on (, b), ii) if I, then the function f + is continuous from the right t the point, iii) if b I, then the function f is continuous from the left t the point b. 3.8 Definitions. Let n N. Let us define function clled (n + )th derivtive of function f by induction f (n+) := ( f (n)). Moreover, let us define function f (0) by f (0) () := f(). 49

50 3.9 Emple. sin (0) = sin, sin = cos, sin = (sin ) = (cos ) = sin, sin = (sin ) = ( sin ) = cos, sin (4) = (sin ) = ( cos ) = sin, sin (5) = (sin (4) ) = (sin ) = cos,

51 G BASIC THEOREMS OF DIFFERENTIAL CALCULUS 4 THEOREMS ON FUNCTION INCREMENT 4. Theorem (Rolle). Let function f be continuous on n intervl [, b] nd differentible on (, b), nd let f() = f(b). Then there is ξ (, b) such tht f (ξ) = 0. PROOF. We shll crry out the proof lter (see 7..6). The mening of Rolle s theorem is illustrted in Fig Fig. 39 Fig Theorem (Lgrnge s Men Vlue Theorem). Let function f be continuous on n intervl [, b] nd differentible on (, b). Then there is ξ (, b) such tht f (ξ) = f(b) f(). b PROOF. Let us define function F by F () := f() f(b) f() b ( ). From the ssumptions it follows tht F is continuous on the intervl [, b] nd differentible on (, b). Moreover, since F () = F (b) (= f()), there is (see Rolle s theorem) ξ (, b) such tht F (ξ) = 0. For ll (, b), the derivtive of F is given by F () = f () Hence nd from F (ξ) = 0 it follows tht f(b) f(). b f (ξ) = f(b) f(). b The mening of the ltter theorem is illustrted in Fig