Lecture Notes in Calculus. Raz Kupferman Institute of Mathematics The Hebrew University

Lecture Notes in Clculus Rz Kupfermn Institute of Mthemtics The Hebrew University July 10, 2013

2

Contents 1 Rel numbers 1 1.1 Axioms of field........................... 1 1.2 Axioms of order (s tught in 2009)................ 7 1.3 Axioms of order (s tught in 2010, 2011)............. 10 1.4 Absolute vlues........................... 13 1.5 Specil sets of numbers....................... 16 1.6 The Archimeden property..................... 19 1.7 Axiom of completeness....................... 23 1.8 Rtionl powers........................... 37 1.9 Rel-vlued powers......................... 42 1.10 Addendum.............................. 42 2 Functions 43 2.1 Bsic definitions........................... 43 2.2 Grphs................................ 49 2.3 Limits................................ 50 2.4 Limits nd order........................... 63 2.5 Continuity.............................. 67 2.6 Theorems bout continuous functions............... 70 2.7 Infinite limits nd limits t infinity................. 79 2.8 Inverse functions.......................... 81 2.9 Uniform continuity......................... 85

ii CONTENTS 3 Derivtives 91 3.1 Definition.............................. 91 3.2 Rules of differentition....................... 98 3.3 Another look t derivtives..................... 102 3.4 The derivtive nd extrem..................... 105 3.5 Derivtives of inverse functions................... 118 3.6 Complements............................ 120 3.7 Tylor s theorem.......................... 123 4 Integrtion theory 133 4.1 Definition of the integrl...................... 133 4.2 Integrtion theorems........................ 143 4.3 The fundmentl theorem of clculus................ 153 4.4 Riemnn sums............................ 157 4.5 The trigonometric functions.................... 158 4.6 The logrithm nd the exponentil................. 163 4.7 Integrtion methods......................... 167 5 Sequences 171 5.1 Bsic definitions........................... 171 5.2 Limits of sequences......................... 172 5.3 Infinite series............................ 186

CONTENTS iii Foreword 1. Texbooks: Spivk, Meizler, Hochmn.

iv CONTENTS

Chpter 1 Rel numbers In this course we will cover the clculus of rel univrite functions, which ws developed during more thn two centuries. The pioneers were Isc Newton (1642-1737) nd Gottfried Wilelm Leibniz (1646-1716). Some of their followers who will be mentioned long this course re Jkob Bernoulli (1654-1705), Johnn Bernoulli (1667-1748) nd Leonhrd Euler (1707-1783). These first two genertions of mthemticins developed most of the prctice of clculus s we know it; they could integrte mny functions, solve mny differentil equtions, nd sum up lrge number of infinite series using welth of sophisticted nlyticl techniques. Yet, they were sometimes very vgue bout definitions nd their theory often lid on shky grounds. The sound theory of clculus s we know it tody, nd s we re going to lern it in this course ws mostly developed throughout the 19th century, notbly by Joseph-Louis Lgrnge (1736-1813), Augustin Louis Cuchy (1789-1857), Georg Friedrich Bernhrd Riemnn (1826-1866), Peter Gustv Lejeune-Dirichlet (1805-1859), Joseph Liouville (1804-1882), Jen- Gston Drboux (1842-1917), nd Krl Weierstrss (1815-1897). As clculus ws being estblished on firmer grounds, the theory of functions needed thorough revision of the concept of rel numbers. In this context, we should mention Georg Cntor (1845-1918) nd Richrd Dedekind (1831-1916). 1.1 Axioms of field Even though we hve been using numbers s n elementry notion since first grde, rigorous course of clculus should strt by putting even such bsic con-

2 Chpter 1 cept on xiomtic grounds. The set of rel numbers will be defined s n instnce of complete, ordered.(שדה סדור שלם ) field Definition 1.1 A field F (ש ד ה ) is non-empty set on which two binry opertions re defined 1 : n opertion which we cll ddition, nd denote by +, nd n opertion which we cll multipliction nd denote by (or by nothing, s in b = b). The opertions on elements of field stisfy nine defining properties, which we list now. 1. Addition is ssocitive :(קיבוצי ) for ll, b, c F, ( + b) + c = + (b + c). 2 It should be noted tht ddition ws defined s binry opertion. As such, there is no -priori mening to + b + c. In fct, the mening is mbiguous, s we could first dd + b nd then dd c to the sum, or conversely, dd b + c, nd then dd the sum to. The first xiom sttes tht in either cse, the result is the sme. Wht bout the ddition of four elements, + b + c + d? Does it require seprte xiom of equivlence? We would like the following dditions (( + b) + c) + d ( + (b + c)) + d + ((b + c) + d) + (b + (c + d)) ( + b) + (c + d) to be equivlent. It is esy to see tht the equivlence follows from the xiom of ssocitivity. Likewise (lthough it requires some non-trivil work), we 1 A binry opertion in F is function tht tkes n ordered pir of elements in F nd returns n element in F. Tht is, to every, b F corresponds one nd only one + b F nd b F. In forml nottion, (, b F)(!c F) : (c = + b). 2 A few words bout the equl sign; we will tke it literlly to mens tht the expressions on both sides re the sme. More precisely, equlity is n equivlence reltion, concept which will be explined further below.

Rel numbers 3 cn show tht the n-fold ddition is defined unmbiguously. 1 + 2 + + n 2. Existence of n dditive-neutrl element נייטרלי לחיבור ) :(איבר there exists n element 0 F such tht for ll F + 0 = 0 + =. 3. Existence of n dditive inverse נגדי ) :(איבר for ll F there exists n element b F, such tht + b = b + = 0. The would like to denote the dditive inverse of, s customry, ( ). There is however one problem. The xiom ssumes the existence of n dditive inverse, but it does not ssume its uniqueness. Suppose there were two dditive inverses: which one would be denote ( )? With the first three xioms there re few things we cn show. For exmple, tht + b = + c implies b = c. The proof requires ll three xioms. Denote by d n dditive inverse of. Then, + b = + c d + ( + b) = d + ( + c) (existence of inverse) (d + ) + b = (d + ) + c (ssocitivity) 0 + b = 0 + c (property of inverse) b = c. (0 is the neutrl element) It follows t once tht every element F hs unique dditive inverse, for if b nd c were both dditive inverses of, then + b = 0 = + c, from which follows tht b = c. Hence we cn refer to the dditive inverse of, which justifies the nottion ( ). We cn lso show tht there is unique element in F stisfying the neutrl property (it is not n priori fct). Suppose there were, b F such tht + b =.

4 Chpter 1 Then dding ( ) (on the left!) nd using the lw of ssocitivity we get tht b = 0. Comment: We only postulte the opertions of ddition nd multipliction. Subtrction is short-hnd nottion for the ddition of the dditive inverse, b def = + ( b)..(חבורה ) Comment: A set stisfying the first three xioms is clled group By themselves, those three xioms hve mny implictions, which cn fill n entire course. 4. Addition is commuttive :(חילופי ) for ll, b F + b = b +. With this lw we finlly obtin tht ny (finite!) summtion of elements of F cn be re-rrnged in ny order. Exmple: Our experience with numbers tells us tht b = b implies tht = b. There is no wy we cn prove it using only the first four xioms! Indeed, ll we obtin from it tht + ( b) = b + ( ) (given) ( + ( b)) + (b + ) = (b + ( )) + ( + b) (commuttivity) + (( b) + b) + = b + (( ) + ) + b + 0 + = b + 0 + b + = b + b, but how cn we deduce from tht tht = b? (ssocitivity) (property of dditive inverse) (property of neutrl element) With very little comments, we cn stte now the corresponding lws for multipliction: 5. Multipliction is ssocitive: for ll, b, c F, (b c) = ( b) c.

Rel numbers 5 6. Existence of multiplictive-neutrl element: there exists n element 1 F such tht for ll F, 1 = 1 =. 7. Existence of multiplictive inverse הופכי ) :(איבר for every 0 there exists n element 1 F such tht 3 1 = 1 = 1. The condition tht 0 hs strong implictions. For exmple, from the fct tht b = c we cn deduce tht b = c only if 0. Comment: Division is defined s multipliction by the inverse, /b def = b 1. 8. Multipliction is commuttive: for every, b F, b = b. The lst xiom reltes the ddition nd multipliction opertions: 9. Distributive lw הקיבוץ ) :(חוק for ll, b, c F, (b + c) = b + c. We cn revisit now the b = b exmple. We proceed, + = b + b 1 + 1 = 1 b + 1 b (property of neutrl element) (1 + 1) = (1 + 1) b (distributive lw). We would be done if we knew tht 1 + 1 0, for we would multiply both sides by (1 + 1) 1. However, this does not follow from the xioms of field! 3 Here gin, the uniqueness of the multiplictive inverse hs to be proved, without which there is no justifiction to the nottion 1.

6 Chpter 1 Exmple: Consider the following set F = {e, f } with the following properties, + e f e e f f f e e f e e e f e f It tkes some explicit verifiction to check tht this is indeed field (in fct the smllest possible field), with e being the dditive neutrl nd f being the multiplictive neutrl (do you recognize this field?). Note tht e f = e + ( f ) = e + f = f + e = f + ( e) = f e, nd yet e f, which is then no wonder tht we cn t prove, bsed only on the field xioms, tht f e = e f implies tht e = f. With 9 xioms t hnd, we cn strt proving theorems tht re stisfied by ny field. Proposition 1.1 For every F, 0 = 0. Proof : Using sequentilly the property of the neutrl element nd the distributive lw: 0 = (0 + 0) = 0 + 0. Adding to both sides ( 0) we obtin 0 = 0. Comment: Suppose it were the cse tht 1 = 0, i.e., tht the sme element of F is both the dditive neutrl nd the multiplictive neutrl. It would follows tht for every F, = 1 = 0 = 0, which mens tht 0 is the only element of F. This is indeed field ccording to the xioms, however very boring one. Thus, we rule it out nd require the field to stisfy 0 1.

Rel numbers 7 Comment: In principle, every lgebric identity should be proved from the xioms of field. In prctice, we will ssume henceforth tht ll known lgebric mnipultions re vlid. For exmple, we will not bother to prove tht b = (b ) (even though the proof is very esy). The only exception is (of course) if you get to prove n lgebric identity s n ssignment. Exercise 1.1 Let F be field. Prove, bsed on the field xioms, tht ➀ If b = for some field element 0, then b = 1. ➁ 3 b 3 = ( b)( 2 + b + b 2 ). ➂ If c 0 then /b = (c)/(bc). ➃ If b, d 0 then Justify every step in your proof! b + c d d + bc =. bd 1.2 Axioms of order (s tught in 2009) The rel numbers contin much more structure thn just being field. They lso form n ordered set. The property of being ordered cn be formlized by three xioms: Definition 1.2 ס דוּר ) (שׁ ד ה A field F is sid to be ordered if it hs distinguished subset P F, which we cll the positive elements, such tht the following three properties re stisfied: 1. Trichotomy: every element F stisfies one nd only one of the following properties. Either (i) P, or (ii) ( ) P, or (iii) = 0. 2. Closure under ddition: if, b P then + b P. 3. Closure under multipliction: if, b P then b P.

8 Chpter 1 We supplement these xioms with the following definitions: < b mens b P > b mens b < b mens < b or = b b mens > b or = b. Comment: > 0 mens, by definition, tht 0 = P. Similrly, < 0 mens, by definition, tht 0 = ( ) P. We cn then show number of well-known properties of rel numbers: Proposition 1.2 For every, b F either (i) < b, or (ii) > b, or (iii) = b. Proof : This is n immedite consequence of the trichotomy lw, long with our definitions. We re sked to show tht either (i) b P, or (ii) b = ( b) P, or (iii) tht b = 0. Tht is, we re sked to show tht b stisfies the trichotomy, which is n xiom. Proposition 1.3 If < b then + c < b + c. Proof : (b + c) ( + c) = b P. Proposition 1.4 (Trnsitivity) If < b nd b < c then < c. Proof : c = (c b) + (b ) } {{ }} {{ }. in P in P } {{ } in P by closure under ddition

Rel numbers 9 Proposition 1.5 If < 0 nd b < 0 then b > 0. Proof : We hve seen bove tht, b < 0 is synonymous to ( ), ( b) P. By the closure under multipliction, it follows tht ( )( b) > 0. It remins to check tht the xioms of field imply tht b = ( )( b). Corollry 1.1 If 0 then 2 > 0. Proof : If 0, then by the trichotomy either > 0, in which cse 2 > 0 follows from the closure under multipliction, or < 0 in which cse 2 > 0 follows from the previous proposition, with = b. Corollry 1.2 1 > 0. Proof : This follows from the fct tht 1 = 1 2. Another corollry is tht the field of complex numbers (yes, this is not within the scope of the present course) cnnot be ordered, since i 2 = ( 1), which implies tht ( 1) hs to be positive, hence 1 hs to be negtive, which violtes the previous corollry. Proposition 1.6 If < b nd c > 0 then c < bc. Proof : It is given tht b P nd c P. Then, bc c = (b ) } {{ } in P c }{{} in P } {{ } in P by closure under multipliction Exercise 1.2 Let F be n ordered field. Prove bsed on the xioms tht.

10 Chpter 1 ➀ < b if nd only if ( b) < ( ). ➁ If < b nd c > d, then c < b d. ➂ If > 1 then 2 >. ➃ If 0 < < 1 then 2 <. ➄ If 0 < b nd 0 c < d then c < bd. ➅ If 0 < b then 2 < b 2. Exercise 1.3 Prove using the xioms of ordered fields nd using induction on n tht ➀ If 0 x < y then x n < y n. ➁ If x < y nd n is odd then x n < y n. ➂ If x n = y n nd n is odd then x = y. ➃ If x n = y n nd n is even then either x = y of x = y. ➄ Conclude tht x n = y n if nd only if x = y or x = y. Exercise 1.4 Let F be n ordered field, with multiplictive neutrl term 1; we lso denote 1 + 1 = 2. Prove tht ( + b)/2 is well defined for ll, b F, nd tht if < b, then < + b 2 < b. Exercise 1.5 Show tht there cn be no ordered field tht hs finite number of elements. 1.3 Axioms of order (s tught in 2010, 2011) The rel numbers contin much more structure thn just being field. They lso form n ordered set. The property of being ordered cn be formlized by four xioms: Definition 1.3 ס דוּר ) (שׁ ד ה A field F is sid to be ordered if there exists reltion < ( reltion is property tht every ordered pir of elements either stisfy or not), such tht the following four properties re stisfied:

Rel numbers 11 O1 Trichotomy: every pir of elements, b F stisfies one nd only one of the following properties. Either (i) < b, or (ii) b <, or (iii) = b. O2 Trnsitivity: if < b nd b < c then < c. O3 Invrince under ddition: if < b then + c < b + c for ll c F. O4 Invrince under multipliction by positive numbers: if < b then c < bc for ll 0 < c. Comment: The set of numbers for which 0 < re clled the positive numbers. The set of numbers for which < 0 re clled the negtive numbers. We supplement these xioms with the following definitions: (2 hrs, 2013) > b mens b < b mens < b or = b b mens > b or = b. Proposition 1.7 The set of positive numbers is closed under ddition nd multipliction, nmely If 0 <, b then 0 < + b nd 0 < b. Proof : Closure under ddition follows from the fct tht 0 < implies tht b = 0 + b (neutrl element) < + b, (O3) nd b < + b follows from trnsitivity. Closure under multipliction follows from the fct tht 0 = 0 (proved) < b. (O4) (2 hrs, 2010) (2 hrs, 2011)

12 Chpter 1 Proposition 1.8 0 < if nd only if ( ) < 0, i.e., number is positive if nd only if its dditive inverse is negtive. Proof : Suppose 0 <, then ( ) = 0 + ( ) (neutrl element) < + ( ) (O3) = 0 (dditive inverse) i.e., ( ) < 0. Conversely, if ( ) < 0 then 0 = ( ) + (dditive inverse) < 0 + (O3) = (neutrl element) Corollry 1.3 The set of positive numbers is not empty. Proof : Since by ssumption 1 0, it follows by O1 tht either 0 < 1 or 1 < 0. In the ltter cse 0 < ( 1), hence either 1 or ( 1) is positive. Proposition 1.9 0 < 1. Proof : By trichotomy, either 0 < 1, or 1 < 0, or 0 = 1. The third possibility is ruled out by ssumption. Suppose then, by contrdiction, tht 1 < 0. Then for every 0 < (nd we know tht t lest one such exists), = 1 < 0 = 0, i.e., < 0 which violtes the trichotomy xiom. Hence 0 < 1.

Rel numbers 13 Proposition 1.10 0 < if nd only if 0 < 1, i.e., number is positive if nd only if its multiplictive inverse is lso positive. Proof : Let 0 < nd suppose by contrdiction tht 1 becuse 1 = 1). Then, < 0 (it cn t be zero 1 = 1 < 0 = 0, which is contrdiction. Proposition 1.11 If < 0 nd b < 0 then b > 0. Proof : We hve seen bove tht, b < 0 implies tht 0 < ( ), ( b). By the closure under multipliction, it follows tht ( )( b) > 0. It remins to check tht the xioms of field imply tht b = ( )( b). Corollry 1.4 If 0 then 2 > 0. Proof : If 0, then by the trichotomy either > 0, in which cse 2 > 0 follows from the closure under multipliction, or < 0 in which cse 2 > 0 follows from the previous proposition, with = b. Another corollry is tht the field of complex numbers (which is not within the scope of the present course) cnnot be ordered, since i 2 = ( 1), which implies tht ( 1) hs to be positive, hence 1 hs to be negtive, which is violtion. 1.4 Absolute vlues Definition 1.4 For every element in n ordered field F we define the bsolute,(ערך מוחלט ) vlue 0 = ( ) < 0.

14 Chpter 1 We see right wy tht = 0 if nd only if = 0, nd otherwise > 0. Proposition 1.12 (Tringle inequlity שיוויון המשולש ) ((אי For every, b F, + b + b. Proof : We need to exmine four cses (by trichotomy): 1. 0 nd b 0. 2. 0 nd b 0. 3. 0 nd b 0. 4. 0 nd b 0. In the first cse, = nd b = b, hence + b = + b = + b. (Hey, does this men tht + b + b?) In the fourth cse, = ( ), b = ( b), nd 0 < ( + b), hence + b = ( + b) = ( ) + ( b) = + b. It remins to show the second cse, s the third cse follows by interchnging the roles of nd b. Let 0 nd b 0, i.e., = ( ) nd b = b. We need to show tht + b ( ) + b. We cn divide this cse into two sub-cses. Either + b 0, in which cse If + b < 0, then + b = + b ( ) + b = + b. + b = ( + b) = ( ) + ( b) ( ) + b = + b, which completes the proof 4. (3 hrs, 2013) 4 We hve used the fct tht 0 implies ( ) nd 0 implies ( ). This is very esy to show. For exmple, nd ( ) < follows from trnsitivity. > 0 = + ( ) > 0 + ( ) = 0 > ( ),

Rel numbers 15 Comment: By replcing b by ( b) we obtin tht b + b. The following version of the tringle inequlity is lso useful: Proposition 1.13 (Reverse tringle inequlity) For every, b F, b + b. Proof : By the tringle inequlity,, c F, c + c. Set c = b +, in which cse b + b +, or b b +. By interchnging the roles of nd b, b + b. It follows tht 5 b + b, Proposition 1.14 For every, b, c F, + b + c + b + c. 5 Here we used the fct tht ny property stisfied by nd ( ) is lso stisfied by.

16 Chpter 1 Proof : Use the tringle inequlity twice, + b + c = + (b + c) + b + c + b + c. (2 hrs, 2009) Exercise 1.6 Let min(x, y) nd mx(x, y) denote the smllest nd the lrgest of the two rguments, respectively. Show tht mx(x, y) = 1 2 (x + y + x y ) nd min(x, y) = 1 (x + y x y ). 2 1.5 Specil sets of numbers Wht kind of sets cn hve the properties of n ordered field? We strt by introducing the nturl numbers הטבעיים ),(המספרים by tking the element 1 (which exists by the xioms), nd nming the numbers 1 + 1, 1 + 1 + 1, 1 + 1 + 1 + 1, etc. Tht these numbers re ll distinct follows from the xioms of order (ech number is greter thn its predecessor becuse 0 < 1). Tht we give them nmes, such s 2, 3, 4, nd my represent them using deciml system is immteril. We denote the set of nturl numbers by N. Clerly the nturl numbers do not form field (for exmple, there re no dditive inverses becuse ll the numbers re positive). We cn then ugment the set of nturl numbers by dding the numbers 0, 1, (1 + 1), (1 + 1 + 1), etc. This forms the set of integers,(השלמים ) which we denote by Z. The integers form commuttive group with respect to ddition, but re not field since multiplictive inverses do not exist 6. 6 How do we know tht (1 + 1) 1, for exmple, is not n integer? We know tht it is positive nd non-zero. Suppose tht it were true tht 1 < (1 + 1) 1, then which is violtion of the xioms of order. 1 + 1 = (1 + 1) 1 < (1 + 1) (1 + 1) 1 = 1,

Rel numbers 17 The set of numbers cn be further ugmented by dding ll integer quotients, m/n, n 0. This forms the set of rtionl numbers, which we denote by Q. It should be noted tht the rtionl numbers re the set of integer quotients modulo n equivlence reltion 7. A rtionl number hs infinitely mny representtions s the quotient of two integers. Proposition 1.15 Let b, d 0. Then b = c d if nd only if d = bc. Proof : Multiply/divide both sides by bd (which cnnot be zero if b, d 0). It cn be checked tht the rtionl numbers form n ordered field. (4 hrs, 2010) (4 hrs, 2011) In this course we will not tech (xiomticlly) mthemticl induction, nor inductive definitions (lso clled recursive definitions); we will ssume these concepts to be understood. As n exmple of n inductive definition, we define for ll Q nd n N, the n-th power, by setting, 1 = nd k+1 = k. As n exmple of n inductive proof, consider the following proposition: Proposition 1.16 Let, b > 0 nd n N. Then > b if nd only if n > b n. 7 A few words bout equivlence reltions שקילות ).(יחס An equivlence reltion on set S is property tht ny two elements either hve or don t. If two elements, b hve this property, we sy tht is equivlent to b, nd denote it by b. An equivlence reltion hs to be symmetric ( b implies b ), reflexive ( for ll ) nd trnsitive ( b nd b c implies c). Thus, with every element S we cn ssocite n equivlence clss שקילות ),(מחלקת which is {b : b }. Every element belongs to one nd only one equivlence clss. An equivlence reltion prtitions S into collection of disjoint equivlence clsses; we cll this set S modulo the equivlence reltion, nd denote it S/.

18 Chpter 1 Proof : We proceed by induction. This is certinly true for n = 1. Suppose this were true for n = k. Then > b implies k > b k implies k+1 = k > b k > b b k = b k+1. Conversely, if n > b n then b would imply tht n b n, hence > b. Another inequlity tht we will need occsionlly is: Proposition 1.17 (Bernoulli inequlity) For ll x > ( 1) nd n N, (1 + x) n 1 + nx. Proof : The proof is by induction. For n = 1 both sides re equl. Suppose this were true for n = k. Then, (1 + x) k+1 = (1 + x)(1 + x) k (1 + x)(1 + kx) = 1 + (k + 1)x + kx 2 1 + (k + 1)x, where in the left-most inequlity we hve used explicitly the fct tht 1 + x > 0. Exercise 1.7 The following exercise del with logicl ssertions, nd is in preprtion to the type of ssertions we will be deling with ll long this course. For ech of the following sttements write its negtion in hebrew, without using the word no. Then write both the sttement nd its negtion, using logicl quntifiers like nd. ➀ For every integer n, n 2 n holds. ➁ There exists nturl number M such tht n < M for ll nturl numbers n. ➂ For every integer n, m, either n m or n m. ➃ For every nturl number n there exist nturl numbers, b,, such tht b < n, 1 < nd n = b. (3 hrs, 2009)

1.6 The Archimeden property Rel numbers 19 We re iming t constructing n entity tht mtches our notions nd experiences ssocited with numbers. Thus fr, we defined n ordered field, nd showed tht it must contin set, which we clled the integers, long with ll numbers expressible t rtios of integers the rtionl numbers Q. Whether it contins dditionl elements is left for the moment unnswered. The point is tht Q lredy hs the property of being n ordered field. Before proceeding, we will need the following definitions: חסום) Definition 1.5 A set of elements A F is sid to be bounded from bove (מלעיל if 8 there exists n element M F, such tht M for ll A, i.e., A is bounded from bove ( M F) : ( A)( M). Such n element M is clled n upper bound מלעיל ) (חסם for A. A is sid to be bounded from below מלרע ) (חסום if ( m F) : ( A)( m). Such n element m is clled lower bound מלרע ) (חסם for A. A set is sid to be bounded (חסום ) if it is bounded both from bove nd below. Note tht if set is upper bounded, then the upper bound is not unique, for if M is n upper bound, so re M + 1, M + 2, nd so on. Proposition 1.18 A set A is bounded if nd only if ( M F) : ( A)( M). Proof : 9 Suppose A is bounded, then ( M 1 F) : ( A)( M 1 ) ( M 2 F) : ( A)( M 2 ). 8 In definitions if lwys stnds for if nd only if. 9 Once nd for ll: P if Q mens tht if Q holds then P holds, or Q implies P. P only if Q mens tht if Q does not hold, then P does not hold either, which implies tht P implies Q. Thus, if nd only if mens tht ech one implies the other.

20 Chpter 1 Tht is, ( M 1 F) : ( A)( M 1 ) ( M 2 F) : ( A)(( ) ( M 2 )). By tking M = mx(m 1, M 2 ) we obtin ( A)( M ( ) M), i.e., ( A)( M). The other direction isn t much different. A few nottionl conventions: for < b we use the following nottions for n open פתוח ),(קטע closed סגור ),(קטע nd semi-open חצי סגור ) (קטע segment, (, b) = {x F : < x < b} [, b] = {x F : x b} (, b] = {x F : < x b} [, b) = {x F : x < b}. Ech of these sets is bounded (bove nd below). We lso use the following nottions for sets tht re bounded on one side, (, ) = {x F : x > } [, ) = {x F : x } (, ) = {x F : x < } (, ] = {x F : x }. The first two sets re bounded from below, wheres the lst two re bounded from bove. It should be emphsized tht ± re not members of F! The bove is purely nottion. Thus, being less thn infinity does not men being less thn field element clled infinity. (5 hrs, 2011) Consider now the set of nturl numbers, N. Does it hve n upper bound? Our geometricl picture of the number xis clerly indictes tht this set is unbounded, but cn we prove it? We cn esily prove the following:

Rel numbers 21 Proposition 1.19 The nturl numbers re not upper bounded by nturl number. Proof : Suppose n N were n upper bound for N, i.e., ( k N) : (k n). However, m = n + 1 N nd m > n, i.e., ( m N) : (n < m), which is contrdiction. Similrly, Corollry 1.5 The nturl numbers re not bounded from bove by ny rtionl number Proof : Suppose tht p/q Q ws n upper bound for N. Since, by the xioms of order, p/q p, it would imply the existence of nturl number p tht is upper bound for N. 10 (5 hrs, 2010) It my however be possible tht n irrtionl element in F be n upper bound for N. Cn we rule out this possibility? It turns out tht this cnnot be proved from the xioms of n ordered field. Since, however, the boundedness of the nturls is so bsic to our intuition, we my impose it s n dditionl xiom, known s the xiom of the Archimeden field: the set of nturl numbers is unbounded from bove 11. (5 hrs, 2013) This xiom hs number of immedite consequences: 10 Convince yourself tht p/q p for ll p, q N. 11 This dditionl ssumption is temporry. The Archimeden property will be provble once we dd our lst xiom.

22 Chpter 1 Proposition 1.20 In n Archimeden ordered field F: ( F)( n N) : ( < n). Proof : If this weren t the cse, then N would be bounded. Indeed, the negtion of the proposition is ( F) : ( n N)(n ), i.e., is n upper bound for N. Corollry 1.6 In n Archimeden ordered field F: ( ɛ > 0)( n N) : (1/n < ɛ). Proof : By the previous proposition, ( ɛ > 0)( n N) : (0 < 1/ɛ < n). } {{ } (0<1/n<ɛ) Corollry 1.7 In n Archimeden ordered field F: ( x, y > 0)( n N) : (y < nx). Comment: This is relly wht is ment by the Archimeden property. For every x, y > 0, segment of length y cn be covered by finite number of segments of length x. y x x x x x x

Rel numbers 23 Proof : By the previous corollry with ɛ = y/x, ( x, y > 0)( n N) : (y/x < n). } {{ } (y<nx) 1.7 Axiom of completeness The Greeks knew lredy tht the field of rtionl numbers is incomplete, in the sense tht there is no rtionl number whose squre equls 2 (wheres they knew by Pythgors theorem tht this should be the length of the digonl of unit squre). In fct, let s prove it: Proposition 1.21 There is no r Q such tht r 2 = 2. Proof : Suppose, by contrdiction, tht r = n/m, where n, m N (we cn ssume tht r is positive) stisfies r 2 = 2. Although it requires some number theoreticl knowledge, we will ssume tht it is known tht ny rtionl number cn be brought into form where m nd n hve no common divisor. By ssumption, m 2 /n 2 = 2, i.e., m 2 = 2n 2. This mens tht m 2 is even, from which follows tht m is even, hence m = 2k for some k N. Hence, 4k 2 = 2n 2, or n 2 = 2k 2, from which follows tht n is even, contrdicting the ssumption tht m nd n hve no common divisor. Proof : Another proof: suppose gin tht r = n/m is irreducible nd r 2 = 2. Since, n 2 = 2m 2, then m 2 < n 2 < 4m 2, it follows tht nd Consider now the rtio m < n < 2m < 2n, 0 < n m < m nd 0 < 2m n < n. q = 2m n n m,

24 Chpter 1 whose numertor nd denomintor re both smller thn the respective numertor nd denomintor of r. By elementry rithmetic q 2 = 4m2 4mn + n 2 n 2 2nm + m 2 = 4 4n/m + n2 /m 2 n 2 /m 2 2n/m + 1 = 6 4n/m 3 2n/m = 2, which is contrdiction. Exercise 1.8 Prove tht there is no r Q such tht r 2 = 3. Exercise 1.9 Wht fils if we try to pply the sme rguments to prove tht there is no r Q such tht r 2 = 4 (n ssertion tht hppens to be flse)? A wy to cope with this missing number would be to dd 2 by hnd to the set of rtionl numbers, long with ll the numbers obtined by field opertions involving 2 nd rtionl numbers (this is clled in lgebr field extension) (שדה הרחבה ) 12. But then, wht bout 3? We could dd ll the squre roots of 3 ll positive rtionl numbers. And then, wht bout 2? Shll we dd ll n-th roots? But then, wht bout solution to the eqution x 5 + x + 1 = 0 (it cnnot be expressed in terms of roots s result of Glois theory)? It turns out tht single dditionl xiom, known s the xiom of completeness השלמות ),(אכסיומת completes the set of rtionl numbers in one fell swoop, such to provide solutions to ll those questions. Let us try to look in more detil in wht sense is the field of rtionl numbers incomplete. Consider the following two sets, A = {x Q : 0 < x, x 2 2} B = {y Q : 0 < y, 2 y 2 }. Every element in B is greter or equl thn every element in A (by trnsitivity nd by the fct tht 0 < x y implies x 2 y 2 ). Formlly, ( A b B)( b). Does there exist rtionl number tht seprtes the two sets, i.e., does there exist n element c Q such tht ( A b B)( c b)? 12 It cn be shown tht this field extension of Q consists of ll elements of the form { + b 2 :, b Q}.

Rel numbers 25 It cn be shown tht if there existed such c it would stisfy c 2 = 2, hence such c does not exist. This observtion motivtes the following definition: Definition 1.6 An ordered field F is sid to be complete if for every two nonempty sets, A, B F stisfying there exists n element c F such tht ( A b B)( b). ( A b B)( c b). We will soon see how this xiom completes the field Q. We next introduce more definitions. Let s strt with motivting exmple: Exmple: For ny set [, b) = {x : x < b}, b is n upper bound, but so is ny lrger element, e.g., b + 1. In fct, it is cler tht b is the lest upper bound. Definition 1.7 Let A F be set. An element M F is clled lest upper bound עליון ) (חסם for A if (i) it is n upper bound for A, nd (ii) if M is lso n upper bound for A then M M. Tht is, M is lest upper bound for A ( A)( M) ( M F)( if ( A)( M ) then (M M )). We cn see right wy tht lest upper bound, if it exists, is unique: Proposition 1.22 Let A F be set nd let M be lest upper bound for A. If M is lso lest upper bound for A then M = M. Proof : It follows immeditely from the definition of the lest upper bound. If M nd M re both lest upper bounds, then both re in prticulr upper bounds, hence M M nd M M, which implies tht M = M.

26 Chpter 1 We cll the lest upper bound of set (if it exists!) supremum, nd denote it by sup A. Similrly, the gretest lower bound תחתון ) (חסם of set is clled the infimum, nd it is denoted by inf A. 13 (7 hrs, 2011) We cn provide n equivlent definition of the lest upper bound: Proposition 1.23 Let A be set. A number M is lest upper bound if nd only if (i) it is n upper bound, nd (ii) ( ɛ > 0)( A) : ( > M ɛ). A M!! Proof : There re two directions to be proved. 1. Suppose first tht M were lest upper bound for A, i.e., ( A)( M) nd ( M F)(if ( A)( M ) then (M M )). Suppose, by contrdiction, tht there exists n ɛ > 0 such tht M ɛ for ll A, i.e., tht ( ɛ > 0) : ( A)( < M ɛ), then M ɛ is n upper bound for A, smller thn the lest upper bound, which is contrdiction. 13 In some books the nottions lub (lest upper bound) nd glb (gretest lower bound) re used insted of sup nd inf.

Rel numbers 27 2. Conversely, suppose tht M is n upper bound nd ( ɛ > 0)( A) : ( > M ɛ). Suppose tht M ws not lest upper bound. Then there exists smller upper bound M < M. Tke ɛ = M M. Then, M = M ɛ for ll A, or in forml nottion, which is contrdiction. ( ɛ > 0) : ( A)( < M ɛ), Exmples: Wht re the lest upper bounds (if they exist) in the following exmples: ➀ [ 5, 15] (nswer: 15). ➁ [ 5, 15) (nswer: 15). ➂ [ 5, 15] {20} (nswer: 20). ➃ [ 5, 15] (17, 18) (nswer: 18). ➄ [ 5, ) (nswer: none). ➅ {1 1/n : n N} (nswer: 1). Exercise 1.10 Let F be n ordered field, nd A F. Prove tht M = inf A if nd only if M is lower bound for A, nd ( ɛ > 0)( A) : ( < M + ɛ). Exercise 1.11 Show tht the xiom of completeness is equivlent to the following sttement: for every two non-empty sets A, B F, stisfying tht A nd b B implies b, there exists n element c F, such tht c b for ll A nd b B. Exercise 1.12 For ech of the following sets, determine whether it is upper bounded, lower bounded, nd if they re find their infimum nd/or supremum: ➀ {( 2) n : n N}.

28 Chpter 1 ➁ {1/n : n Z, n 0}. ➂ {1 + ( 1) n : n N}. ➃ {1/n + ( 1) n : n N}. ➄ {n/(2n + 1) : n N}. ➅ {x : x 2 1 < 3}. Exercise 1.13 Let A, B be non-empty sets of rel numbers. We define A + B = { + b : A, b B} A B = { b : A, b B} A B = { b : A, b B}. Find A + B, A B nd A A in ech of the following cses: ➀ A = {1, 2, 3} nd B = { 1, 2, 1/2, 1/2}. ➁ A = B = [0, 1]. Exercise 1.14 Let A be non-empty set of rel numbers. Is ech of the following sttements necessrity true? Prove it or give counter exmple: ➀ A + A = {2} A. ➁ A A = {0}. ➂ x A A, x 0. Exercise 1.15 In ech of the following items, A, B re non-empty subsets of complete ordered field. For the moment, you cn only use the xiom of completeness for the lest upper bound. ➀ Suppose tht A is lower bounded nd set B = {x : x A}. Prove tht B is upper bounded, tht A hs n infimum, nd sup B = inf A. ➁ Show tht if A is upper bounded nd B is lower bounded, then sup(a B) = sup A inf B. ➂ Show tht if A, B re lower bounded then inf(a + B) = inf A + inf B. ➃ Suppose tht A, B re upper bounded. It is necessrily true tht sup(a B) = sup A sup B? ➄ Show tht if A, B only contin non-negtive terms nd re upper bounded, then sup(a B) = sup A sup B.

Rel numbers 29 Definition 1.8 Let A F be subset of n ordered field. It is sid to hve mximum if there exists n element M A which is n upper bound for A. We denote M = mx A. It is sid to hve minimum if there exists n element m A which is lower bound for A. We denote m = min A. (7 hrs, 2010) (7 hrs, 2013) Proposition 1.24 If set A hs mximum, then the mximum is the lest upper bound, i.e., sup A = mx A. Similrly, if A hs minimum, then the minimum is the gretest lower bound, inf A = min A. Proof : Let M = mx A. 14 Then M is n upper bound for A, nd for every ɛ > 0, A M > M ɛ, tht is ( ɛ > 0)( A)( > M ɛ), which proves tht M is the lest upper bound. Exmples: Wht re the mxim (if they exist) in the following exmples: ➀ [ 5, 15] (nswer: 15). ➁ [ 5, 15) (nswer: none). ➂ [ 5, 15] {20} (nswer: 20). ➃ [ 5, 15] (17, 18) (nswer: none). ➄ [ 5, ) (nswer: none). ➅ {1 1/n : n N} (nswer: none). 14 We emphsize tht not every (non-empty) set hs mximum. The sttement is tht if mximum exists, then it is lso the supremum.

30 Chpter 1 (5 hrs, 2009) Proposition 1.25 Every finite set in n ordered field hs minimum nd mximum. Proof : The proof is by induction on the size of the set. Corollry 1.8 Every finite set in n ordered field hs lest upper bound nd gretest lower bound. Consider now the field Q of rtionl numbers, nd its subset A = {x Q : 0 < x, x 2 < 2}. This set is bounded from bove, s 2, for exmple, is n upper bound. Indeed, if x A, then x 2 < 2 < 2 2, which implies tht x < 2. A n upper bound 0 1 2 But does A hve lest upper bound? Proposition 1.26 Suppose tht A hs lest upper bound, then (sup A) 2 = 2.

Rel numbers 31 Proof : We ssume tht A hs lest upper bound, nd denote Clerly, 1 < α < 2. α = sup A. Suppose, by contrdiction, tht α 2 < 2. Relying on the Archimeden property, we choose n to be n integer lrge enough, such tht ( ) 1 n > mx α, 6. 2 α 2 We re going to show tht there exists rtionl number r > α, such tht r 2 < 2, which mens tht α is not even n upper bound for A contrdiction: (α + 1/n) 2 = (α) 2 + 2 n α + 1 n 2 < α 2 + 3 α (use 1/n < α) n < α 2 + 6 n (use α < 2) which mens tht α + 1/n A. < α 2 + 2 α 2 = 2, (use 6/n < 2 α 2 ) Similrly, suppose tht α 2 > 2. This time we set n > 2α α 2 2. We re going to show tht there exists rtionl number r < α, such tht r 2 > 2, which mens tht r is n upper bound for A smller thn α gin, contrdiction: (α 1/n) 2 = α 2 2 n α + 1 n 2 > α 2 2 n α (omit 1/n2 ) > α 2 + 2 α 2 = 2, (use 2α/n > 2 α 2 ) which mens tht, indeed, α 1/n is n upper bound for A. 15 15 You my sk yourself: how could I hve ever guessed such proof? Tke for exmple the ssumption tht α 2 > 2. We wnt to show tht there exists n upper bound for A tht is smller

32 Chpter 1 (8 hrs, 2013) Since we sw tht there is no r Q such tht r 2 = 2, it follows tht there is no supremum to this set in Q. We will now see tht completeness gurntees the existence of lowest upper bound: Proposition 1.27 In complete ordered field every non-empty set tht is upperbounded hs lest upper bound. Proof : Let A F be non-empty set tht is upper-bounded. Define B = {b F : b is n upper bound for A}, which by ssumption is non-empty. Clerly, By the xiom of completeness, ( A b B)( b). ( c F) : ( A b B)( c b). Clerly c is n upper bound for A smller or equl thn every other upper bound, hence c = sup A. (8 hrs, 2010) The following proposition is n immedite consequence: thn α, contrdicting the fct tht it is the lest upper bound. Let s try to find n n sufficiently lrge such tht α 1/n is n upper bound for A, nmely, such tht (α 1/n) 2 > 2. We expnd: ( α 1 ) 2 = α 2 2α n n + 1 n > 2 + ( α 2 2 ) 2α. 2 } {{ n} needs to be positive Similrly, when we tke the ssumption α 2 < 2 we wnt to show tht there exists n element of A tht is greter thn α. Let s try to find n n sufficiently lrge such tht α + 1/n is in A, nmely, such tht (α + 1/n) 2 < 2. We expnd: ( α + 1 n ) 2 = α 2 + 2α n + 1 n 2 < 2 + ( α 2 2 ) + 2α n + 1 n 2 } {{ } needs to be negtive.

Rel numbers 33 Proposition 1.28 In complete ordered field every non-empty set tht is bounded from below hs gretest lower bound (n infimum). Proof : Let m be lower bound for A, i.e., Consider the set Since ( A)(m ). B = {( ) : A}. ( A)(( m) ( )), it follows tht ( m) is upper bound for B. By the xiom of completeness, B hs lest upper bound, which we denote by M. Thus, ( A)(M ( )) nd ( ɛ > 0)( A) : (( ) > M ɛ). This mens tht ( A)(( M) ) nd ( ɛ > 0)( A) : ( < ( M) + ɛ). This mens tht ( M) = inf A. With tht, we define the set of rel numbers s complete ordered field, which we denote by R. It turns out tht this defines the set uniquely, up to relbeling of its elements (i.e., up to n isomorphism). 16 The xiom of completeness cn be used to prove the Archimeden property. In other words, the xiom of completeness solves t the sme time the problems pointed out in the previous section. Proposition 1.29 (Archimeden property) In complete ordered field N is not bounded from bove. 16 Strictly speking, we should prove tht such n niml exists. Constructing the set of rel numbers from the set of rtionl numbers is beyond the scope of this course. You re strongly recommended, however, to red bout it. See for exmple the construction of Dedekind cuts.

34 Chpter 1 Proof : Suppose N ws bounded. Then it would hve lest upper bound M, which is, in prticulr, nd upper bound: Since n N implies n + 1 N, i.e., ( n N) (n M). ( n N) ((n + 1) M), ( n N) (n (M 1)). This mens tht M 1 is lso n upper bound for N, contrdicting the fct tht M is the lest upper bound. (9 hrs, 2011) We next prove n importnt fcts bout the density (צפיפות ) of rtionl numbers within the rels. Proposition 1.30 (The rtionl numbers re dense in the rels) Let x, y R, such tht x < y. There exists rtionl number q Q such tht x < q < y. In forml nottion, ( x, y R : x < y)( q Q) : (x < q < y). Proof : Since the nturl numbers re not bounded, there exists n m Z Q such tht m < x. Also there exists nturl number n N such tht 1/n < y x. Consider now the sequence of rtionl numbers, r k = m + k, k = 0, 1, 2,.... n From the Archimeden property follows tht there exists k such tht r k = m + k n > x. Let k be the smllest such number (the exists smllest k becuse every finite set hs minimum), i.e., r k > x nd r k 1 x.

Rel numbers 35 Hence, which completes the proof. x < r k = r k 1 + 1 n x + 1 n < y, 1/" q If we strt #om n integer poin$ on the le% of x nd proceed by rtionl steps of size less thn &y'x(, then w) rech rtionl point between x nd y.! x y The following proposition provides useful fct bout complete ordered fields. We will use it lter in the course. Proposition 1.31 החתכים )) ((למת Let A, B F be two non-empty sets in complete ordered field, such tht ( A b B)( b). Then the following three sttements re equivlent: ➀ (!M F) : ( A b B)( M b). ➁ sup A = inf B. ➂ ( ɛ > 0)( A, b B) : (b < ɛ). Comment: The clim is not tht the three items follow from the given dt. The clim is tht ech sttement implies the two other. Proof : Suppose tht the first sttement holds. This mens tht M is both n upper bound for A nd lower bound for B. Thus, sup A M inf B.

36 Chpter 1 This implies t once tht sup A inf B. If sup A < inf B, then ny number m in between would stisfy < m < b for ll A, b B, contrdicting the ssumption, hence 1 implies 2. By the properties of the infimum nd the supremum, Tht is, i.e., ( ɛ > 0)( A, b B) such tht Thus, 2 implies 3. ( ɛ > 0)( A) : ( > sup A ɛ/2) ( ɛ > 0)( b B) : (b < inf B + ɛ/2). ( ɛ > 0)( A) : ( + ɛ/2 > sup A) ( ɛ > 0)( b B) : (b ɛ/2 < inf B), b < (inf B + ɛ/2) (sup A ɛ/2) = ɛ. It remins to show tht 3 implies 1. Suppose M ws not unique, i.e., there were M 1 < M 2 such tht ( A b B)( M 1 < M 2 b). Set m = (M 1 + M 2 )/2 nd ɛ = (M 2 M 1 ). Then, ( A, b B) ( m ɛ/2, b m + ɛ/2), } {{ } b (m+ɛ/2) (m ɛ/2)=ɛ i.e., ( ɛ > 0) : ( A b B)(b > ɛ), i.e., if 1 does not hold then 3 does not hold. This concludes the proof. (7 hrs, 2009) (10 hrs, 2013) Exercise 1.16 In this exercise you my ssume tht 2 is irrtionl. ➀ Show tht if + b 2 = 0, where, b Q, then = b = 0. ➁ Conclude tht + b 2 = c + d 2, where, b, c, d Q, then = c nd b = d. ➂ Show tht if + b 2 Q, where, b Q, then b = 0.

Rel numbers 37 ➃ Show tht the irrtionl numbers re dense, nmely tht there exists n irrtionl number between every two. ➄ Show tht the set { + b 2 :, b Q} with the stndrd ddition nd multipliction opertions is field. Exercise 1.17 Show tht the set is dense. {m/2 n : m Z, n N} Exercise 1.18 Let A, B R be non-empty sets. Prove or disprove ech of the following sttements: ➀ If A B nd B is bounded from bove then A is bounded from bove. ➁ If A B nd B is bounded from below then A is bounded from below. ➂ If every b B is n upper bound for A then every A is lower bound form B. ➃ A is bounded from bove if nd only if A Z is bounded from bove. Exercise 1.19 Let A R be non-empty set. Determine for ech of the following sttements whether it is equivlent, contrdictory, implied by, or implying the sttement tht s = sup A, or none of the bove: ➀ s for ll A. ➁ For every A there exists r R, such tht < t < s. ➂ For every x R such tht x > s there exists n A such tht s < < x. ➃ For every finite subset B A, s mx B. ➄ s is the supremum of A {s}. 1.8 Rtionl powers We defined the integer powers recursively, x 1 = x x k+1 = x x k. We lso define for x 0, x 0 = 1 nd x n = 1/x n.

38 Chpter 1 Proposition 1.32 (Properties of integer powers) Let x, y 0 nd m, n Z, then ➀ x m x n = x m+n. ➁ (x m ) n = x mn. ➂ (xy) n = x n y n. ➃ 0 < α < β nd n > 0 implies α n < β n. ➄ 0 < α < β nd n < 0 implies α n > β n. ➅ α > 1 nd n > m implies α n > α m. ➆ 0 < α < 1 nd n > m implies α n < α m. Proof : Do it. (10 hrs, 2011) Definition 1.9 Let x > 0. An n-th root of x is positive number y, such tht y n = x. It is esy to see tht if x hs n n-th root then it is unique (since y n = z n, would imply tht y = z). Thus, we denote it by either n x, or by x 1/n. Theorem 1.1 (Existence nd uniqueness of roots) ( x > 0 n N)(!y > 0) : (y n = x). (10 hrs, 2010) Proof : Consider the set S = {z 0 : z n < x}. This is non-empty set (it contins zero) nd bounded from bove, since if x 1 nd z S then z n < x 1, which implies tht z 1 if x > 1 nd z S then z n < x < x n, which implies tht z < x,

Rel numbers 39 It follows tht regrdless of the vlue of x, z S implies tht z mx(1, x), nd the ltter is hence n upper bound for S. As consequence of the xiom of completeness, there exists unique y = sup S, which is the nturl suspect for being n n-th root of x. Indeed, we will show tht y n = x. Clim: y is positive Indeed, 0 < x 1 + x < 1, hence ( x ) n x < 1 + x 1 + x < x. Thus, x/(1 + x) S, which implies tht 0 < x 1 + x y. Clim: y n x Suppose, by contrdiction tht y n < x. We will show by contrdiction tht y is not n upper bound for S by showing tht there exists n element of S tht is greter thn y. We will do it by showing the existence of n ɛ > 0 such tht y ( 1 ɛ S i.e. y ) n < x. 1 ɛ This mens tht we look for n ɛ > 0 stisfying y n x < (1 ɛ)n. Since we cn choose ɛ > 0 t will, we cn tke it smller thn 1, in which cse (1 ɛ) n (1 ɛn) If we choose ɛ sufficiently smll such tht 1 ɛn > yn x, then forteriori (1 ɛ) n > y n /x, nd this will be stisfied if we choose ɛ < 1 yn /x, n which is possible becuse the right hnd side is positive. Thus, we found number greter thn y, whose n-th power is less thn x, i.e., in S. This contrdicts the ssumption tht y is n upper bound for S.

40 Chpter 1 Clim: y n x Suppose, by contrdiction tht y n > x. This time we will show tht there exists positive number less thn y whose n-th power is greter then x, i.e., it is n upper bound for S, contrdicting the ssumption tht y is the lest upper bound for S. To find such number we will show tht there exists n ɛ > 0, such tht [(1 ɛ)y] n > x, i.e., (1 ɛ) n > x y n. Once gin, we cn ssume tht ɛ < 1, in which cse by the Bernoulli inequlity (1 ɛ) n (1 ɛn), so tht if (1 ɛn) > x y n, then forteriori (1 ɛ) n > x/y n. Thus, we need 0 < ɛ < 1 to stisfiy ɛ < 1 x/yn, n which is possible becuse the right hnd side is positive. From the two inequlities follows (by trichotomy) tht y n = x. Hving shown tht every positive number hs unique n-th root, we my proceed to define rtionl powers. Definition 1.10 Let r = m/n Q, then for ll x > 0 x r = (x m ) 1/n. There is one little deliccy with this definition. Recll tht rtionl numbers do not hve unique representtion s the rtio of two integers. We thus need to show tht the bove definition is independent of the representtion 17. In other words, if d = bc, with, b, c, d Z, b, d 0, then for ll x > 0 x /b = x c/d. 17 This is not obvious. Suppose we wnted to define the notion of even/odd for rtionl numbers, by sying tht rtionl number is even if its numertor is even. Such s definition would imply tht 2/6 is even, but 1/3 (which is the sme number!) is odd.

Rel numbers 41 This is non-trivil fct. We need to prove tht d = bc implies tht sup{z > 0 : z b < x } = sup{z > 0 : z d < x c }. The rguments goes s follows: (x /b ) bd = (((x ) 1/b ) b ) d = (x ) d = x d = x bc = (x c ) b = (((x c ) 1/d ) d ) b = (x c/d ) bd, hence x /b = x c/d. (8 hrs, 2009) Proposition 1.33 (Properties of rtionl powers) Let x, y > 0 nd r, s Q, then ➀ x r x s = x r+s. ➁ (x r ) s = x rs. ➂ (xy) r = x r y r. ➃ 0 < α < β nd r > 0 implies α r < β r. ➄ 0 < α < β nd r < 0 implies α r > β r. ➅ α > 1 nd r > s implies α r > α s. ➆ 0 < α < 1 nd r > s implies α r < α s. Proof : We re going to prove only two items. Strt with the first. Let r = /b nd s = c/d. Using the (proved!) lws for integer powers, (x /b x c/d ) bd = (x /b ) bd (x c/d ) bd = = x d x bc = x d+bc = (x /b+c/d ) bd, from which follows tht x /b x c/d = x /b+c/d. Tke then the fourth item. Let r = /b. We lredy know tht α < β. Thus it remins to show α 1/b < β 1/b. This hs to be becuse α 1/b β 1/b would hve implied (from the rules for integer powers!) tht α β. (12 hrs, 2013)

42 Chpter 1 1.9 Rel-vlued powers Delyed to much lter in the course. 1.10 Addendum Existence of non-archimeden ordered fields Does there exist ordered fields tht do not stisfy the Archimeden property? The nswer is positive. We will give one exmple, bering in mind tht some of the rguments rely on lter mteril. Consider the set of rtionl functions. It is esy to see tht this set forms field with respect to function ddition nd multipliction, with f (x) 0 for dditive neutrl element nd f (x) 1 for multiplictive neutrl element. Ignore the fct tht rtionl function my be undefined t finite collection of points. The nturl elements re the constnt functions, f (x) 1, f (x) 2, etc. We endow this set with n order reltion by defining the set P of positive elements s the rtionl functions f tht re eventully positive for x sufficiently lrge. We hce f > g if f (x) > g(x) for x sufficiently lrge. It is esy to see tht the set of nturl elements re bounded, sy, by the element f (x) = x (which obviously belongs to the set). Connected set A set A R will be sid to be connected (קשיר ) if x, y A nd x < z < y implies z A (tht is, every point between two elements in the set is lso n the set). We stte tht (non-empty) connected set cn only be of one of the following forms: {}, (, b), [, b), (, b], [, b] (, ), [, ), (, b), (, b], (, ). Tht is, it cn only be single point, n open, closed or semi-open segments, n open or closed ry, or the whole line. Exercise 1.20 Prove it.

Chpter 2 Functions 2.1 Bsic definitions Wht is function? There is stndrd wy of defining functions, but we will delibertely be little less forml thn perhps we should, nd define function s mchine, which when provided with number, returns number (only one, nd lwys the sme for the sme input). In other words, function is rule tht ssigns rel numbers to rel numbers. A function is defined by three elements: ➀ A domin :(תחוּם ) subset of R. The numbers which my be fed into the mchine. ➁ A rnge :(טווח ) nother subset of R. Numbers tht my be emitted by the mchine. We do not exclude the possibility tht some of these numbers my never be returned. We only require tht every number returned by the function belongs to its rnge. ➂ A trnsformtion rule.(העתקה ) The crucil point is tht to every number.(חד ערכיוּת ) in its domin corresponds one nd only one number in its rnge We normlly denote functions by letters, like we do for rel numbers (nd for ny other mthemticl entity). To void mbiguities, the function hs to be defined properly. For exmple, we my denote function by the letter f. If A R is its domin, nd B R is its rnge, we write f : A B ( f mps the set A into the set B). The trnsformtion rule hs to specify wht number in B is ssigned by the

44 Chpter 2 function to ech number x A. We denote the ssignment by f (x) (the function f evluted t x). Tht the ssignment rule is ssign f (x) to x is denoted by f : x f (x) (pronounced f mps x to f (x) ) 1. Definition 2.1 Let f : A B be function. Its imge (תמונה ) is the subset of B of numbers tht re ctully ssigned by the function. Tht is, Imge( f ) = {y B : x A : f (x) = y}. The function f is sid to be onto B (על ) if B is its imge. f is sid to be one-toone חד ערכית ) (חד if to ech number in its imge corresponds unique number in its domin, i.e., Exmples: ( y Imge( f ))(!x A) : ( f (x) = y). ➀ A function tht ssigns to every rel number its squre. If we denote the function by f, then We my lso write f (x) = x 2. f : R R nd f : x x 2. We should not sy, however, tht the function f is x 2. In prticulr, we my use ny letter other thn x s n rgument for f. Thus, the functions f : R R, defined by the trnsformtion rules f (x) = x 2, f (t) = t 2, f (α) = α 2 nd f (ξ) = ξ 2 re identicl 2. Also, it turns out tht the function f only returns non-negtive numbers. There is however nothing wrong with the definition of the rnge s the whole of R. We could limit the rnge to be the set [0, ), but not to the set [1, 2]. ➁ A function tht ssigns to every w ±1 the number (w 3 + 3w + 5)/(w 2 1). If we denote this function by g, then g : R \ {±1} R nd g : w w3 + 3w + 5. w 2 1 1 Progrmmers: think of f : A B s defining the type or syntx of the function, nd of f : x f (x) s defining the ction of the function. 2 It tkes gret skill not to confuse the Greek letters ζ (zet) nd ξ (xi).

Functions 45 ➂ A function tht ssigns to every 17 x π/3 its squre. This function differs from the function in the first exmple becuse the two functions do not hve the sme domin (different syntx but sme routine ). ➃ A function tht ssigns to every rel number the vlue zero if it is irrtionl nd one if it is rtionl 3. This function is known s the Dirichlet function. We hve f : R {0, 1}, with 0 x is irrtionl f : x 1 x is rtionl. ➄ A function defined on the domin such tht A = {2, 17, π 2 /17, 36/π} { + b 2 :, b Q}, 5 x = 2 36/π x = 17 x 28 x = π 2 /17 or 36/π 16 otherwise. The rnge my be tken to be R, but the imge is {6, 16, 28, 36/π}. ➅ A function defined on R \ Q (the irrtionl numbers), which ssigns to x the number of 7 s in its deciml expnsion, if this number is finite. If this number is infinite, then it returns π. This exmple differs from the previous ones in tht we do not hve n ssignment rule in closed form (how the heck do we compute f (x)?). Nevertheless it provides legl ssignment rule 4. ➆ For every n N we my define the n-th power function f n : R R, by f n : x x n. Here gin, we will void referring to the function x n. The function f 1 : x x is known s the identity function, often denoted by Id, nmely Id : R R, Id : x x. 3 This function is going to be the course s fvorite to disply counter exmples. 4 We limited the rnge of the function to the irrtionls becuse rtionl numbers my hve non-unique deciml representtion, e.g., 0.7 = 0.6999999....

46 Chpter 2 ➇ There re mny functions tht you ll know since high school, such s the sine, the cosine, the exponentil, nd the logrithm. These function require creful, sometimes complicted, definition. It is our choice in the present course to ssume tht the mening of these functions (long with their domins nd rnges) re well-understood. Given severl functions, they cn be combined together to form new functions. For exmple, functions form vector spce over the rels. Let f : A R nd g : B R be given functions, nd, b R. We my define new function 5 f + b g : A B R, f + b g : x f (x) + b g(x). This is vector field whose zero element is the zero function, x 0; given function f, its inverse is f = ( 1) f. Moreover, functions form n lgebr. We my define the product of two functions, f g : A B R, f g : x f (x)g(x), s well s their quotient, f /g : A {z B : g(z) 0} R, f /g : x f (x)/g(x), (12 hrs, 2011) (13 hrs, 2013) A third opertion tht combines two functions is composition. Let f : A B nd g : B C. We define g f : A C, g f : x g( f (x)). For exmple, if f is the sine function nd g is the squre function, then 6 g f : ξ sin 2 ξ nd f g : ζ sin ζ 2, i.e., composition is non-commuttive. On the other hnd, composition is ssocitive, nmely, ( f g) h = f (g h). 5 This is not trivil. We re dding mchines, not numbers. 6 It is definitely hbit to use the letter x s generic rgument for functions. Bewre of fixtions! Any letter including Ξ is s good.

Functions 47 Note tht for every function f, Id f = f Id = f, so tht the identity is the neutrl element with respect to function composition. This should not be confused with the fct tht x 1 is the neutrl element with respect to function multipliction. Exmple: Consider the function f tht ssigns the rule This function cn be written s f : x x + x2 + x sin x 2 x sin x + x sin 2 x. f = Id + Id Id + Id sin (Id Id). Id sin + Id sin sin Exmple: Recll the n-th power functions f n. A function P is clled polynomil of degree n if there exist rel numbers ( i ) n i=0, with n 0, such tht P = n k f k. k=0 The union over ll n s of polynomils of degree n is the set of polynomils. A function is clled rtionl if it is the rtio of two polynomils. (10 hrs, 2009) Exercise 2.1 The following exercise dels with the lgebr of functions. Consider the three following functions, ll three defined on R. s(x) = sin x P(x) = 2 x nd S (x) = x 2, ➀ Express the following functions without using the symbols s, P, nd S : (i) (s P)(y), (ii) (S s)(ξ), nd (iii) (S P s)(t) + (s P)(t).

48 Chpter 2 ➁ Write the following functions using only the symbols s, P, nd S, long with the binry opertors +,, nd : (i) f (x) = 2 sin x, (ii) f (t) = 2 2x, (iii) f (u) = sin(2 u + 2 u2 ), (iv) f (y) = sin(sin(sin(2 2sin y ))), nd (v) f () = 2 sin2 + sin( 2 ) + 2 sin(2 +sin ). Exercise 2.2 An open segment I is clled symmetric if there exists n > 0 such tht I = (, ). Let f : I R where I is symmetric segment; the function f is clled even if f ( x) = f (x) for every x I, nd odd if f ( x) = f (x) for every x I. Let f, g : I R, where I is symmetric segments, nd consider the four possibilities of f, g being either even or odd. In ech cse, determine whether the specified function is even, odd, or not necessrily either two. Explin your sttement, nd where pplicble give counter exmple: ➀ f + g. ➁ f g. ➂ f g (here we ssume tht g : I J nd f : J R, where I nd J re symmetric segments). Exercise 2.3 Show tht ny function f : R R cn be written s f = E + O, where E is even nd O is odd. Prove then tht this decomposition is unique. Tht is, if f = E + O = E + O, then E = E nd O = O. Exercise 2.4 The following exercise dels with polynomils. ➀ Prove tht for every polynomil of degree 1 nd for every rel number there exists polynomil g nd rel number b, such tht f (x) = (x )g(x) + b (hint: use induction on the degree of the polynomil). ➁ Prove tht if f is polynomil stisfying f () = 0, then there exists polynomil g such tht f (x) = (x )g(x). ➂ Conclude tht if f is polynomil of degree n, then there re t most n rel numbers, such tht f () = 0 (t most n roots). Exercise 2.5 Let, b, c R nd 0, nd let f : A B be given by f : x x 2 + bx + c.

Functions 49 ➀ Suppose B = R. Find the lrgest segment A for which f is one-to-one. ➁ Suppose A = R. Find the lrgest segment B for which f is onto. ➂ Find segments A nd B s lrge s possible for which f is one-to-one nd onto. Exercise 2.6 Let f stisfy the reltion f (x + y) = f (x) + f (y) for ll x, y R. ➀ Prove inductively tht for ll x 1,..., x n R. f (x 1 + x 2 + + x n ) = f (x 1 ) + f (x 2 ) + + f (x n ) ➁ Prove the existence of rel number c, such tht f (x) = cx for ll x Q. (Note tht we clim nothing for irrtionl rguments. ) 2.2 Grphs As you know, we cn ssocite with every function grph. Wht is grph? A drwing? A grph hs to be thought of s subset of the plne. For function f : A B, we define the grph of f to be the set Grph( f ) = {(x, y) : x A, y = f (x)} A B. The defining property of the function is tht it is uniquely defined, i.e., (x, y) Grph( f ) nd (x, z) Grph( f ) implies y = z. The function is one-to-one if (x, y) Grph( f ) nd (w, y) Grph( f ) implies x = w. It is onto B if ( y B)( x A) : ((x, y) Grph( f )). The definition we gve to function s n ssignment rule is, strictly speking, not forml one. The stndrd wy to define function is vi its grph. A function is grph subset of the Crtesin product spce A B.

50 Chpter 2 Comment: In fct, this is the generl wy to define functions. Let A nd B be two sets (of nything!). A function f : A B is subset of the Crtesin product Grph( f ) A B = {(, b) : A, b B}, stisfying ( A)(!b B) : ((, b) Grph( f )). (12 hrs, 2010) 2.3 Limits Definition 2.2 Let x R. A neighborhood of x (סביבה ) is n open segment (, b) tht contins the point x (note tht since the segment is open, x cnnot be boundry point). A punctured neighborhood of x מנוּק ב ת ) (סביבה is set (, b) \ {x} where < x < b. punctured neighborhood neighborhood נקוּדה) Definition 2.3 Let A R. A point A is clled n interior point of A (פנימית if it hs neighborhood contined in A. Nottion: We will mostly del with symmetric neighborhoods (whether punctured or not), i.e., neighborhoods of of the form {x : x < δ} for some δ > 0. We will introduce the following nottions for neighborhoods, B(, δ) = ( δ, + δ) B (, δ) = {x : 0 < x < δ}.

Functions 51 We will lso define one-sided neighborhoods, B + (, δ) = [, + δ) B +(, δ) = (, + δ) B (, δ) = ( δ, ] B (, δ) = ( δ, ). Exmple: For A = [0, 1], the point 1/2 is n interior point, but not the point 0. Exmple: The set Q R hs no interior points, nd neither does its complement, R \ Q. In this section we explin the mening of the limit of function t point. Informlly, we sy tht: The limit of function f t point is l, if we cn mke f ssign vlue s close to l s we wish, by mking its rgument sufficiently close to (excluding the vlue itself). Note tht the function does not need to be equl to l t the point ; in fct, it does not even need to be defined t the point. Exmple: Consider the function f : R R f : x 3x. We clim tht the limit of this function t the point 5 is 15. This mens tht we cn mke f (x) be s close to 15 s we wish, by mking x sufficiently close to 5, with 5 itself being excluded. Suppose you wnt f (x) to differ from 15 by less thn 1/100. This mens tht you wnt 15 1 100 This requirement is gurnteed if < f (x) = 3x < 15 + 1 100. 5 1 300 < x < 5 + 1 300.

52 Chpter 2 Thus, if we tke x to differ from 5 by less thn 1/300 (but more thn zero!), then we re gurnteed to hve f (x) within the desired rnge. Since we cn repet this construction for ny number other thn 1/100, then we conclude tht the limit of f t 5 is 15. We write, lthough the more common nottion is 7 lim 5 f = 15. lim f (x) = 15. x 5 We cn be more precise. Suppose you wnt f (x) to differ from 15 by less thn ɛ, for some ɛ > 0 of your choice. In other words, you wnt f (x) 15 = 3x 15 < ɛ. This is gurnteed if x 5 < ɛ/3, thus given ɛ > 0, choosing x within symmetric punctured neighborhood of 5 of rdius ɛ/3 gurntees tht f (x) is within distnce of ɛ from 15. This exmple insinutes wht would be forml definition of the limit of function t point: Definition 2.4 Let f : A B with A n interior point. We sy tht the limit of f t is l, if for every ɛ > 0, there exists δ > 0, such tht 8 In forml nottion, f (x) l < ɛ for ll x B (, δ). ( ɛ > 0)( δ > 0) : ( x B (, δ))( f (x) l < ɛ). 7 The choice of not using the common nottion becomes cler when you relize tht you could s well write lim f (ℵ) = 15 or lim f (ξ) = 15. ℵ 5 ξ 5 8 The gme is you give me ɛ nd I give you δ in return.

Functions 53 Plyer A: This is my!. Plyer B: No problem. Here is my ". l!! l Since Plyer B cn find " for every choice of! mde by Plyer A,it fo!ows th" the limit of f t is l. " Exmple: Consider the squre function f : R R, f : x x 2. We will show tht lim 3 f = 9. By definition, we need to show tht for every ɛ > 0 we cn find δ > 0, such tht f (x) 9 < ɛ whenever x B (3, δ). Thus, the gme it to respond with n pproprite δ for every ɛ. To find the pproprite δ, we exmine the condition tht needs to be stisfied: If we impose tht x 3 < δ then f (x) 9 = x 3 x + 3 < ɛ. f (x) 9 < x + 3 δ. If the x + 3 wsn t there we would hve set δ = ɛ nd we would be done. Insted, we note tht by the tringle inequlity, so tht for ll x 3 < δ, x + 3 = x 3 + 6 x 3 + 6, f (x) 9 x 3 ( x 3 + 6) < (6 + δ)δ.

54 Chpter 2 Now recll: given ɛ we hve the freedom to choose δ > 0 such to mke the righthnd side less or equl thn ɛ. We my freely require for exmple tht δ 1, in which cse f (x) 9 < (6 + δ)δ 7δ. If we furthermore tke δ ɛ/7, then we rech the desired gol. To summrize, given ɛ > 0 we tke δ = min(1, ɛ/7), in which cse ( x : 0 < x 3 < δ), f (x) 9 = x 3 x + 3 x 3 ( x 3 + 6) < (6 + δ)δ (6 + 1) ɛ 7 = ɛ. This proves (by definition) tht lim 3 f = 9. We cn show, in generl, tht lim f = 2 for ll R. [do it!] (13 hrs, 2010) (15 hrs, 2013) Exmple: The next exmple is the function f : (0, ), f : x 1/x. We re going to show tht for every > 0, lim f = 1. First fix ; it is not vrible. Let ɛ > 0 be given. We first observe tht f (x) 1 = 1 x 1 x = x. We need to be creful tht the domin of x does not include zero. We strt by requiring tht x < /2, which t once implies tht x > /2, hence f (x) 1 2 x <. 2 If we further require tht x < 2 ɛ/2, then f (x) 1/ < ɛ. To conclude, f (x) 1 < ɛ whenever x B (, δ), for δ = min(/2, 2 ɛ/2). (14 hrs, 2011)

Functions 55 Exmple: Consider next the Dirichlet function, f : R R, 0 x is irrtionl f : x 1 x is rtionl. We will show tht f does not hve limit t zero. To show tht for ll l, we need to show the negtion of i.e., lim 0 f l, ( l R) : ( ɛ > 0)( δ > 0) : ( x B (0, δ))( f (x) B(l, ɛ)), ( l R)( ɛ > 0) : ( δ > 0)( x B (0, δ)) : ( f (x) l ɛ). Indeed, tke ɛ = 1/4. Then no mtter wht δ is, there exist points 0 < x < δ for which f (x) = 0 nd points for which f (x) = 1, i.e., No l cn stisfy both ( δ > 0)( x, y B (0, δ)) : ( f (x) = 0, f (y) = 1). 0 l < 1 4 nd 1 l < 1 4. I.e., (12 hrs, 2009) ( l R, δ > 0)( x B (0, δ)) : ( f (x) B(l, 1 4 )). Comment: It is importnt to stress wht is the negtion tht the limit f t is l: There exists n ɛ > 0, such tht for ll δ > 0, there exists n x, which stisfies 0 < x < δ, but not f (x) l < ɛ. Exmple: Consider, in contrst, the function f : R R, 0 x is irrtionl f : x x x is rtionl.

56 Chpter 2 Here we show tht lim 0 f = 0. Indeed, let ɛ > 0 be given, then by tking δ = ɛ, x B (0, δ) implies f (x) 0 < ɛ. Hving forml definition of limit, nd hving seen number of exmple, we re in mesure to prove generl theorems bout limits. The first theorem sttes tht limit, if it exists, is unique. Exercise 2.7 Prove tht if ( ɛ > 0)( δ > 0) : ( x : 0 < x 3 < δ)( f (x) 14 < ɛ), then nd lso ( ɛ > 0)( δ > 0) : ( x : 0 < x 3 < δ)( f (x) 14 < ɛ/2), ( ɛ > 0)( δ > 0) : ( x : 0 < x 3 < δ)( f (x) 14 < ɛ). Theorem 2.1 (Uniqueness of the limit) A function f : A B hs t most one limit t ny interior point A. Proof : Suppose, by contrdiction, tht lim f = l nd lim f = m, with l m. By definition, there exists δ 1 > 0 such tht f (x) l < 1 4 l m whenever x B (, δ 1 ), nd there exists δ 2 > 0 such tht f (x) m < 1 4 l m whenever x B (, δ 2 ).

Functions 57 Tke δ = min(δ 1, δ 2 ). Then, whenever x B (, δ), l m = ( f (x) m) ( f (x) l) f (x) l + f (x) m < 1 l m, 2 which is contrdiction. Plyer A: This is my!. Plyer B: I give up. No " fits. # l!! # l!! Since Plyer B cnnot find " for prticulr choice of! mde by Plyer A,it fo!ows th" the limit of f t cnnot be both l nd #. " Exercise 2.8 In ech of the following items, determine whether the limit exists, nd if it does clculte its vlue. You hve to clculte the limit directly from its definition (without using, for exmple, limits rithmetic see below). ➀ lim 2 f, where f (x) = 7x. ➁ lim 0 f, where f (x) = x 2 + 3. ➂ lim 0 f, where f (x) = 1/x. ➃ lim f, where nd R. ➄ lim f, where f (x) = x. ➅ lim f, where f (x) = x. 1 x > 0 f (x) = 0 x = 0 1 x < 0,

58 Chpter 2 Exercise 2.9 Let I be n open intervl in R, with I. Let f, g : I \ {} R (i.e., defined on punctured neighborhood of ). Let J I be n open intervl contining the point, such tht f (x) = g(x) for ll x J \ {}. Show tht lim f exists if nd only if lim g exists, nd if they do exist they re equl. (This exercise shows tht the limit of function t point only depends on its properties in neighborhood of tht point). Exercise 2.10 The Riemnn function R : R R is defined s 1/q x = p/q Q R(x) = 0 x Q, where the representtion p/q is in reduced form (for exmple R(4/6) = R(2/3) = 1/3). Note tht R(0) = R(0/1) = 1. Show tht lim R exists everywhere nd is equl to zero. (17 hrs, 2013) We hve seen bove vrious exmples of limits. In ech cse, we hd to work hrd to prove wht the limit ws, by showing tht the definition ws stisfied. This becomes imprcticl when the functions re more complex. Thus, we need to develop theorems tht will mke our tsk esier. But first, lemm: Lemm 2.1 Let l, m R nd let ɛ > 0. Then, ➀ x l < ɛ 2 nd y m < ɛ 2 imply (x + y) (l + m) < ɛ. ➁ ( ) ɛ x l < min 1, 2( m + 1) nd ( ) ɛ y m < min 1, 2( l + 1) imply xy lm < ɛ.

Functions 59 ➂ If m 0 nd ( ) m y m < min 2, m 2 ɛ, 2 then y 0 nd 1 y 1 m < ɛ. (15 hrs, 2011) Proof : The first item is obvious (the tringle inequlity). Consider the second item. If the two conditions re stisfied then hence, x = x l + l < x l + l < l + 1, xy lm = x(y m) + m(x l) x y m + m x l < ɛ l + 1 2( l + 1) + ɛ m 2( m + 1) < ɛ. For the third item it is cler tht y m < m /2 implies tht y > m /2, nd in prticulr, y 0. Then 1 y 1 m = y m y m < m 2 ɛ 2 m ( m /2) = ɛ. (13 hrs, 2009) (15 hrs, 2010) Theorem 2.2 (Limits rithmetic) Let f nd g be two functions defined in neighborhood of, such tht lim f = l nd lim g = m.

60 Chpter 2 Then, lim ( f + g) = l + m nd lim ( f g) = l m. If, moreover, l 0, then lim ( ) 1 = 1 f l. Proof : This is n immedite consequence of the bove lemm. Since, given ɛ > 0, we know tht there exists δ > 0, such tht 9 f (x) l < ɛ 2 nd g(x) m < ɛ 2 whenever x B (, δ). It follows tht ( f (x) + g(x)) (l + m) < ɛ whenever x B (, δ). There lso exists δ > 0, such tht ( ) ɛ f (x) l < min 1, 2( m + 1) whenever x B (, δ). It follows tht nd ( ) ɛ g(x) m < min 1, 2( l + 1) f (x)g(x) lm < ɛ whenever x B (, δ). Finlly, there exists δ > 0, such tht ( ) l f (x) l < min 2, l 2 ɛ 2 whenever x B (, δ). It follows tht 1 f (x) 1 l < ɛ whenever x B (, δ). Comment: It is importnt to point out tht the fct tht f + g hs limit t does not imply tht either f or g hs limit t tht point; tke for exmple the functions f (x) = 5 + 1/x nd g(x) = 6 1/x t x = 0. 9 This δ > 0 is lredy the smllest of the two δ s corresponding to ech of the functions.

Functions 61 Exmple: Wht does it tke now to show tht 10 We need to show tht for ll c R x 3 + 7x 5 lim x x 2 + 1 = 3 + 7 5 2 + 1. lim c = c, nd tht for ll k N, lim x k = k, x which follows by induction once we show tht for k = 1. Exercise 2.11 Show tht lim x n = n, R, n N, x directly from the definition of the limit, i.e., without using limits rithmetic. Exercise 2.12 Let f, g be defined in punctured neighborhood of point. Prove or disprove: ➀ If lim f nd lim g do not exist, then lim ( f + g) does not exist either. ➁ If lim f nd lim g do not exist, then lim ( f g) does not exist either ➂ If both lim f nd lim ( f + g) exist, then lim g lso exists. ➃ If both lim f nd lim ( f g) exist, then lim g lso exists. ➄ If lim f exists nd lim ( f +g) does not exist, then lim ( f +g) does no exist. ➅ If g is bounded in punctured neighborhood of nd lim f = 0, then lim ( f g) = 0. We conclude this section by extending the notion of limit to tht of one-sided limit. 10 This is the stndrd nottion to wht we write in these notes s f : ξ ξ3 + 7ξ 5 ξ 2 + 1 nd lim f = 3 + 7 5 2 + 1.

62 Chpter 2 Definition 2.5 Let f : A B with A n interior point. We sy tht the limit on the right מימין ) (גבול of f t is l, if for every ɛ > 0, there exists δ > 0, such tht f (x) l < ɛ whenever x B +(, δ). We write, lim f = l. +.(גבול משמאל ) An nlogous definition is given for the limit on the left Plyer A: This is my!. Plyer B: No problem. Here is my ". l!! l Since Plyer B cn find " for every choice of! mde by Plyer A,it fo!ows th" the right#limit of f t is l. " Exercise 2.13 Prove tht lim f exists if nd only lim f = lim + Exercise 2.14 We denote by f : x x the lower integer vlue of x. ➀ Drw the grph of this function. ➁ Find lim f nd lim f + for R. Prove it bsed on the definition of one-sided limits. Exercise 2.15 Show tht f : (0, ) R, f : x x stisfies for ll > 0. lim f =, f.

Functions 63 2.4 Limits nd order In this section we will prove number of properties pertinent to limits nd order. First definition: Definition 2.6 Let f : A B nd A n interior point. We sy tht f is loclly bounded מקומית ) (חסומה ner if there exists punctured neighborhood U of, such tht the set { f (x) : x U} is bounded. Equivlently, there exists δ > 0 such tht the set is bounded. { f (x) : x B (, δ)} Comment: As in the previous section, we consider punctured neighborhoods of point, nd we don t even cre if the function is defined t the point. Proposition 2.1 Let f nd g be two functions defined in punctured neighborhood U of point. Suppose tht f (x) < g(x) x U, nd tht lim f = l nd lim g = m. Then l m. Comment: The very fct tht f nd g hve limits t is n ssumption. Comment: Note tht even though f (x) < g(x) is strict inequlity, the resulting inequlity of the limits is in wek sense. To see wht this must be the cse, consider the exmple f (x) = x nd g(x) = 2 x. Even though f (x) < g(x) in n open neighborhood of 0, lim 0 f = lim 0 g = 0.

64 Chpter 2 Proof : Suppose, by contrdiction, tht l > g nd set ɛ = 1 (l m). Thus, there 2 exists δ > 0 such tht ( x B (, δ)) : ( f (x) l < ɛ nd g(x) m < ɛ). In prticulr, for every such x, which is contrdiction. (19 hrs, 2013) f (x) > l ɛ = l l m 2 = m + l m 2 = m + ɛ > g(x), Proposition 2.2 Let f nd g be two functions defined in punctured neighborhood U of point. Suppose tht lim f = l nd lim g = m, with l < m. Then there exists δ > 0 such tht ( x B (, δ)) : ( f (x) < g(x)). Comment: This time both inequlities re strong. Proof : Let ɛ = 1 (m l). There exists δ > 0 such tht 2 For every such x, ( x B (, δ)) : ( f (x) l < ɛ nd g(x) m < ɛ). f (x) < l + e = l + 1 2 (m l) = 1 (l + m) 2 g(x) > m e = m 1 2 (m l) = 1 (l + m) 2 hence, f (x) < g(x).

Functions 65 Proposition 2.3 Let f, g, h be defined in punctured neighborhood U of. Suppose tht ( x U) : ( f (x) g(x) h(x)), nd lim f = lim h = l. Then lim g = l. Proof : It is given tht for every ɛ > 0 there exists δ > 0 such tht ( x B (, δ)) : (l ɛ < f (x) nd h(x) < l + ɛ). Then for every such x, l ɛ < f (x) g(x) h(x) < l + ɛ, i.e., (17 hrs, 2011) ( x B (, δ)) : ( g(x) l < ɛ). Proposition 2.4 Let f be defined in punctured neighborhood U of. If the limit lim f = l exists, then f is loclly bounded ner. Proof : Immedite from the definition of the limit. Proposition 2.5 Let f be defined in punctured neighborhood of point. Then, lim f = l if nd only if lim ( f l) = 0.

66 Chpter 2 Proof : Very esy. Proposition 2.6 Let f nd g be defined in punctured neighborhood of point. Suppose tht lim f = 0 wheres g is loclly bounded ner. Then, lim ( f g) = 0. Proof : Very esy. Comment: Note tht we do not require g to hve limit t. With the bove tools, here is nother wy of proving the product property of the rithmetic of limits (this time without ɛ nd δ). Proposition 2.7 (Arithmetic of limits, product) Let f, g be defined in punctured neighborhood of point, nd lim f = l nd lim g = m. Then lim ( f g) = lm. Proof : Write (17 hrs, 2010) f g lm = ( f l) g l (g m). } {{ }}{{}} {{ } limit zero loc. bdd limit zero } {{ }} {{ } limit zero limit zero

Functions 67 2.5 Continuity Definition 2.7 A function f : A R is sid to be continuous t n inner point A, if it hs limit t, nd lim f = f (). Comment: If f hs limit t nd the limit differs from f () (or tht f () is (אי רציפות סליקה ) undefined), then we sy tht f hs removble discontinuity t. Exmples: ➀ We sw tht the function f (x) = x 2 hs limit t 3, nd tht this limit ws equl 9. Hence, f is continuous t x = 3. ➁ We sw tht the function f (x) = 1/x hs limit t ny > 0 nd tht this limit equls 1/. By similr clcultion we could hve shown tht it hs limit t ny < 0 nd tht this limit lso equls 1/. Hence, f is continuous t ll x 0. Since f is not even defined t x = 0, then it is not continuous t tht point. ➂ The function f (x) = x sin 1/x (it ws seen in the tutoring session) is continuous t ll x 0. At zero it is not defined, but if we define f (0) = 0, then it is continuous for ll x R (by the bounded times limit zero rgument). ➃ The function x x Q f : x 0 x Q, is continuous t x = 0, however it is not continuous t ny other point, becuse the limit of f t 0 does not exist. ➄ The elementry functions sin nd cos re continuous everywhere. To prove tht sin is continuous, we need to prove tht for every ɛ > 0 there exists δ > 0 such tht sin x sin < ɛ whenever x B (, δ). We will tke it for grnted t the moment. Our theorems on limit rithmetic imply right wy similr theorems for continuity:

68 Chpter 2 Theorem 2.3 If f : A R nd g : A R re continuous t A then f + g nd f g re continuous t. Moreover, if f () 0 then 1/ f is continuous t. Proof : Obvious. Comment: Recll tht in the definition of the limit, we sid tht the limit of f t is l if for every ɛ > 0 there exists δ > 0 such tht f (x) l < ɛ whenever x B (, δ). There ws n emphsis on the fct tht the point x itself is excluded. Continuity is defined s tht for every ɛ > 0 there exists δ > 0 such tht f (x) f () < ɛ whenever x B (, δ). Note tht we my require tht this be true whenever x B(, δ). There is no need to exclude the point x =. (19 hrs, 2010) With tht, we hve ll the tools to show tht function like, sy, f (x) = sin2 x + x 2 + x 4 sin x 1 + sin 2 x is continuous everywhere in R. But wht bout function like sin x 2. Do we hve the tools to show tht it is continuous. No. We don t yet hve theorem for the composition of continuous functions. Theorem 2.4 Let g : A B nd f : B C. Suppose tht g is continuous t n inner point A, nd tht f is continuous t g() B, which is n inner point. Then, f g is continuous t. Proof : Let ɛ > 0. Since f is continuous t g(), then there exists δ 1 > 0, such tht f (y) f (g()) < ɛ whenever y B(g(), δ 1 ).

Functions 69 Since g is continuous t, then there lso exists δ 2 > 0, such tht g(x) g() < δ 1 or g(x) B(g(), δ 1 ) whenever x B(, δ 2 ). Combining the two we get tht f (g(x)) f (g()) < ɛ whenever B(, δ 2 ). Definition 2.8 A function f is sid to be right-continuous מימין ) (רציפה t if lim + f = f ()..(רציפה משמאל ) A similr definition is given for left-continuity So fr, we hve only delt with continuity t points. Usully, we re interested in continuity on intervls. Definition 2.9 A function f is sid to be continuous on n open intervl (, b) if it is continuous t ll x (, b). It is sid to be continuous on closed intervl [, b] if it is continuous on the open intervl, nd in ddition it is right-continuous t nd left-continuous t b. Comment: We usully think of continuous function s well-behved. should be creful with such interprettions; see for exmple x sin 1 x 0 x f (x) = 0 x = 0. One Exercise 2.16 Suppose f : R R stisfies f ( + b) = f () + f (b) for ll, b R. )We hve lredy seen tht this implies tht f (x) = cx, for some constnt c for ll x Q.) Suppose tht in ddition f is continuous t zero. Prove tht f is continuous everywhere. Exercise 2.17 ➀ Prove tht if f : A B is continuous in A, then so is f (where f (x) def = f (x) ).

70 Chpter 2 ➁ Prove tht if f : A R nd g : A R re continuous then so re mx( f, g) nd min( f, g). Exercise 2.18 Suppose tht g, h : R R re continuous t nd stisfy g() = h(). We define new function tht glues g nd h t : g(x) x f (x) = h(x) x >. Prove tht f is continuous t. Exercise 2.19 Here is nother gluing problem: Suppose tht g is continuous in [, b], tht h is continuous is [b, c], nd g(b) = h(b). Define g(x) x [, b] f (x) = h(x) x [b, c]. Prove tht f is continuous in [, c]. Exmple: Here is nother crzy function due to Johnnes Krl Thome (1840-1921): 1/q x = p/q r(x) = 0 x Q, where x = p/q ssumes tht x is rtionl in reduced form. This function hs the wonderful property of being continuous t ll x Q nd discontinuous t ll x Q (this is becuse its limit is everywhere zero). It took some more time until Vito Volterr proved in 1881 tht there cn be no function tht is continuous on Q nd discontinuous on R \ Q. (21 hrs, 2013) 2.6 Theorems bout continuous functions Theorem 2.5 Suppose tht f is continuous t nd f () > 0. Then there exists neighborhood of in which f (x) > 0. Tht is, there exists δ > 0 such tht f (x) > 0 whenever x B(, δ).

Functions 71 Comment: An nlogous theorem holds if f () < 0. Also, one-sided version cn be proved, whereby if f is right-continuous t nd f () > 0, then f (x) > 0 whenever x B + (, δ). Comment: Note tht continuity is only required t the point. Proof : Since f is continuous t, there exists δ > 0 such tht f (x) f () < f () 2 But, f (x) f () < f () 2 implies whenever x B(, δ). f () f (x) f () f (x) < f () 2, i.e., f (x) > f ()/2 > 0. 3/2 f!" f!" 1/2 f!"! This theorem cn be viewed s lemm for the following importnt theorem: (15 hrs, 2009) (20 hrs, 2010)

72 Chpter 2 Theorem 2.6 (Intermedite vlue theorem ערך הביניים (משפט Suppose f is continuous on n intervl [, b], with f () < 0 nd f (b) > 0. Then, there exists point c (, b), such tht f (c) = 0. (19 hrs, 2011) Comment: Continuity is required on the whole intervls, for consider f : [0, 1] R: 1 0 x < 1 2 f (x) = 1 +1 x 1. 2 Comment: If f () > 0 nd f (b) < 0 the sme conclusion holds for just replce f by ( f ). Proof : Consider the set A = {x [, b] : f (y) < 0 for ll y [, x]}. The set is non empty for it contins. Actully, the previous theorems gurntees the existence of δ > 0 such tht [, +δ) A. This set is lso bounded by b. The previous theorem gurntees even the existence of δ > 0 such tht b δ is n upper bound for A. By the xiom of completeness, there exists number c (, b) such tht c = sup A. f!b" A c b f!"

Functions 73 Suppose it were true tht f (c) < 0. By the previous theorem, there exists δ > 0 such tht f (x) < 0 whenever c x c + δ, Becuse c is the lest upper bound of it follows tht lso f (x) < 0 whenever x < c, i.e., f (x) < 0 for ll x [, c + δ], which mens tht c + δ A, contrdicting the fct tht c is n upper bound for A. Suppose it were true tht f (c) > 0. By the previous theorem, there exists δ > 0 such tht f (x) > 0 whenever c δ x c, which implies tht c δ is n upper bound for A, i.e., c cnnot be the lest upper bound of A. By the trichotomy property, we conclude tht f (c) = 0. Corollry 2.1 If f is continuous on closed intervl [, b], such tht f () < α < f (b), then there exists point c (, b) t which f (c) = α. Proof : Apply the previous theorem for g(x) = f (x) α. (23 hrs, 2013) Exercise 2.20 Suppose tht f is continuous on [, b] with f (x) Q for ll x [, b]. Wht cn be sid bout f? Exercise 2.21 (i) Suppose tht f, g : [, b] R re continuous such tht f () < g() nd f (b) > g(b). Show tht there exists point c [, b] such tht f (c) = g(c). (ii) A fly flies from Tel-Aviv to Or Akiv (long the shortest pth). He strts its wy t 10:00 nd finishes t 16:00. The following dy re mkes his wy bck long the very sme pth, strting gin t 10:00 nd finishing t 16:00. Show tht there will be point long the wy tht he will cross the exct sme time on both dys.

74 Chpter 2 Exercise 2.22 (i) Let f : [0, 1] R be continuous, f (0) = f (1). Show tht there exists n x 0 [0, 1], such tht f (x 2 0) = f (x 0 + 1 ). (Hint: consider 2 g(x) = f (x + 1) f (x).) (ii) Show tht there exists n x 2 0 [0, 1 1 ], such tht n f (x 0 ) = f (x 0 + 1). n Lemm 2.2 If f is continuous t, then there is δ > 0 such tht f hs n upper bound on the intervl ( δ, + δ). Proof : This is obvious, for by the definition of continuity, there exists δ > 0, such tht f (x) B( f (), 1) whenever x B(, δ), i.e., f (x) < f () + 1 on this intervl. Theorem 2.7 (Krl Theodor Wilhelm Weierstrß) If f is continuous on [, b] then it is bounded from bove of tht intervl, tht is, there exists number M, such tht f (x) < M for ll x [, b]. Comment: Here too, continuity is needed on the whole intervl, for look t the counter exmple f : [ 1, 1] R, 1/x x 0 f (x) = 0 x = 0. Comment: It is crucil tht the intervl [, b] be closed. The function f : (0, 1] R, f : x 1/x is continuous on the semi-open intervl, but it is not bounded from bove. Proof : Let A = {x [, b] : f is bounded from bove on [, x]}. By the previous lemm, there exists δ > 0 such tht + δ A. Also A is upper bounded by b, hence there exists c (, b], such tht c = sup A.

Functions 75 We clim tht c = b, for if c < b, then by the previous lemm, there exists neighborhood of c in which f is upper bounded, nd c cnnot be n upper bound for A. CORRECT THE PROOF. NEED TO CONNECT TO CONTINUITY AT b. Comment: In essence, this theorem is bsed on the fct tht if continuous function is bounded up to point, then it is bounded up to little frther. To be ble to tke such increments up to b we need the xiom of completeness. (17 hrs, 2009) (22 hrs, 2010) (20 hrs, 2011) Theorem 2.8 (Weierstrß, Mximum principle המקסימום (עקרון If f is continuous on [, b], then there exists point c [, b] such tht f (x) f (c) for ll x [, b]. Comment: Of course, there is corresponding minimum principle. Proof : We hve just proved tht f is upper bounded on [, b], i.e., the set A = { f (x) : x b} is upper bounded. This set is non-empty for it contins the point f (). By the xiom of completeness it hs lest upper bound, which we denote by α = sup A. We need to show tht this supremum is in fct mximum; tht there exists point c [, b], for which f (c) = α. Suppose, by contrdiction, tht this were not the cse, i.e., tht f (x) < α for ll x [, b]. We define then new function g : [, b] R, g = 1 α f.

76 Chpter 2 This function is defined everywhere on [, b] (since we ssumed tht the denomintor does not vnish), it is continuous nd positive. We will show tht g is not upper bounded on [, b], contrdicting thus the previous theorem. Indeed, since α = sup A: For ll M > 0 there exists y [, b] such tht f (y) > α 1/M. For this y, 1 g(y) > α (α 1/M) = M, i.e., for every M > 0 there exists point in [, b] t which f tkes vlue greter thn M. Comment: Here too we needed continuity on closed intervl, for consider the counter exmples f : [0, 1) R f (x) = x 2, nd x 2 x < 1 f : [0, 1] R f (x) = 0 x = 1. These functions do not ttin mximum in [0, 1]. Exercise 2.23 Let f : [, b] R be continuous, with f () = f (b) = 0 nd M = sup{ f (x) : x b} > 0. Prove tht for every 0 < c < M, the set {x [, b] : f (x) = c} hs t lest two elements. Exercise 2.24 Let f : [, b] R be continuous. Prove or disprove ech of the following sttements: ➀ It is possible tht f (x) > 0 for ll x [, b], nd tht for every ɛ > 0 there exists n x [, b] for which f (x) < ɛ. ➁ There exists n x 0 [, b] such tht for every [, b] x x 0, f (x) < f (x 0 ). Theorem 2.9 If n N is odd, then the eqution f (x) = x n + n 1 x n 1 +... 1 x + 0 = 0 hs root ( solution) for ny set of constnts 0,..., n 1.

Functions 77 Proof : The ide it to show tht existence of points, b for which f (b) > 0 nd f () < 0, nd pply the intermedite vlue theorem, bsed on the fct tht polynomils re continuous. The only technicl issue is to find such points, b in wy tht works for ll choices of 0,..., n 1. Let Then for x > M, M = mx(1, 2n 0,..., 2n n 1 ). f (x) x n = 1 + n 1 x (u + v u v ) 1 n 1 x ( x n > x ) 1 n 1 x + + 0 x n 0 x n 0 x ( x > i ) 1 n 1 2n n 1 0 2n 0 1 = 1 n 2n > 0. Since the sign of x n is the sme s the sign of x, it follows tht f (x) is positive for x M nd negtive for x M, hence there exists < c < b such tht f (c) = 0. (18 hrs, 2009) The next theorem dels with the cse where n is even: Theorem 2.10 Let n be even. Then the function f (x) = x n + n 1 x n 1 +... 1 x + 0 = 0 hs minimum. Nmely, there exists c R, such tht f (c) f (x) for ll x R. Proof : The ide is simple. We re going to show tht there exists b > 0 for which f (x) > f (0) for ll x b. Then, looking t the function f in the intervl [ b, b], we know tht it ssumes minimum in this intervl, i.e., tht there exists c [ b, b] such tht f (c) f (x) for ll x [ b, b]. In prticulr, f (c) f (0),

78 Chpter 2 which in turn is less thn f (x) for ll x [ b, b]. It follows tht f (c) f (x) for ll x R. It remins to find such b. Let M be defined s in the previous theorem. Then, for ll x > M, using the fct tht n is even, ( f (x) = x n 1 + n 1 + + ) 0 x x n 0 x n ( 1 n 1 x x n ( x n 1 n 1 ) 0 x x x n ( 1 n 1 2n n 1 0 2n 0 = 1 2 xn. ) ) Let then b > mx(m, n 2 f (0) ), from which follows tht f (x) > f (0) for x > b. This concludes the proof. Corollry 2.2 Consider the eqution f (x) = x n + n 1 x n 1 +... 1 x + 0 = α with n even. Then there exists number m such tht this eqution hs solution for ll α m but hs no solution for α < m. Proof : According to the previous theorem there exists c R such tht f (c) is the minimum of f. If α < f (c) then for ll x, f (x) f (c) > α, i.e., there is no solution. For α = f (c), c is solution. For α > f (c) we hve f (c) < α nd for lrge enough x, f (x) > α, hence, by the intermedite vlue theorem root exists. (25 hrs, 2013)

Functions 79 2.7 Infinite limits nd limits t infinity In this section we extend the notions of limits discussed in previous sections to two cses: (i) the limit of function t point is infinite, nd (ii) the limit of function t infinity. Before we strt recll: infinity is not rel number!. Definition 2.10 Let f : A B be function. We sy tht the limit of f t is infinity, denoted lim f =, if for every M there exists δ > 0, such tht f (x) > M whenever 0 < x < δ. Similrly, lim f =, if for every M there exists δ > 0, such tht f (x) < M whenever 0 < x < δ. Exmple: Consider the function 1/ x x 0 f : x 17 x = 0. The limit of f t zero is infinity. Indeed given M we tke δ = 1/M, then 0 < x < δ implies f (x) > M. Definition 2.11 Let f : R R. We sy tht lim f = l, if for every ɛ > 0 there exists n M, such tht f (x) l < ɛ whenever x > M. Similrly, lim f = l, if for every ɛ > 0 there exists n M, such tht f (x) l < ɛ whenever x < M.

80 Chpter 2 Figure 2.1: Infinite limit nd limit t infinity. Exmple: Consider the function f : x 3 + 1/x 2. Then the limit of f t infinity is 3, since given ɛ > 0 we tke M = 1/ ɛ, nd x > M implies f (x) 3 < ɛ. Comment: These definitions re in full greement with ll previous definitions of limits, if we dopt the ide tht neighborhoods of infinity re sets of the form (M, ). Exercise 2.25 Prove directly from the definition tht x 3 + 2x + 1 lim x 3 + x 3 =. Exercise 2.26 Clculte the following limit using limit rithmetic, Exercise 2.27 Prove tht lim x ( x2 + 2x x ). lim f (x) exists if nd only if lim f (1/x) x x 0 + in which cse they re equl. exist,

Functions 81 Exercise 2.28 Let g be defined in punctured neighborhood of R, with lim g =. Show tht lim 1/g = 0. Exercise 2.29 ➀ Let f, g be defined in punctured neighborhood of R, with lim f = lim g =. Show tht lim ( f g) =. ➁ Let f, g be defined in punctured neighborhood of R, with lim f = K < 0 nd lim g =. Show tht lim ( f /g) = 0. (24 hrs, 2010) 2.8 Inverse functions Suppose tht f : A B is one-to-one nd onto. This mens tht for every b B there exists unique A, such tht f () = b. This property defines function from B to A. In fct, this function is lso one-to-one nd onto. We cll this function the function inverse to f הופכית ),(פונקציה nd denote it by f 1 (not to be mistken with 1/ f ). Thus, f 1 : B A, f 1 (y) = {x A : f (x) = y}. 11 Note lso tht f 1 f is the identity in A wheres f f 1 is the identity in B. Comment: Recll tht function f : A B cn be defined s subset of A B, Grph( f ) = {(, b) : f () = b}. The inverse function cn be defined s subset of B A: it is ll the pirs (b, ) for which (, b) Grph( f ), or Grph( f 1 ) = {(b, ) : f () = b}. 11 Strictly speking, the right-hnd side is set of rel numbers, nd not rel number. However, the one-to-one nd onto properties of f gurntee tht it contins one nd only one number, hence we identify this singleton (יחידון ) with the unique element it contins.

82 Chpter 2 Exmple: Let f : [0, 1] R be given by f : x x 2. This function is one-toone nd its imge is the segment [0, 1]. Thus we cn define n inverse function f 1 : [0, 1] [0, 1] by f 1 (y) = {x [0, 1] : x 2 = y}, i.e., it is the (positive) squre root. (22 hrs, 2011) Theorem 2.11 If f : I R is continuous nd one-to-one (I is n intervl), then f is either monotoniclly incresing, or monotoniclly decresing. Comment: Continuity is crucil. The function f : [0, 2] R, x 0 x 1 f (x) = 6 x 1 < x 2 is one-to-one with imge [0, 1] [4, 5), but it is not monotonic. Proof : We first prove tht for every three points < b < c, either f () < f (b) < f (c), or f () > f (b) > f (c). Suppose tht f () < f (c) nd tht, by contrdiction, f (b) < f () (see Figure 2.2). By the intermedite vlue theorem, there exists point x (b, c) such tht f (x) = f (), contrdicting the fct tht f is one-to-one. Similrly, if f (b) > f (c), then there exists point y [, b] such tht f (y) = f (c). Thus, f () < f (c) implies tht f () < f (b) < f (c). We proceed similrly if f () > f (c). It follows s once tht for every four points < b < c < d, either f () < f (b) < f (c) < f (d) or f () > f (b) > f (c) > f (d). Fix now nd b, nd suppose w.l.o.g tht f () < f (b) (they cn t be equl). Then, for every x, y such tht x < y, we pply the bove rguments to the four points, b, x, y (whtever their order is) nd conclude tht f (x) < f (y).

Functions 83 f!c" f!" f!b" b c Figure 2.2: Illustrtion of proof Suppose tht the function f : [, b] R is one-to-one nd monotoniclly incresing (without loss of generlity). Then, its imge is the intervl [ f (), f (b)]. Indeed, f () nd f (b) re in the rnge, nd by the intermedite vlue theorem, so is ny point in this intervl. By monotonicity, no point outside this intervl cn be in the rnge of f. It follows tht the inverse function hs rnge nd imge f 1 : [ f (), f (b)] [, b]. We hve proved, en pssnt, tht continuous one-to-one functions mp closed segments into closed segments 12. (26 hrs, 2013) Theorem 2.12 If f : [, b] R is continuous nd one-to-one then so is f 1. Proof : Without loss of generlity, let us ssume tht f is monotoniclly incresing, i.e., x < y implies f (x) < f (y). Then, so is f 1. We need to show tht for every y 0 ( f (), f (b)), lim y0 f 1 = f 1 (y 0 ). 12 More generlly, continuous one-to-one function mps connected sets into connected sets, nd retins the open/closed properties.

84 Chpter 2 Equivlently, we need to show tht for every ɛ > 0 there exists δ > 0, such tht f 1 (y) f 1 (y 0 ) < ɛ whenever y y 0 < δ. Now, every such y 0 is equl to f (x 0 ) for unique x 0. Since f (nd f 1 ) re both monotoniclly incresing, it is cler how to set δ. Let y 1 = f (x 0 ɛ) nd y 2 = f (x 0 + ɛ). By monotonicity, y 1 < y 0 < y 2. We let then δ = min(y 0 y 1, y 2 y 0 ). Every y 0 δ < y < y 0 + δ is in the intervl (y 1, y 2 ), nd therefore, We cn do it more formlly. Let f 1 (y 0 ) ɛ < f 1 (y) < f 1 (y 0 ) + ɛ. δ = min( f ( f 1 (y 0 ) + ɛ) y 0, y 0 f ( f 1 (y 0 ) ɛ)). Then, y y 0 < δ implies or, y 0 δ < y < y 0 + δ, f ( f 1 (y 0 ) ɛ) = y 0 [y 0 f ( f 1 (y 0 ) ɛ)] < y < y 0 +[ f ( f 1 (y 0 )+ɛ) y 0 ] = f ( f 1 (y 0 )+ɛ). Since f 1 is monotoniclly incresing, we my pply f 1 on ll terms, giving f 1 (y 0 ) ɛ < f 1 (y) < f 1 (y 0 ) + ɛ, or f 1 (y 0 ) f 1 (y) < ɛ. f!x 0 "=y 0 f!x 0 +!" f!x 0 #!" "! f #1!y 0 "=x 0

Functions 85 (20 hrs, 2009) (26 hrs, 2010) 2.9 Uniform continuity Recll the definition of continuous function: function f : (, b) R is continuous on the intervl (, b), if it is continuous t every point in the intervl. Tht is, for every x (, b) nd every ɛ > 0, there exists δ < 0, such tht f (y) f (x) < ɛ whenever y x < δ nd y (, b). In generl, we expect δ to depend both on x nd on ɛ. In fct, the wy we set it here, this definition pplies for continuity on closed intervl s well (i.e, it includes one-sided continuity s well). Let us re-exmine two exmples: Exmple: Consider the function f : (0, 1) R, f : x x 2. This function is continuous on (0, 1). Why? Becuse for every x (0, 1), nd every y (0, 1), f (y) f (x) = y 2 x 2 = y x y + x (x + 1) y x. Thus, given x nd ɛ > 0, if we choose δ = δ(ɛ, x) = ɛ/(x + 1), then f (y) f (x) < ɛ whenever x y < δ(ɛ, x) nd y (0, 1). Thus, δ depends both on ɛ nd x. However, it is lwys legitimte to replce δ by smller number. If we tke δ = ɛ/2, then the sme δ fits ll points x. Exmple: Consider next the function f : (0, 1) R, f : x 1/x. This function is lso continuous on (0, 1). Why? Given x (0, 1) nd y (0, 1), f (x) f (y) = x y xy = x y x[x + (y x)]. If we tke δ(ɛ, x) = min(x/2, ɛx 2 /2), then x y < δ implies f (x) f (y) < δ x(x δ) < ɛ.

86 Chpter 2 Here, given ɛ, the closer x is to zero, the smller is δ(ɛ, x). There is no wy we cn find δ = δ(ɛ) tht would fit ll x. These two exmples motivte the following definition: רציפה) Definition 2.12 f : A B is sid to be uniformly continuous on A שווה (במידה if for every ɛ > 0 corresponds δ = δ(ɛ) > 0, such tht f (y) f (x) < ɛ whenever y x < δ nd x, y A. Note tht x nd y ply here symmetric roles 13. Wht is the essentil difference between the two bove exmples? The function x x 2 could hve been defined on the closed intervl [0, 1] nd it would hve been continuous there too. In contrst, there is no wy we could hve defined the function x 1/x s continuous function on [0, 1], even if we took cre of its vlue t zero. We hve lredy seen exmples where continuity on closed intervl hd strong implictions (ensures boundedness nd the existence of mximum). This is lso the cse here. We will prove tht continuity on closed intervl implies uniform continuity. We proceed by steps: Lemm 2.3 Let f : [, c] R be continuous. Let < b < c, nd suppose tht for given ɛ > 0 there exist δ 1, δ 2 > 0, such tht x, y [, b] nd x y < δ 1 implies f (x) f (y) < ɛ x, y [b, c] nd x y < δ 2 implies f (x) f (y) < ɛ. Then there exists δ > 0 such tht x, y [, c] nd x y < δ implies f (x) f (y) < ɛ. Proof : We hve to tke cre t the gluing t the point b. Since the function is continuous t b, then there exists δ 3 > 0 such tht f (x) f (b) < ɛ 2 whenever x b < δ 3. 13 The term uniform mens tht the sme number cn be used for ll points.

Functions 87 Hence, x b < δ 3 nd y b < δ 3 implies f (x) f (y) < ɛ. (2.1) Tke now δ = min(δ 1, δ 2, δ 3 ). We clim tht f (x) f (y) < ɛ whenever x y < δ nd x, y [, c]. Why is tht true? One of three: either x, y [, b], in which cse it follows from the fct tht δ δ 1, or x, y [b, c], in which cse it follows from the fct tht δ δ 2. Remins the third cse, where, without loss of generlity, x [, b] nd y [b, c]. Since x y < δ 3, it follows tht x b < δ 3 nd y b < δ 3 nd we use (2.1). (24 hrs, 2011) Theorem 2.13 If f is continuous on [, b] then it is uniformly continuous on tht intervl. Proof : Given ɛ > 0, we will sy tht f is ɛ-good on [, b] if there exists δ > 0, such tht f (x) f (y) < ɛ whenever x y < δ nd x, y [, b] (i.e., if for tht prticulr ɛ there exists corresponding δ). We need to show tht f is ɛ-good on [, b] for ll ɛ > 0. Note tht we proved in Lemm 2.3 tht if f is ɛ-good on [, b] nd on [b, c], then it is ɛ-good on [, c]. Given ɛ, define A = {x [, b] : f is ɛ-good on [, x]}. The set A is non-empty, s it includes the point (trivil) nd it is bounded by b. Hence it hs supremum, which we denote by α. Suppose tht α < b. Since f is continuous t α, then there exists δ 0 > 0 such tht f (y) f (α) < ɛ 2 whenever y α < δ 0 nd y [, b]. Thus, y, z B(α, δ 0 ), f (y) f (z) f (y) f (α) + f (z) f (α) < ɛ,

88 Chpter 2 from which follows tht f is ɛ-good in (α δ 0, α+δ 0 ) [, b] nd hence in the closed intervl [α δ 0 /2, α + δ 0 /2] [, b]. By the definition of α, it is lso ɛ-good on [, α δ 0 /2] [, b], nd by the previous lemm, f is ɛ-good on [, α+δ 0 /2] [, b], contrdicting the fct tht α is n upper bound for A. Thus f is ɛ-good on [, b]. Since this rgument holds independently of ɛ we conclude tht f is ɛ-good in [, b] for ll ɛ > 0. (28 hrs, 2013) Comment: In this course we will mke use of uniform continuity only once, when we study integrtion. (27 hrs, 2010) Exmples: ➀ Consider the function f (x) = sin(1/x) defined on (0, 1). Even though it is continuous, it is not uniformly continuous. ➁ Consider the function Id : R R. It is esy to see tht it is uniformly continuous. The function Id Id, however, is continuous on R, but not uniformly continuous. Thus, the product of uniformly continuous functions is not necessrily uniformly continuous. ➂ Finlly the function f (x) = sin x 2 is continuous nd bounded on R, but it is not uniformly continuous. (22 hrs, 2009) Exercise 2.30 Let f : R R be continuous. Prove or disprove: ➀ If lim then f ttins mximum vlue. f = lim f = 0, ➁ If lim f = nd lim f =, then f ssumes ll rel vlues; tht is f is on R. ➂ If on ny intervl [n, n + 1], n Z, f is uniformly continuous, then f is uniformly continuous on R. Exercise 2.31 Suppose A R is non-empty.

Functions 89 ➀ Prove tht if f, g re uniformly continuous on A then α f + βg is lso uniformly continuous on A, where α, β R. ➁ Is in the bove cse f g necessrily uniformly continuous? Exercise 2.32 Let f : (0, ) R be given by f : x 1/ x: ➀ Let > 0. Prove tht f is uniformly continuous on [, ). ➁ Show tht f is not uniformly continuous on (0, ). Exercise 2.33 Let I be n open connected domin (i.e., n open segment, n open ry, or the whole line) nd let f, g : I R be uniformly continuous on I. Prove tht f g is uniformly continuous on I. (Hint: red the proof bout the limit of product of functions being the product of their limits.) Exercise 2.34 Let f : A B nd g : B C be uniformly continuous. Show tht g f is uniformly continuous. Exercise 2.35 Let > 0. Show tht f (x) = sin(1/x) is not uniformly continuous on (0, ). Exercise 2.36 For which vlues of α > 0 is the function f (x) = x α uniformly continuous on [0, )? Exercise 2.37 Let α > 0. We sy tht f : A B is Hölder continuous of order α if there exists constnt M > 0, such tht f (x) f (y) < M x y α, x, y A. ➀ Show tht f (x) = x is Hölder continuous of order 1/2 on (0, ). ➁ Prove tht if f is Hölder continuous of order α on A then it is uniformly continuous on A. ➂ Prove tht if f is Hölder continuous of order α > 1, then it is constnt function.

90 Chpter 2

Chpter 3 Derivtives 3.1 Definition In this chpter we define nd study the notion of differentible functions. Derivtives were historiclly introduced in order to nswer the need of mesuring the rte of chnge of function. We will ctully dopt this pproch s prelude to the forml definition of the derivtive. Before we strt, one technicl clrifiction. In the previous chpter, we introduced the limit of function f : A B t n (interior) point, Consider the function g defined by on the rnge, lim f. g(x) = f ( + x), {x : + x A}. In prticulr, zero is n interior point of this set. We clim tht lim 0 g = lim f, provided, of course, tht the right hnd side exists. It is good exercise to prove it formlly. Suppose tht the right hnd side equls l. This mens tht ( ɛ > 0)( δ > 0) : ( x B (, δ))( f (x) l < ɛ).

92 Chpter 3 Setting x = y, nd by the definition of g, ( ɛ > 0)( δ > 0) : ( y B (0, δ))( f ( + y) l < ɛ), ( ɛ > 0)( δ > 0) : ( y B (0, δ))( g(y) l < ɛ), i.e., lim 0 g = l. Often, one does not bother to define the function g nd simply writes lim x 0 f ( + x) = lim f, lthough this in not in greement with our nottions; f ( + x) stnds for the composite function, f (x + x). Let now f be function defined on some intervl I, nd let nd x be two points inside this intervl. The vrition in the vlue of f between these two points is f (x) f (). We define the men rte of chnge שינוי ממוצע ) (קצב of f between the points nd x to be the rtio f (x) f (). x This quntity cn be ttributed with number of interprettions. First, if we consider the grph of f, then the men rte of chnge is the slope of the secnt line (מיתר ) tht intersects the grph t the points nd x (see Figure 3.1). Second, it hs mening in mny physicl situtions. For exmple, f cn be the position long line (sy, in meters reltive to the origin) s function of time (sy, in seconds reltive to n origin of time). Thus, f () is the distnce from the origin in meters seconds fter the time origin, nd f (x) is the distnce from the origin in meters x seconds fter the time origin. Then f (x) f () is the displcement (העתק ) between time nd time x, nd [ f (x) f ()]/(x ) is the men displcement per unit time, or the men velocity. This physicl exmple is good preliminry towrd the definition of the derivtive. The men velocity, or men rte of displcement, is n verge quntity between two instnts, but there is nothing to gurntee tht within this time intervl, the body ws in fixed stte. For exmple, we could intersect this time intervl into two equl sub-intervls, nd mesure the men velocity in ech hlf. Nothing gurntees tht these men velocities will equl the men velocity over

Derivtives 93 f!+h" f!"! +h Figure 3.1: Reltion between the men rte of chnge of function nd the slope of the corresponding secnt. the whole intervl. Physicists imed to define n instntneous velocity, nd the wy to do it ws to mke the length of the intervl very smll. Of course, hving h smll (wht does smll men?) does not chnge the verge nture of the mesured velocity. An instntneous rte of chnge cn be defined by using limits. Fixing the point, we define function f (x) f () f, : x. x The function f, cn be defined on the sme domin s f, but excluding the point. We sy tht f is differentible t או דיפרנציאבילית ) (גזירה if f, hs limit t. We write lim f, = f (). We cll the limit f () the derivtive (נגזרת ) of f t. Comment: In the more trditionl nottion, f f (x) f () () = lim. x x At this stge, f () is nothing but nottion. It is not (yet!) function evluted

94 Chpter 3 t the point. Another stndrd nottion, due to Leibniz, is d f dx (). Hving identified the men rte of chnge s the slope of the secnt line, the derivtive hs simple geometricl interprettion. As x tends to, the corresponding fmily of secnts tends to line which is tngent to f t the point. For us, these re only hnd-wving rguments, s we hve never ssigned ny mening to the limit of fmily of lines. Exmple: Consider the constnt function, f : R R, f : x c. We re going to clculte its derivtive t the point. We construct the function f, (y) = f (y) f () y = c c y = 0. The limit of this function t is clerly zero, hence f () = 0. In the bove exmple, we could hve computed the derivtive t ny point. In other words, we cn define function tht given point x, returns the derivtive of f t tht point, i.e., returns f (x). A function f : A B is clled differentible in subset U A of its domin if it hs derivtive t every point x U. We then define the derivtive function, f : U R, s the function, Agin, the more trditionl nottion is, f (x) = lim x f,x. f f (y) f (x) (x) = lim. y x y x Exmple: Consider the function f : x x 2. We clculte its derivtive t point x by first observing tht f,x (y) = y2 x 2 y x = x + y or f,x = Id +x, so tht f (x) = lim x f,x = 2x.

Derivtives 95 (23 hrs, 2009) (25 hrs, 2011) Exmple: Consider now the function f : x x. Let s first clculte its derivtive t point x > 0. For x > 0, f,x (y) = y x y x. Since we re interested in the limit of f,x t x > 0 we my well ssume tht y > 0, in which cse f,x (y) = 1. Hence f (x) = lim x f,x = 1. For negtive x, x = x nd we my consider y < 0 s well, so tht so tht Remins the point 0 itself, f,x (y) = ( y) ( x) y f (x) = lim x f,x = 1. f,0 (y) = y 0 y 0 = 1, = sgn(y). The limit t zero does not exist since every neighborhood of zero hs points where this function equls one nd points where this function equls minus one. Thus f is differentible everywhere except for the origin 1. Comment: If f is differentible t, then f, hs limit t, hence it hs removble discontinuity t tht point. The function f, (x) x x f () x = is continuous. (29 hrs, 2010) 1 We use here generl principle, whereby lim f does not exist if there exist l 1 l 2, such tht every neighborhood of hs points x, y, such tht f (x) = l 1 nd f (y) = l 2.

96 Chpter 3 One-sided differentibility This lst exmple shows cse where limit does not exist, but one-sided limits do exist. This motivtes the following definitions: (גזירה מימין ) Definition 3.1 A function f is sid to be differentible on the right t if the one-sided limit lim f, + exists. We denote the right-hnd derivtive ימנית ) (נגזרת by f ( + ). A similr definition holds for left-hnd derivtives. Exmple: Here is one more exmple tht prctices the clcultion of the derivtive directly from the definition. Consider first the function x sin 1 x 0 x f : x 0 x = 0. We hve lredy seen tht tht this function is continuous t zero, but is it differentible t zero? We construct the function f,0 (y) = f (y) f (0) y 0 = sin 1 y. The function f,0 does not hve limit t 0, becuse every neighborhood of zero hs both points where f,0 = 1 nd points where f,0 = 1. Exmple: In contrst, consider the function x 2 sin 1 x 0 x f : x 0 x = 0. We construct f,0 (y) = f (y) f (0) y 0 Now lim 0 f,0 = 0, hence f is differentible t zero. = y sin 1 y. Differentibility nd continuity The following theorem shows tht differentible functions re subclss ( better behved ) of the continuous functions.

Derivtives 97 Theorem 3.1 If f is differentible t then it is continuous t. Proof : Note tht for y, f (y) = f () + (y ) f (y) f (). (y ) Viewing f () s constnt, we hve functionl identity, f = f () + (Id ) f, vlid in punctured neighborhood of. By limit rithmetic, lim f = f () + lim (Id ) lim f, = f () + 0 f () = f (). Exercise 3.1 Show, bsed on the definition of the derivtive, tht if f is differentible t R with f (x) 0, then f is lso differentible t tht point. (Hint, use the fct tht function differentible t point is lso continuous t tht point.) Exercise 3.2 Let I be connected domin, nd suppose tht f is differentible t every interior point of I nd continuous on I. Suppose furthermore tht f (x) < M for ll x I. Show tht f is uniformly continuous on I. Higher order derivtives If function f is differentible on n intervl A, we cn construct new function its derivtive, f. The derivtive f cn hve vrious properties, For exmple, it my be continuous, or not, nd in prticulr, it my be differentible on A, or on subset of A. Then, we cn define new function the derivtive of the derivtive, or the second derivtive שניה ),(נגזרת which we denote by f. By definition f (x) = lim x f,x. (Note tht this is limit of limits.) Likewise, the second derivtive my be differentible, in which cse we my define the third derivtive f, nd so on. For

98 Chpter 3 derivtives higher thn the third, it is customry to use, for exmple, the nottion f (4) rther thn f. The k-th derivtive, f (k), is defined recursively, f (k) (x) = lim x f (k 1),x. For the recursion to hold from k = 1, we lso set f (0) = f. (30 hrs, 2013) 3.2 Rules of differentition Recll tht when we studied limits, we first clculted limits by using the definition of the limit, but very soon this becme imprcticl, nd we proved number of theorems (rithmetic of limits), with which we were ble to esily clculte lrge vriety of limits. The sme exctly will hppen with derivtives. After hving clculted the derivtives of smll number of functions, we develop tools tht will enble us to (esily) compute derivtives, without hving to go bck to the definitions. Theorem 3.2 If f is constnt function, then f = 0 (by this we men tht f : x 0). Proof : We proved this is in the previous section. Theorem 3.3 If f = Id then f = 1 (by this we men tht f : x 1). Proof : Immedite from the definition. Theorem 3.4 Let f, g be functions defined on the sme domin (it suffices tht the domins hve non-empty intersection). If both f nd g re differentible t then f + g is differentible t, nd ( f + g) () = f () + g ().

Derivtives 99 Proof : We consider the function nd by the rithmetic of limits, ( f + g) ( f + g)() f +g, = Id f + g f () g() = Id = f, + g,, ( f + g) () = f () + g (). Theorem 3.5 (Leibniz rule) Let f, g be functions defined on the sme domin. If both f nd g re differentible t then f g is differentible t, nd ( f g) () = f () g() + f () g (). Proof : We consider the function, f g, = f g ( f g)() Id = f g f g() + f g() f ()g() Id nd it remins to pply the rithmetic rules of limits. = f g, + g() f,, Corollry 3.1 The derivtive is liner opertor, (α f + βg) = α f + βg. Proof : We only need to verify tht if f is differentible t then so is α f nd (α f ) = α f. This follows from the derivtive of the product. (31 hrs, 2013)

100 Chpter 3 Exmple: Let n N nd consider the functions f n : x x n. Then, f n(x) = n x n 1 or equivlently f n = n f n 1. We cn show this inductively. We know lredy tht this is true for n = 0, 1. Suppose this were true for n = k. Then, f k+1 = f k Id, nd by the differentition rule for products, f k+1 = f k Id + f k Id, i.e., f k+1 (x) = k xk 1 x + x k 1 = (k + 1) x k. Theorem 3.6 If g is differentible t nd g() 0, then 1/g is differentible t nd (1/g) () = g () g 2 (). Proof : Since g is continuous t (it is continuous since it is differentible) nd it is non-zero, then there exists neighborhood U of in which g does not vnish (by Theorem 2.5). In U \ {} we consider the function 1/g, = 1/g (1/g)() Id = 1 ( 1 Id g 1 ) = 1 g() g g() g,. It remins to tke the limit t nd pply the rithmetic lws of limits. (31 hrs, 2010) Theorem 3.7 If f nd g re differentible t nd g() 0, then the function f /g is differentible t nd ( f /g) = f g g f g 2.

Derivtives 101 Proof : Apply the lst two theorems. (25 hrs, 2009) With these theorems in hnd, we cn differentite lrge number of functions. In prticulr there is no difficulty in differentiting product of more thn two functions. For exmple, ( f gh) = (( f g)h) = ( f g) h + ( f g)h = ( f g + f g )h = ( f g)h = f gh + f g h + f gh. Suppose we tke for grnted tht sin = cos nd cos = sin. Then we hve no problem clculting the derivtive of, sy, x sin 3 x, s product of three functions. But wht bout the derivtive of x sin x 3? Here we need rule for how to differentite compositions. Consider the composite function g f, i.e., (g f )(x) = g( f (x)). Let s try to clculte its derivtive t point, ssuming for the moment tht both f nd g re differentible everywhere. We then need to look t the function g f, (y) = (g f )(y) (g f )() y = g( f (y)) g( f ()), y nd clculte its limit t. When y is close to, we expect the rguments f (y) nd f () of g to be very close. This suggest the following tretment, g f, (y) = g( f (y)) g( f ()) f (y) f () f (y) f () y = g( f (y)) g( f ()) f (y) f () f, (y). It looks tht s y, since f (y) f () 0, this product tends to g ( f ()) f (). The problem is tht while the limit y mens tht the cse y = is not to be considered, there is nothing to prevent the denomintor f (y) f () from vnishing, rendering this expression meningless. Yet, the result is correct, nd it only tkes little more subtlety to prove it. (27 hrs, 2011)

102 Chpter 3 Theorem 3.8 Let f : A B nd g : B R. If f is differentible t A, nd g is differentible t f (), then (g f ) () = g ( f ()) f (). Proof : We introduce the following function, defined in neighborhood of f (), g, f () (z) z f () ψ : z g ( f ()) z = f (). The fct tht g is differentible t f () implies tht ψ is continuous t f (). We next clim tht for y g f, = (ψ f ) f,. Why tht? If f (y) f (), then this eqution reds g( f (y)) g( f ()) y = g( f (y)) g( f ()) f (y) f () which holds indeed, wheres if f (y) = f () it reds, g( f (y)) g( f ()) y = g ( f ()) f (y) f (), y f (y) f (), y nd both sides re zero. Consider now the limit of the right hnd side t. By limits rithmetic, lim [ (ψ f ) f, ] = ψ( f ()) lim f, = g ( f ()) f (). (32 hrs, 2010) 3.3 Another look t derivtives In this short section we provide nother (equivlent) chrcteriztion of differentibility nd the derivtive. Its purpose is to offer slightly different ngle of view on the subject, nd how clever definitions cn sometimes gretly simplify proofs.

Derivtives 103 Our definition of derivtives sttes tht function f is differentible t n interior point, if the function f, hs limit t, nd we denote this limit by f (). The function f, is not defined t, hence not continuous t, but this is removble discontinuity. We could sy tht f is differentible t if there exists rel number, l, such tht the function f, (y) y S f,x (y) = l y = is continuous t, nd S f, () = l is clled the derivtive of f t. For y, or equivlently, S f, (y) = f, (y) = f (x) f (), x f (y) = f () + S f, (y)(y ). This eqution holds lso for y =. This suggests the following lterntive definition of the derivtive: Definition 3.2 A function f defined in on open neighborhood of point is sid to be differentible t if there exists function S f, continuous t, such tht f (y) = f () + S f, (y)(y ). S f, () is clled the derivtive of f t nd is denoted by f (). Of course, S f, coincides with f, in some punctured neighborhood of. A first exmple Tke the function f : x x 2. Then or f (y) f () = (y + )(y ), f = f () + (Id +)(Id ). Since the function Id + is continuous t, it follows tht f is differentible t nd f () = lim (Id +) = 2.

104 Chpter 3 Leibniz rule Let s now see how this lterntive definition simplifies certin proofs. Suppose for exmple tht both f nd g re differentible t. This implies the existence of two continuous functions, S f, nd S g,, such tht f (y) = f () + S f, (y)(y ) g(y) = g() + S g, (y)(y ) in some neighborhood of nd f () = S f, () nd g () = S g, (). Then, f (y)g(y) = ( f () + S f, (y)(y ) ) ( g() + S g, (y)(y ) ) = f ()g() + ( f ()S g, (y) + g() S f, (y) + S f, (y)s g, (y)(y ) ) (y ). By limit rithmetic, the function in the brckets, F = f ()S g, + g() S f, + S f, S g, (Id ) is continuous t, hence f g is differentible t nd ( f g) () = lim F = f ()g () + g() f (). Derivtive of composition Suppose now tht f is differentible t nd g is differentible t f (). Then, there exist continuous functions, S f, nd S g, f (), such tht f (y) = f () + S f, (y)(y ) Now, or, g(z) = g( f ()) + S g, f () (z)(z f ()). g( f (y)) = g( f ()) + S g, f () ( f (y))( f (y) f ()) = g( f ()) + S g, f () ( f (y))s f, (y)(y ), (g f )(y) = (g f )() + S g, f () ( f (y))s f, (y)(y ). } {{ } [S g, f () f] S f, (y) By the properties of continuous functions, the function in the squre brckets is continuous t, nd (34 hrs, 2010) (28 hrs, 2011) (g f ) () = lim [ Sg, f () f ] S f, = g ( f ()) f ().

3.4 The derivtive nd extrem Derivtives 105 In high-school clculus, one of the min uses of differentil clculus is to find extrem of functions. We re going to put this prctice on solid grounds. First, recll the definition, Definition 3.3 Let f : A B. A point A (not necessrily n internl point) is sid to be mximum point of f in A, if We define similrly minimum point. f (x) f () x A. Comment: By no mens mximum point hs to be unique, nor to exist. For exmple, in constnt function ll points re minim nd mxim. On the other hnd, the function f : (0, 1) R, f : x x 2 does not hve mximum in (0, 1). Also, we proved tht continuous function defined on closed intervl lwys hs mximum (nd minumum, both, not necessrily unique). Here is first connection between mximum (nd minimum) points nd derivtives: Theorem 3.9 Let f : (, b) R. If x (, b) is mximum point of f nd f is differentible t x, then f (x) = 0. Proof : Since x is, by ssumption, mximum point, then for every y (, b), f (y) f (x) 0. In prticulr, for y x, stisfies f,x (y) = f (y) f (x), y x f,x (y) 0 y > x f,x (y) 0 y < x. It is given tht f,x hs limit t x (it is f (x)). We will show tht this limit is necessrily zero.

106 Chpter 3 Suppose the limit, l, were positive. Then, tking ɛ = l/2, there hs to be δ > 0, such tht 0 < l 2 < f,x(y) < 3l 2 whenever 0 < y x < δ, but this cn t hold, since every neighborhood of x hs points y where f,x (y) 0. Similrly, l cn t be negtive, hence l = 0. (33 hrs, 2013) Comment: This is uni-directionl theorem. It does not imply tht if f (x) = 0 then x is mximum point (nor minimum point). Comment: The open intervl cnnot be replced by the closed intervl [, b], becuse t the end points we cn only consider one-sided limits. Definition 3.4 Let f : A B. An interior point A is clled locl mximum of f מקומי ),(מקסימום if there exists δ > 0 such tht is mximum of f in ( δ, + δ) ( locl lwys mens in sufficiently smll neighborhood ). Theorem 3.10 (Pierre de Fermt) If is locl mximum (or minimum) of f in some open intervl nd f is differentible t then f () = 0. Proof : There is ctully nothing to prove, s the previous theorem pplies verbtim for f restricted to the intervl ( δ, + δ). Comment: The converse is not true. Tke for exmple the function f : x x 3. If is differentible in R nd f (0) = 0, but zero is not locl minimum, nor locl mximum of f. Comment: Although this theorem is due to Fermt, its proof is slightly shorter thn the proof of Fermt s lst theorem. Definition 3.5 Let f : A B be differentible. A point A is clled criticl point of f if f () = 0. The vlue f () is clled criticl vlue.

Derivtives 107 Locting extreml points Let f : [, b] R nd suppose we wnt to find mximum point of f. If f is continuous, then mximum is gurnteed to exist. There re three types of cndidtes: (1) criticl points of f in (, b), (ii) the end points nd b, nd (iii) points where f is not differentible. If f is differentible, then the tsk reduces to finding ll the criticl points nd compring ll the criticl vlues, { f (x) : f (x) = 0}. Finlly, the lrgest criticl vlue hs to be compred to f () nd f (b). Exmple: Consider the function f : [ 1, 2] R, f : x x 3 x. This function is differentible everywhere, nd its criticl points stisfy f (x) = 3x 2 1 = 0, i.e., x = ±1/ 3, which re both in the domin. It is esily checked tht f (1/ 3) = 2 3 3 nd f ( 1/ 3) = 2 3 3. Finlly, f ( 1) = 0 nd f (2) = 6, hence 2 is the (unique) mximum point. So fr, we hve derived properties of the derivtive given the function. Wht bout the reverse direction? Tke the following exmple: we know tht the derivtive of constnt function is zero. Is it lso true tht if the derivtive is zero then the function is constnt? A priori, it is not cler how to show it. How cn we go from the knowledge tht f (y) f (x) lim = 0, y x y x to showing tht f is constnt function. The following two theorems will provide us with the necessry tools: Theorem 3.11 (Michel Rolle, 1691) If f is continuous on [, b], differentible in (, b), nd f () = f (b), then there exists point c (, b) where f (c) = 0. Proof : It follows from the continuity of f tht it hs mximum nd minimum point in [, b]. If the mximum occurs t some c (, b) then f (c) = 0 nd we re done. If the minimum occurs t some c (, b) then f (c) = 0 nd we re done. The only remining lterntive is tht nd b re both minim nd mxim, in which cse f is constnt nd its derivtive vnishes t some interior point (well, t ll of them).

108 Chpter 3 Comment: The requirement tht f be differentible everywhere in (, b) is impertive, for consider the function x 0 x 1 2 f (x) = 1 1 x < x 1. 2 Even though f is continuous in [0, 1] nd f (0) = f (1) = 0, there is no interior point where f (x) = 0. Theorem 3.12 (Men-vlue theorem הערך הממוצע ) ((משפט If f is continuous on [, b] nd differentible in (, b), then there exists point c (, b) where f (c) = f (b) f (), b tht is, point t which the derivtive equls to the men rte of chnge of f between nd b. Comment: Rolle s theorem is prticulr cse. Proof : This is lmost direct consequence of Rolle s theorem. Define the function g : [, b] R, f (b) f () g(x) = f (x) (x ). b g is continuous on [, b] nd differentible in (, b). Moreover, g() = g(b) = f (). Hence by Rolle s theorem there exists point c (, b), such tht 0 = g (c) = f (c) f (b) f (). b Corollry 3.2 Let f : [, b] R. If f (x) = 0 for ll x (, b) then f is constnt.

Derivtives 109 Proof : Let c, d (, b), c < d. By the men-vlue theorem there exists some e (c, d), such tht f f (d) f (c) (e) =, d c however f (e) = 0, hence f (d) = f (c). Since this holds for ll pir of points, then f is constnt. Comment: This is only true if f is defined on n intervl. Tke domin of definition which is the union of two disjoint sets, nd this is no longer true ( f is.((רכיב קשירות ) only constnt in every connected component (27 hrs, 2009) Corollry 3.3 If f nd g re differentible on n intervl with f (x) = g (x), then there exists rel number c such tht f = g + c on tht intervl. Proof : Apply the previous corollry for f g. Corollry 3.4 If f is continuous on closed intervl [, b] nd f (x) > 0 in (, b), then f is incresing on tht intervl. Proof : Let x < y belong to tht intervl. By the men-vlue theorem there exists c (x, y) such tht f f (y) f (x) (c) =, y x however f (c) > 0, hence f (y) > f (x). Comment: The converse is not true. If f is incresing nd differentible then f (x) 0, but equlity my hold, s in f (x) = x 3 t zero. We hve seen tht t locl minimum (or mximum) the derivtive (if it exists) vnishes, but tht the opposite is not true. The following theorem gives sufficient condition for point to be locl minimum (with corresponding theorem for locl mximum).

110 Chpter 3 Theorem 3.13 If f () = 0 nd f () > 0 then is locl minimum of f. (30 hrs, 2011) Comment: The fct tht f exists t implies: 1. f exists in some neighborhood of. 2. f is continuous t. 3. f is continuous in some neighborhood of. Proof : By definition, f () = lim f, > 0 where f,(y) = f (y) f () y = f (y) y. Thus there exists δ > 0 such tht or, f (y) y > 0 whenever 0 < y < δ, f (y) > 0 whenever < y < + δ f (y) < 0 whenever δ < y <. It follows tht f is incresing in right-neighborhood of nd decresing in left-neighborhood of, i.e., is locl minimum. Another wy of completing the rgument is s follows: for every y (, + δ): f (y) f () y = f (c y ) > 0 i.e., f (y) > f (), where c y (, y). Similrly, for every y ( δ, ): f (y) f () y = f (c y ) < 0 i.e., f (y) > f (), where here c y (y, ).

Derivtives 111 Exmple: Consider the function f (x) = x 3 x. There re three criticl points 0, ±1/ 3: one locl mximum, one locl minimum, nd one which is neither. Exmple: Consider the function f (x) = x 4. Zero is minimum point, even though f (0) = 0. (36 hrs, 2010) Theorem 3.14 If f hs locl minimum t nd f () exists, then f () 0. Proof : Suppose, by contrdiction, tht f () < 0. By the previous theorem this would imply tht is both locl minimum nd locl mximum. This in turn implies tht there exists neighborhood of in which f is constnt, i.e., f () = f () = 0, contrdiction. The following theorem sttes tht the derivtive of continuous function cnnot hve removble discontinuity. Theorem 3.15 Suppose tht f is continuous t nd lim f exists, then f () exists nd f is continuous t. Proof : The ssumption tht f hs limit t implies tht f exists in some punctured neighborhood U of. Thus, for y U \ {} the function f is continuous on the closed segment tht connects y nd nd differentible on the corresponding open segment. By the men-vlue theorem, there exists point c y between y nd, such tht f f (y) f () (c y ) = = f, (y). y Since f hs limit t, ( ɛ > 0)( δ > 0) : ( y B (, δ)) ( ) f (y) lim f < ɛ.

112 Chpter 3 Since y B (, δ) implies tht c y B (, δ), ( ɛ > 0)( δ > 0) : ( y B (, δ))( f (c y ) lim }{{} f, (y) f < ɛ), which precisely mens tht f () = lim f. Exmple: The function x 2 x < 0 f (x) = x 2 + 5 x 0, is differentible in neighborhood of 0 nd lim 0 f exists, but it is not continuous, nd therefore not differentible t zero. Exmple: The function x 2 sin 1/x x 0 f (x) = 0 x = 0, is continuous nd differentible in neighborhood of zero. However, the derivtive does not hve limit t zero, so tht the theorem does not pply. Note, however, tht f is differentible t zero. (29 hrs, 2009) (37 hrs, 2010) Theorem 3.16 (Augustin Louis Cuchy; intermedite vlue theorem) Suppose tht f, g re continuous on [, b] nd differentible in (, b). Then, there exists c (, b), such tht [ f (b) f ()]g (c) = [g(b) g()] f (c).

Derivtives 113 Comment: If g(b) g() then this theorem sttes tht f (b) f () g(b) g() = f (c) g (c) (provided lso tht g (c) 0). This my seem like corollry of the men-vlue theorem. We know tht there exist c,d, such tht f (c) = f (b) f () b nd g (d) = g(b) g(). b The problem is tht there is no reson for the points c nd d to coincide. Proof : Define h(x) = f (x)[g(b) g()] g(x)[ f (b) f ()], nd pply Rolle s theorem. Nmely, we verify tht h is continuous on [, b] nd differentible in (, b), nd tht h() = h(b), hence there exists point c (, b), such tht h (c) = f (c)[g(b) g()] g (c)[ f (b) f ()] = 0. Theorem 3.17 (L Hôpitl s rule) Suppose tht lim f = lim g = 0, nd tht Then the limit lim ( f /g) exists nd lim f g exists. lim f g = lim f g. Comment: This is very convenient tool for obtining limit of frction, when both numertor nd denomintor vnish in the limit 2. 2 L Hôpitl s rule ws in fct derived by Johnn Bernoulli who ws giving to the French noblemn, le Mrquis de l Hôpitl, lessons in clculus. L Hôpitl ws the one to publish this theorem, cknowledging the help of Bernoulli.

114 Chpter 3 Proof : It is implicitly ssumed tht f nd g re differentible in neighborhood of (except perhps t ) nd tht g is non-zero in neighborhood of (except perhps t ); otherwise, the limit lim ( f /g ) would not hve existed. Note tht f nd g re not necessrily defined t ; since they hve limit there, there will be no hrm ssuming tht they re continuous t, i.e., f () = g() = 0 3. First, we clim tht g does not vnish in some neighborhood U of, for by Rolle s theorem, it would imply tht g vnishes somewhere in this neighborhood. More formlly, if g (x) 0 whenever 0 < x < δ, then g(x) 0 whenever 0 < x < δ, for if g(x) = 0, with, sy, 0 < x < δ, then there exists 0 < c x < x < δ, where g (c x ) = 0. Both f nd g re continuous nd differentible in some neighborhood U tht contins, hence by Cuchy s men-vlue theorem, there exists for every x U, point c x between x nd, such tht f (x) 0 g(x) 0 = f (x) g(x) = f (c x ) g (c x ). (3.1) Since the limit of f /g t exists, ( ɛ > 0)( δ > 0) : x B (, δ) implies ( f (x) g (x) lim f g ) < ɛ, nd since x B (, δ) implies c x B (, δ), ( ɛ > 0)( δ > 0) : x B (, δ) implies ( f (c x ) g (c x ) }{{} f (x)/g(x) f lim ), < ɛ g 3 Wht we relly do is to define continuous functions f (x) = f (x) x 0 x = g(x) x nd g(x) = 0 x =, nd prove the theorem for the corrected functions f nd g. At the end, we observe tht the result must pply for f nd g s well.

Derivtives 115 which mens tht lim f g = lim f g. Exmple: ➀ The limit ➁ The limit sin x lim x 0 x = 1. sin 2 x lim = 1, x 0 x 2 with two consecutive pplictions of l Hôpitl s rule. (32 hrs, 2011) Comment: Purposely, we only prove one vrint mong mny other of l Hôpitl s rule. Another vrints is: Suppose tht nd tht Then Yet nother one is: Suppose tht lim f = lim g = 0, lim lim f g exists. f g = lim f g. lim f = lim g =, nd tht Then lim lim f g exists. f g = lim f g.

116 Chpter 3 The following theorem is very reminiscent of l Hôpitl s rule, but note how different it is: Theorem 3.18 Suppose tht f nd g re differentible t, with f () = g() = 0 nd g () 0. Then, lim f g = f () g (). Proof : Without loss of generlity, ssume g () > 0. Tht is, g(y) g() lim y y g(y) = lim y y = g () > 0, hence there exists punctured neighborhood of where g(x)/(x ) > 0, nd in prticulr, in which the numertor g(x) is non-zero. Thus, we cn divide by g(x) in punctured neighborhood of, nd f (x) g(x) f (x) f () f (x) f () = g(x) g() = x g(x) g() x = f,(x) g, (x). Using the rithmetic of limits, we obtin the desired result. (33 hrs, 2011) (30 hrs, 2009) Exercise 3.3 Show tht the function 1 f (x) = x + 2 x2 sin 1 x 0 x 0 x = 0 stisfies f (0) > 0, nd yet, is not incresing in ny neighborhood of 0 (i.e., there is no neighborhood of zero in which x < y implies f (x) < f (y)).

Derivtives 117 Exercise 3.4 Let I be n open segment. Any suppose tht f (x) = f (x) for ll x I. Show tht there exists constnt c, such tht f (x) = c e x. (Hint: consider the function g(x) = f (x)/e x.) Exercise 3.5 ➀ Let f : R R be differentible. Show tht if f vnishes t m points, then f vnishes t t lest m 1 points. ➁ Let f be n times continuously differentible in [, b], nd n + 1 times differentible in (, b). Suppose furthermore tht f (b) = 0 nd f (k) () = 0 for k = 0, 1,..., n. Show tht there exists c (, b) such tht f (n+1) (c) = 0. Exercise 3.6 Let sin(ωx) x < 0 f (x) = γ x = 0 αx 2 + βx x > 0. Find ll the vlues of α, β, γ, ω for which f is (i) continuous t zero, (ii) differentible t zero, (iii) twice differentible t zero. Exercise 3.7 Clculte the following limits: ➀ lim x 0 sin(x) sin(bx) cos(cx). ➁ lim x 0 tn(x) x x sin(x). ➂ lim x 0 x 2 sin(1/x) sin(x). ➃ lim x 1 (1 x) tn ( πx 2 ). Exercise 3.8 Suppose tht f : (0, ) R is differentible on (0, ) nd lim f = L. Prove, without using l Hopitl s theorem tht lim x f (x)/x = L. Hint: write f (x) x = f (x) f (x 0) x x 0, x x 0 x nd choose x 0 ppropritely in order to use Lgrnge s theorem.

118 Chpter 3 Figure 3.2: The derivtive of the inverse function 3.5 Derivtives of inverse functions We hve lredy shown tht if f : A B is one-to-one nd onto, then it hs n inverse f 1 : B A. We then sy tht f is invertible.(הפיכה ) We hve seen tht n invertible function defined on n intervl must be monotonic, nd tht the inverse of continuous function is continuous. In this section, we study under wht conditions is the inverse of differentible function differentible. Recll lso tht f 1 f = Id, with Id : A A. If both f nd f 1 were differentible t nd f (), respectively, then by the composition rule, ( f 1 ) ( f ()) f () = 1. Setting f () = b, or equivlently, = f 1 (b), we get ( f 1 ) (b) = 1 f ( f 1 (b)). There is one little flw with this rgument. We hve ssumed ( f 1 ) to exist. If this were indeed the cse, then we hve just derived n expression for the derivtive of the inverse. Corollry 3.5 If f ( f 1 (b)) = 0 then f 1 is not differentible t b.

Derivtives 119 Exmple: The function f (x) = x 3 + 2 is continuous nd invertible, with f 1 (x) = 3 x 2. Since, f ( f 1 (2)) = 0, then f 1 is not differentible t 2. (39 hrs, 2010) Theorem 3.19 Let f : I R be continuous nd invertible. If f is differentible t f 1 (b) nd f ( f 1 (b)) 0, then f 1 is differentible t b. Proof : Let b = f (), i.e., = f 1 (b). It is given lim f, = f () 0. To ech x corresponds y b such tht Then, f, (x) = which we cn rewrite s y = f (x) nd x = f 1 (y). f (x) f () x f 1,b(y) = = y b f 1 (y) f 1 (b) = 1 f 1,b(y), 1 f, (x) = 1 f, ( f 1 (y)). Since f 1 is continuous t b nd f, is continuous t = f 1 (b), it follows from limit rtithmetic tht 1 lim f 1,b = b f ( f 1 (b)). Exmple: Consider the function f (x) = tn x defined on ( π/2, π/2). We denote its inverse by rctn x. Then 4, ( f 1 ) (x) = 1 f ( f 1 (x)) = cos2 (rctn x) = 1 1 + tn 2 (rctn x) = 1 1 + x. 2 (35 hrs, 2013) 4 We used the identity cos 2 = 1/(1 + tn 2 ).

120 Chpter 3 Exmple: Consider the function f (x) = x n, with n integer. This proves tht the derivtive of f 1 (x) = x 1/n is ( f 1 ) (x) = 1 n x1/n 1, nd together with the composition rule gives the derivtive for ll rtionl powers. Comment: So fr, we hve not defined rel-vlued powers. We will get to it fter we study integrtion theory. Exercise 3.9 Let f : R R be one-to-one nd onto, nd furthermore differentible with f (x) 0 for ll x R. Suppose tht there exists function F : R R such tht F = f. Define G(x) = x f 1 (x) F( f 1 (x)). Prove tht G (x) = f 1 (x). Note tht this sttement sys tht if we know function whose derivtive is one-to-one nd onto, then we know the function whose derivtive is its inverse. Exercise 3.10 Let f be invertible nd twice differentible in (, b). Furthermore, f (x) 0 in (, b). Prove tht ( f 1 ) exists nd find n expression for it in terms of f nd its derivtives. Exercise 3.11 Suppose tht f : R R is continuous, one-to-one nd onto, nd stisfies f = f 1. (i) Show tht f hs fixed point, i.e., there exists n x R such tht f (x) = x. (ii) Find such n f tht hs unique fixed point. Hint: strt by figuring out how do such functions look like. 3.6 Complements Theorem 3.20 Suppose tht f : A B is monotonic (sy, incresing), then it hs one-sided limits t every point tht hs one-sided neighborhood tht belongs to A. (In prticulr, if A is connected set then f hs one-sided limits everywhere.) Proof : Consider segment (b, ] A. We need to show tht f hs left-limit t. The set K = { f (x) : b < x < }

Derivtives 121 is non-empty nd upper bounded by f () (since f is incresing). Hence we cn define l = sup K. We re going to show tht l = lim Let ɛ > 0. By the definition of the supremum, there exists n m K such tht m > l ɛ. Let y be such f (y) = m. By the monotonicity of f, l ɛ < f (x) l whenever y < x <. f. This concludes the proof. We proceed similrly for right-limits. Theorem 3.21 Let f : [, b] R be monotonic (sy, incresing). Then the imge of f is segment if nd only if f is continuous. Proof : (i) Suppose first tht f is continuous. By monotonicity, Imge( f ) [ f (), f (b)]. Tht every point of [ f (), f (b)] is in the imge of f follows from the intermeditevlue theorem. (ii) Suppose then tht Imge( f ) = [ f (), f (b)]. Suppose, by contrdiction, tht f is not continuous. This implies the existence of point c [, b], where f (c) > lim c f nd/or f (c) < lim c + (We rely on the fct tht one-sided derivtives exist.) Sy, for exmple, tht f (c) < lim c + f. Then, f (c) < f (c) + lim c + f 2 < lim c + is not in the imge of f, which contrdicts the fct tht Imge( f ) = [ f (), f (b)]. f f. Theorem 3.22 (Jen-Gston Drboux) Let f : A B be differentible on (, b), including right-derivtive t nd left-derivtive t b. Then, f hs the intermedite-vlue property, whereby for every t between f ( + ) nd f (b ) there exists n x [, b], such tht f (x) = t.

122 Chpter 3 Comment: The sttement is trivil when f is continuously differentible (by the intermedite-vlue theorem). It is nevertheless correct even if f hs discontinuities. This theorem thus limits the functions tht re derivtives of other functions not every function cn be derivtive. Proof : Without loss of generlity, let us ssume tht f ( + ) > f (b ). Let f ( + ) > t > f (b ), nd consider the function g(x) = f (x) tx. (Here t is fixed prmeter; for every t we cn construct such function.) We hve g ( + ) = f ( + ) t > 0 nd g (b ) = f (b ) t < 0, nd we wish to show the existence of n x [, b] such tht g (x) = 0. Since g is continuous on [, b] then it must ttin mximum. The point cnnot be mximum point becuse g is incresing t. Similrly, b cnnot be mximum point becuse g is decresing t b. Thus, there exists n interior point c which is mximum, nd by Fermt s theorem, g (c) = f (c) t = 0. (32 hrs, 2009) Exmple: The function x 2 sin 1/x x 0 f (x) = 0 x = 0 is differentible everywhere, nd its derivtive is f 2x sin 1/x cos 1/x x 0 (x) = 0 x = 0. The derivtive is not continuous t zero, but nevertheless stisfes the Drboux theorem. Exmple: How crzy cn continuous function be? It ws Weierstrss who shocked the world by constructing continuous function tht it nowhere differentible! As instnce of his clss of exmples is cos(21 k πx) f (x) =. 3 k k=0

Derivtives 123 3.7 Tylor s theorem Recll the men-vlue theorem. If f : [, b] R is continuous on [, b] nd differentible in (, b), then for every point x (, b) there exists point c x, such tht f (x) f () = f (c x ). x We cn write it lterntively s f (x) = f () + f (c x )(x ). Suppose, for exmple, tht we knew tht f (y) < M for ll y. Then, we would deduce tht f (x) cnnot differ from f () by more thn M times the seprtion x. In prticulr, lim [ f f ()] = 0. It turns out tht if we hve more informtion, bout higher derivtives of f, then we cn refine lot the estimtes we hve on the vrition of f s we get wy from the point. Theorem 3.23 Let f be given function, nd n interior point of its domin. Suppose tht the derivtives f (), f (),..., f (n) () ll exist. Let k = f (k) ()/k! nd set the Tylor polynomil of degree n, P f,n, (x) = 0 + 1 (x ) + 2 (x ) 2 + + n (x ) n. Then, lim f P f,n, (Id ) n = 0. Comment: Loosely speking, this theorem sttes tht, s x pproches, P f,n, (x) is closer to f (x) thn (x ) n (or tht f nd P f,n, re equl up to order n ; see below). For n = 0 it sttes tht lim [ f f ()] = 0,

124 Chpter 3 wheres for n = 1, lim ( ) f f () (Id ) f () = lim f, f () = 0, which holds by definition. Proof : Consider the rtio f P f,n, (Id ) n, which is rtio of n-times differentible functions t. Both numertor nd denomintor vnish t, so we will ttempt to use l Hôpitl s rule. For tht we note tht P f,n, = P f,n 1,. Thus, lim f P f,n, (Id ) n = lim f P f,n 1, n (Id ) n 1, provided tht the right hnd side exists. We then proceed to consider the right hnd side, which is the limit of the rtio of two differentible functions tht vnish t. Hence, lim f P f,n 1, n (Id ) n 1 = lim f P f,n 2, n(n 1) (Id ) n 2, provided tht the right hnd side exists. We proceed inductively, until we get tht lim f P f,n, f (n 1) P f = lim (n 1),1,, (Id ) n n! (Id ) provided tht the right hnd side exists. Now write the right hnd side explicitly, 1 n! lim f (n 1) f (n 1) () f (n) ()(Id ) Id = 1 n! [ lim ] f (n 1), f (n) (), which vnishes by the definition of f (n) (). This concludes the proof. Comment: Why couldn t we use Tylor s theorem one more time t the end? Becuse it is not given tht the n-th derivtive hs limit t. It is only given tht the n-derivtive exists t tht point. (35 hrs, 2011)

Derivtives 125 The polynomil P f,n, (x) is known s the Tylor polynomil of degree n of f bout. Its existence relies on f being differentible n times t. If we denote by Π n the set of polynomils of degree t most n, then the defining property of P f,n, is: 1. P f,n, Π n. 2. P (k) f,n, () = f (k) () for k = 0, 1,..., n. Definition 3.6 Let f, g be defined in neighborhood of. We sy tht f nd g re equl up to order n t if lim f g (Id ) n = 0. It is esy to see tht equl up to order n is n equivlence reltion, which defines n equivlence clss. Thus, the bove theorem proves tht f (x) nd P f,n, (provided tht the ltter exists) re equl up to order n t. (36 hrs, 2013) The next tsk is to obtin n expression for the difference between the function f nd its Tylor polynomil. We define the reminder (שארית ) of order n, R n (x), by f (x) = P f,n, (x) + R f,n, (x). Proposition 3.1 Let P, Q Π n be equl up to order n t. Then P = Q. Proof : We cn lwys express these polynomils s P(x) = 0 + 1 (x ) + + n (x ) n Q(x) = b 0 + b 1 (x ) + + b n (x ) n. We know tht For k = 0 lim P Q = 0 for k = 0, 1,..., n. (Id ) k 0 = lim (P Q) = 0 b 0, i.e., 0 = b 0. We proceed similrly for ech k to show tht k = b k for k = 0, 1,..., n.

126 Chpter 3 Corollry 3.6 If f is n times differentible t nd f is equl to Q Π n up to order n t, then P f,n, = Q. Exmple: We know from high-school mth tht for x < 1 Thus, f (x) = 1 1 x = n k=0 x k + xn+1 1 x. f (x) n k=0 x k x n = x 1 x. Since the right hnd side tends to zero s x 0 it follows tht f nd n k=0 x k re equl up to order n t 0, which mens tht P f,n,0 (x) = n x k. k=0 Exmple: Similrly for x < 1 f (x) = 1 n 1 + x = ( 1) k x 2k n+1 x2n+2 + ( 1) 2 1 + x. 2 k=0 Thus, f (x) n k=0( 1) k x 2k x 2n = x2 1 + x 2. Since the right hnd side tends to zero s x 0 it follows tht P f,2n,0 (x) = n ( 1) k x 2k. k=0

Derivtives 127 Exmple: Let f = rctn. Then, x dt x f (x) = 0 1 + t = 2 0 n = ( 1) k x2k+1 2k + 1 + Since it follows tht (36 hrs, 2011) k=0 n ( 1) k t 2k + ( 1) k=0 x 0 n+1 t2n+2 ( 1) 1 + t dt. 2 n f (x) ( 1) k x2k+1 2k + 1 x 2n+3, k=0 P f,2n+2,0 = n ( 1) k x2k+1 2k + 1. k=0 n+1 t2n+2 1 + t dt 2 Theorem 3.24 Let f, g be n times differentible t. Then, P f +g,n, = P f,n, + P g,n,, nd P f g,n, = [P f,n, P g,n, ] n, where [] n stnds for trunction of the polynomil in (Id )t the n-th power. Proof : We use gin the fct tht if polynomil looks like the Tylor polynomil then it is. The sum P f,n, + P g,n, belongs to Π n nd ( f + g) (P f,n, + P g,n, ) f P f,n, lim = lim (Id ) n (Id ) + lim n which proves the first sttement. Second, we note tht P f,n, P g,n, Π 2n, nd tht g P g,n, (Id ) n = 0, f g P f,n, P g,n, (Id ) n = f g P f,n,g + P f,n, g P f,n, P g,n, (Id ) n = g f P f,n, (Id ) + P g P g,n, n f,n, (Id ), n

128 Chpter 3 whose limit t is zero. This proves tht P f,n, P g,n, nd f g re equivlent up to order n t. All tht remins it to truncte the product t the n-th power of (Id ). (38 hrs, 2013) Theorem 3.25 (Tylor) Suppose tht f is (n + 1)-times differentible on the intervl [, x]. Then, f (x) = P f,n, (x) + R n, f, (x), where the reminder R f,n, (x) cn be represented in the Lgrnge form, R f,n, (x) = f (n+1) (c) (n + 1)! (x )n+1, for some c (, x). It cn lso be represented in the Cuchy form, for some ξ (, x). R f,n, (x) = f (n+1) (ξ) (x ξ) n (x ), n! Comment: For n = 0 this is simply the men-vlue theorem. Proof : Fix x, nd consider the function φ : [, x] R, φ(z) = f (x) P f,n,z (x) We note tht = f (x) f (z) f (z)(x z) f (z) 2! φ() = R f,n, (x) nd φ(x) = 0. (x z) 2 f (n) (z) (x z) n. n! Moreover, φ is differentible, with φ (z) = f (z) [ f (z)(x z) f (z) ] ( ) f (z) (x z) 2 f (z)(x z) 2! ( f (n+1) (z) (x z) n f ) (n) (z) (x z)n 1 n! (n 1)! = f (n+1) (z) (x z) n. n!

Derivtives 129 Let now ψ be ny function defined on [, x], differentible on (, x) with nonvnishing derivtive. By Cuchy s men-vlue theorem, there exists point < ξ < x, such tht φ(x) φ() ψ(x) ψ() = φ (ξ) ψ (ξ). Tht is, R f,n, (x) = ψ(x) ψ() ψ (ξ) Set for exmple ψ(z) = (x z) p for some p. Then, f (n+1) (ξ) (x ξ) n n! R f,n, (x) = (x ) p f (n+1) (ξ) (x ξ) n p+1 p n! For p = 1 we retrieve the Cuchy form. For p = n + 1 we retrieve the Lgrnge form. Comment: Lter in this course we will study series nd sk questions like does the Tylor polynomil P f,n, (x) tend to f (x) s n?. This is the notion of converging series. Here we del with different best: the degree of the polynomil n is fixed nd we consider how does P f,n, (x) pproch f (x) s x. The fct tht P f,n, pproches f fster thn (x ) n s x mens tht P f,n, is n symptotic series of f. (34 hrs, 2009) Exercise 3.12 Write the polynomil p(x) = x 3 3x 2 + 3x 6 s polynomil in x + 1. Tht is, find coefficients 0, 1,..., n, such tht p(x) = n k=0 k (x + 1) k. Solve this problem using two methods: (i) Newton s binomil, nd (ii) Tylor s polynomil. Exercise 3.13 Let f : R R be n times differentible, such tht f (n) (x) = 0 for ll x R. Prove tht f is polynomil whose degree is t most n 1. Exercise 3.14 Write the Tylor polynomils of degree n t the point x 0 for the following functions, s well s the corresponding reminders, both in the Lgrnge nd Cuchy forms: ➀ f (x) = 1/(1 + x), x 0 = 0, n N. ➁ f (x) = cosh(x), x 0 = 0, n N.

130 Chpter 3 ➂ f (x) = x 5 x 4 + 7x 3 2x 2 + 3x + 1, x 0 = 1, n = 5. ➃ f (x) = (x 2) 6 + 9(x + 1) 4 + (x 2) 2 + 8, x 0 = 0, n = 6. ➄ f (x) = (1 + x) α, α N, x 0 = 0, n N. Use the nottion ( ) α def α(α 1)... (α n + 1) =, n n! which generlizes Newton s binomil. ➅ f (x) = sin(x), x 0 = π/2, n = 2k, k N. Exercise 3.15 Show using different methods tht of f () exists, then ➀ Using l Hopitl s rule. f f ( + h) + f ( h) 2 f () () = lim. h 0 h ➁ Using Tylor s polynomil of order two f t. Note tht you cn t use neither Lgrnge or Cuchy s form of the reminder, since it is only given tht f is twice differentible t. ➂ Show tht if x 2 x 0 f (x) = x 2 x < 0, then f (0) does not exist, but the limit of the bove frction does exist. Why isn t it contrdiction? Exercise 3.16 Evlute, using Tylor s polynomil, the following expressions, up to n ccurcy of t lest 0.01: ➀ e 1/2. ➁ cos(1/2). Exercise 3.17 ➀ For every polynomil P(x) = n k=0 k x k nd evert q N, we denote q k=0 P q (x) = kx k q < n P(x) q n.

Derivtives 131 Suppose tht f, g hve Tylor polynomils of order n t zero, P nd Q. Suppose furthermore tht g(0) = 0, nd tht g does not vnish in punctured neighborhood of zero. Show tht the Tylor polynomil of order n t zero of f g is given by (P Q) n. Hint: write f (g(x)) = f (g(x))/g(x) g(x). ➁ Given the Tylor polynomil P of order n t zero of the function f, write the Tylor polynomil of order n t zero of the function g(x) = f (2x). ➂ Find the Tylor polynomil of order three t zero of the function f (x) = exp(sin(x)). Exercise 3.18 Let f be n times differentible t x 0 nd let P(x) = n k=0 k x k be its Tylor polynomil t zero. Wht is the Tylor polynomil of f of order n 1 t zero. Justify your nswer. Exercise 3.19 Show tht if f nd g re n times differentible t x 0, then ( f g) (n) (x 0 ) = n k=0 ( ) n f (k) (x 0 ) g (n k) (x 0 ). k Hint: think of the Tylor polynomil of product of functions. Exercise 3.20 Suppose tht f nd g hve derivtives of ll orders in R nd tht P n (x) = Q n (x) for ll n N nd every x R, where P n nd Q n re the corresponding Tylor polynomils t zero. Does it imply tht f (x) = g(x)?

132 Chpter 3

Chpter 4 Integrtion theory 4.1 Definition of the integrl Derivtives were bout clculting the rte of chnge of function. Integrls were invented in order to clculte res. The fct tht the two re intimtely relted tht derivtives re in sense the inverse of integrls is wonder. When we lern in school bout the re of region, it is defined s the number of squres with sides of length one tht fit into the region. It is of course hrd to mke sense with such definition. We will lern how to clculte the re of region bounded by the x-xis, the grph of function f, nd two verticl lines pssing though points (, 0) nd (b, 0). For the moment let us ssume tht f 0. This region will be denoted by R( f,, b), nd its re (once defined) will be clled the integrl of f on [, b] 1. 1 There re multiple non-equivlent wys to define integrls. We study integrtion theory in the sense of Riemnn; nother clssicl integrtion theory is tht of Lebesgue (studied in the context of mesure theory). Within the Riemnn integrtion theory, we dopt prticulr derivtion due to Drboux.

134 Chpter 4 f R!f,,b" b Definition 4.1 A prtition (חלוקה ) of the intervl [, b] is prtition (yes, I know, you cn t use the word to be defined in definition) of the intervl into finite number of segments, [, b] = [x 0, x 1 ] [x 1, x 2 ] [x n 1, x n ]. We will identify prtition with finite number of points, = x 0 < x 1 < x 2 < < x n = b. Definition 4.2 Let f : [, b] R be bounded function, nd let P = {x 0, x 1,..., x n } be prtition of [, b]. Let m i = inf{ f (x) : x i 1 x x i } M i = sup{ f (x) : x i 1 x x i }. We define the lower sum תחתון ) (סכום of f for P to be L( f, P) = n m i (x i x i 1 ). i=1 We define the upper sum עליון ) (סכום of f for P to be U( f, P) = n M i (x i x i 1 ). i=1 Clerly, ny sensible definition of the re A of R( f,, b) must stisfy for ll prtitions P. L( f, P) A U( f, P)

Integrtion theory 135 f x 0 x 1 x 2 x 3 b Figure 4.1: The lower nd upper sums of f for P. Comments: ➀ The boundedness of f is essentil in these definitions. ➁ It is lwys true tht L( f, P) U( f, P) becuse m i M i for ll i. ➂ Is it however true tht L( f, P 1 ) U( f, P 2 ) where P 1 nd P 2 re rbitrry prtitions? It ought to be true. (35 hrs, 2009) (38 hrs, 2011) Exercise 4.1 Let f : [ 1, 1] R be given by f : x x 2 nd let P = { 1, 1/2, 1/3, 2/3, 8/9, 1} be prtition of [ 1, 1]. Clculte L( f, P) nd U( f, P). Definition 4.3 A prtition Q is clled refinement (עידון ) of prtition P if P Q. Lemm 4.1 Let Q be refinement of P. Then L( f, Q) L( f, P) nd U( f, Q) U( f, P).

136 Chpter 4 Proof : Suppose first tht Q differs from P by exctly one point. Tht is, Then, nd where Thus, P = {x 0, x 1,..., x n } Q = {x 0, x 1,..., x j 1, y, x j,..., x n }. L( f, P) = n m i (x i x i 1 ) j 1 L( f, Q) = m i (x i x i 1 ) + m j (y x j 1 ) + ˆm j (x j y) + i=1 i=1 m j = inf{ f (x) : x j 1 x y} ˆm j = inf{ f (x) : y x x j }. n m i (x i x i 1 ), i= j+1 L( f, Q) L( f, P) = ( m j m j )(y x j 1 ) + ( ˆm j m j )(x j y). It remins to note tht m j m j nd ˆm j m j, since both re infimums over smller set 2. Thus, L( f, Q) L( f, P) 0. Consider now two prtitions, P Q. There exists sequence of prtitions, P = P 0 P 1 P k = Q, such tht ech P j differs from P j 1 by one point. Hence, L( f, P) L( f, P 1 ) L( f, Q). A similr rgument works for the upper sums. Theorem 4.1 For every two prtitions P 1,P 2 of [, b], L( f, P 1 ) U( f, P 2 ). 2 It is lwys true tht if A B then inf A inf B nd sup A sup B.

Integrtion theory 137 Proof : Let Q = P 1 P 2. Since Q is finer thn both P 1 nd P 2 it follows tht L( f, P 1 ) L( f, Q) U( f, Q) U( f, P 2 ). (40 hrs, 2013) Exercise 4.2 Prove or disprove: ➀ A prtition Q is refinement of prtition P if nd only if every segment in Q is contined in segment of P. ➁ If prtition P is refinement of prtition Q then every segment in Q is segment in P. ➂ If R is prtition of both P nd Q then it is lso refinement of P Q. Exercise 4.3 Chrcterize the functions f for which there exists prtition P such tht L( f, P) = U( f, P). We my now consider the collection of lower nd upper sums. Denote by Prt(, b) = {All prtitions of [, b]}. L ( f ) = {L( f, P) : P Prt(, b)} U ( f ) = {U( f, P) : P Prt(, b)}. Note tht both L ( f ) nd U ( f ) re sets of rel numbers, nd not sets of prtitions. The set L ( f ) is non-empty nd bounded from bove by ny element of U ( f ). Similrly, the set U ( f ) is non-empty nd bounded from below by ny element of L ( f ). Thus, we my define the lest upper bound nd gretest lower bound, L( f ) = sup L ( f ) nd U( f ) = inf U ( f ). The rel number U( f ) is clled the upper integrl עליון ) (אינטגרל of f on [, b] nd L( f ) is clled the lower integrl תחתון ) (אינטגרל of f on [, b] We once proved theorem stting tht the following ssertions re equivlent: ➀ sup L ( f ) = inf U ( f ). ➁ There is unique number c, such tht l c u for ll l L ( f ) nd u U ( f ).

138 Chpter 4 ➂ For every ɛ > 0 there exist n l L ( f ) nd u U ( f ), such tht u l < ɛ. Definition 4.4 If L( f ) = U( f ) we sy tht f is integrble on [, b]. We cll this number the integrl of f on [, b], nd denote it by b f. Tht is, f is integrble on [, b] if the upper nd lower integrls coincide. Comment: The integrl of f is the unique number such tht L( f, P) b f U( f, Q), for ll prtitions P nd Q. Comment: You re ll ccustomed to the nottion b f (x) dx. This nottion is indeed very suggestive bout the mening of the integrl, however, it is very importnt to relize tht the integrl only involves f, nd b, nd it is not function of ny x. Comment: At this stge we only consider integrls of bounded functions over bounded segments so-clled proper integrls. Integrls like f nd 1 0 log x dx, re instnces of improper integrls. We will only briefly tlk bout them (only due to lck of time). Comment: Wht bout the cse where the sign of f is not everywhere positive? The definition remins the sme. The geometric interprettion is tht res under the x-xis count s negtive res. Two questions rise:

Integrtion theory 139 ➀ How do we know whether function is integrble? ➁ How to compute the integrl of n integrble function? (40 hrs, 2011) The following lemm will prove useful in checking whether function is integrble: Theorem 4.2 A bounded function f is integrble if nd only if there exists for every ɛ > 0 prtition P, such tht U( f, P) < L( f, P) + ɛ. Proof : If for every ɛ > 0 there exists prtition P, such tht U( f, P) < L( f, P) + ɛ, then f is integrble, by the equivlent criteri described bove. Conversely, suppose f is integrble. Then, given ɛ > 0, there exist prtitions P, P, such tht U( f, P) L( f, P ) < ɛ. Let P = P P. Since P is refinement of both, then U( f, P ) L( f, P ) U( f, P) L( f, P ) < ɛ. Exmple: Let f : [, b] R, f : x c. Then for every prtition P, n L( f, P) = m i (x i x i 1 ) = c(b ) U( f, P) = i=1 n M i (x i x i 1 ) = c(b ). It follows tht L( f ) = U( f ) = c(b ), hence f is integrble nd i=1 b f = c(b ).

140 Chpter 4 Exmple: Let f : [0, 1] R, 0 x Q f : x 1 x Q. For every prtition nd every j, m j = 0 nd M j = 1, hence L( f, P) = 0 nd U( f, P) = 1, nd L( f ) = 0 nd U( f ) = 1. Since L( f ) U( f ), f is not integrble. (41 hrs, 2010) Exmple: Consider next the function f : [0, 2] R, 0 x 1 f : x 1 x = 1. Let P n, n N denote the fmily of uniform prtitions, For every such prtition, x i = 2i n. L( f, P) = 0 nd U( f, P) = 2/n. Set now, L unif ( f ) = {L( f, P n ) : n N} U unif ( f ) = {U( f, P n ) : n N} Clerly, L unif ( f ) L ( f ) nd U unif ( f ) U ( f ), hence L( f ) = sup L ( f ) sup L unif ( f ) = 0 U( f ) = inf U ( f ) inf U unif ( f ) = 0, thus 0 L( f ) U( f ) 0,

Integrtion theory 141 which implies tht f is integrble, nd 2 0 f = 0. This exmple shows tht integrls re not ffected by the vlue of function t n isolted point. The following exmple motivte the following useful lemm: Lemm 4.2 Let A Prt(, b) be subset of the set of prtitions. Suppose tht sup{l( f, P) : P A } = inf{u( f, P) : P A } = l. Then f is integrble on [, b] nd b f = l. Proof : Since if follows tht {L( f, P) : P A } L ( f ) nd {U( f, P) : P A } U ( f ), l = sup{l( f, P) : P A } sup L ( f ) inf U ( f ) inf{u( f, P) : P A } = l, which completes the proof. We re next going to exmine number of exmples, which on the one hnd, will prctice our understnding of integrbility, nd on the other hnd, will convince us tht we must develop tools to del with more complicted functions. Exmple: Let f : [0, b] R, f = Id. Let P be ny prtition, then m i = x i 1 nd M i = x i, hence L( f, P) = U( f, P) = n x i 1 (x i x i 1 ) i=1 n x i (x i x i 1 ). i=1

142 Chpter 4 Using gin uniform prtition, Since L( f, P n ) = U( f, P n ) = P n = {0, b/n, 2b/n,..., b}, n i=1 n i=1 (i 1)b b n n = b2 n(n 1) n 2 2 ib b n n = b2 n(n + 1) n 2 2 = b2 2 = b2 2 ( 1 + 1 n ( 1 1 ) n ). sup{l( f, P n ) : n N} = b2 2 if follows from Lemm 4.2 tht b 0 nd inf{u( f, P n ) : n N} = b2 2, f = b2 2. (37 hrs, 2009) Exmple: Let f : [0, b] R, f (x) = x 2. Let P be ny prtition, then m i = x 2 i 1 nd M i = x 2 i, hence L( f, P) = U( f, P) = Tke the sme uniform prtition, then 3 L( f, P n ) = U( f, P n ) = n (i 1) 2 b 2 i=1 n i=1 i 2 b 2 n 2 n 2 n x 2 i 1 (x i x i 1 ) i=1 n xi 2 (x i x i 1 ). i=1 b n = b3 (n 1)n(2n 1) n 3 6 b n = b3 n(n + 1)(2n + 1) n 3 6 = b3 3 = b3 3 ( 1 + 1 n ( 1 1 ) ( 1 1 n 2n ) ( 1 + 2 ). n ) 3 n k=1 k 2 = 1 n(n + 1)(2n + 1). 6

Integrtion theory 143 Since sup{l( f, P n ) : n N} = b3 3 if follows from Lemm 4.2 tht nd inf{u( f, P n ) : n N} = b3 3, b 0 f = b3 3. This gme cn t go much further, i.e., we need stronger tools. We strt by verifying which clsses of functions re integrble. (42 hrs, 2010) 4.2 Integrtion theorems Theorem 4.3 If f [, b] R is monotone, then it is integrble. Proof : Assume, without loss of generlity, tht f is monotoniclly incresing. Then, for every prtition P, L( f, P) = n f (x i 1 )(x i x i 1 ) nd U( f, P) = i=1 n f (x i )(x i x i 1 ), i=1 hence, U( f, P) L( f, P) = n [ f (x i ) f (x i 1 )](x i x i 1 ). i=1 Tke now fmily of eqully-spced prtitions, P n = { + j(b )/n} n j=0. For such prtitions, U( f, P n ) L( f, P n ) = b n n [ f (x i ) f (x i 1 )] = i=1 (b )[ f (b) f ()]. n Thus, for every ɛ > 0 we cn choose n > (b )[ f (b) f ()]/ɛ, which gurntees tht U( f, P n ) L( f, P n ) < ɛ (we used the fct tht the sum ws telescopic ).

144 Chpter 4 Theorem 4.4 If f [, b] R is continuous, then it is integrble. Proof : First, continuous function on segment is bounded. We wnt to show tht for every ɛ > 0 there exists prtition P, such tht U( f, P) < L( f, P) + ɛ. We need to somehow control the difference between m i nd M i. Recll tht function tht is continuous on closed intervl is uniformly continuous. Tht is, there exists δ > 0, such tht f (x) f (y) < ɛ b Tke then prtition P such tht whenever x y < δ. x i x i 1 < δ for ll i = 1,..., n. Since continuous functions on closed intervls ssume minim nd mxim in the intervl, it follows tht M i m i < ɛ/(b ), nd U( f, P) L( f, P) = n (M i m i )(x i x i 1 ) < ɛ b i=1 n (x i x i 1 ) = ɛ. i=1 (42 hrs, 2011) Theorem 4.5 Let f be bounded function on [, b], nd let < c < b. Then, f is integrble on [, b] if nd only if it is integrble on both [, c] nd [c, b], in which cse b f = c f + b c f. Proof : (1) Suppose first tht f is integrble on [, b]. It is sufficient to show tht it is integrble on [, c]. For every ɛ > 0 there exists prtition P of [, b], such tht U( f, P) L( f, P) < ɛ.

Integrtion theory 145 If the prtition P does not include the point c then let Q = P {c}, nd U( f, Q) L( f, Q) U( f, P) L( f, P) < ɛ. Suppose tht x j = c, then P = {x 0,..., x j } is prtition of [, c], nd P = {x j,..., x n } is prtition of [c, b]. We hve L( f, Q) = L( f, P ) + L( f, P ) nd U( f, Q) = U( f, P ) + U( f, P ), from which we deduce tht [U( f, P ) L( f, P )] = [U( f, Q) L( f, Q)] [U( f, P } {{ } ) L( f, P )] < ɛ, } {{ } <ɛ 0 hence f is integrble on [, c]. (2) Conversely, suppose tht f is integrble on both [, c] nd [c, b]. Then there exists prtition P of [, c] nd prtition P of [c, b], such tht U( f, P ) L( f, P ) < ɛ 2 nd U( f, P ) L( f, P ) < ɛ 2. Since P = P P is prtition of [, b], summing up, we obtin the desired result. Thus, we hve shown tht f is integrble on [, b] if nd only if it is integrble on both [, c] nd [c, b]. Next, since it follows tht L( f, P ) L( f, P ) c b c f U( f, P ) for every prtition P of [, c] f U( f, P ) for every prtition P of [c, b], L( f, P) c f + b c f U( f, P) for every prtition P of [, b]. (More precisely, for every prtition P tht is union of prtitions of P nd P, but it is esy to see tht this implies for every prtition.) Since there is only one such number, it is equl to b f. (39 hrs, 2009)

146 Chpter 4 Comment: So fr, we hve only defined the integrl on [, b] with b >. We now dd the following conventions, b b f = f nd f = 0, so tht (s cn esily be checked) the ddition rule lwys holds. Theorem 4.6 If f nd g re integrble on [, b], then so is f + g, nd b ( f + g) = b f + b g. Proof : First of ll, we observe tht if f nd g re bounded functions on set A, then 4 inf{ f (x) + g(x) : x A} inf{ f (x) : x A} + inf{g(x) : x A} sup{ f (x) + g(x) : x A} sup{ f (x) : x A} + sup{g(x) : x A}. This implies tht for every prtition P, hence, L( f + g, P) L( f, P) + L(g, P) U( f + g, P) U( f, P) + U(g, P), U( f + g, P) L( f + g, P) [U( f, P) L( f, P)] + [U(g, P) L(g, P)]. (4.1) We re prcticlly done. Given ɛ > 0 there exist prtitions P, P, such tht U( f, P ) L( f, P ) < ɛ 2 nd U(g, P ) L(g, P ) < ɛ 2. 4 This is becuse for every x, f (x) + g(x) inf{ f (x) : x A} + inf{g(x) : x A}, so tht the right-hnd side is lower bound of f + g.

Integrtion theory 147 Tke P = P P, then U( f, P) L( f, P) U( f, P ) L( f, P ) < ɛ 2 U(g, P) L(g, P) U(g, P ) L(g, P ) < ɛ 2, nd it remins to substitute into (4.1), to prove tht f + g is integrble. Now, for every prtition P, L( f, P) + L(g, P) L( f + g, P) nd lso It follows tht b ( f + g) L( f, P) + L(g, P) b f b b b ( f + g) U( f + g, P) U( f, P) + U(g, P), f + b g U( f, P) + U(g, P), g [U( f, P) + U(g, P)] [L( f, P) + L(g, P)]. Since the right-hnd side cn be mde less thn ɛ for every ɛ, it follows tht it is zero 5. Corollry 4.1 If f is integrble on [, b] nd g differs from f only t finite number of points then b f = b g. Proof : We sw tht (g f ), which is non-zero only t finite number of points, is integrble nd its integrl is zero. Then we only need to use the dditivity of the integrl for g = f + (g f ). Exercise 4.4 Denote by the upper integrl. Let f, g be bounded on [, b]. Show tht b ( f + g) b f + b Also, find two functions f, g for which the inequlity is strong. 5 If x < ɛ for every ɛ > 0 then x = 0. g.

148 Chpter 4 Exercise 4.5 Recll the Riemnn function, Let [, b] be gievn intervl. 1/q x = p/q Q R(x) = 0 x Q. ➀ Show tht b R = 0 (the lower integrl). ➁ Let ɛ > 0 nd define Z ɛ = {x [, b] : R(x) > ɛ}. Show tht Z ɛ is finite set. ➂ Show tht there exists prtition P such tht U(R, P) < ɛ. ➃ Conclude tht R is integrble on [, b] nd tht b R = 0. (44 hrs, 2010) Theorem 4.7 If f is integrble on [, b] nd c R, then c f is integrble on [, b] nd b (c f ) = c b f. Proof : Strt with the cse where c > 0. For every prtition P, L(c f, P) = c L( f, P) nd U(c f, P) = c U( f, P). Since f is integrble on [, b], then there exists for every ɛ > 0 prtition P, such tht U( f, P) L( f, P) < ɛ c, hence U(c f, P) L(c f, P) < ɛ, which proves tht c f is integrble on [, b]. Then, we note tht for every prtition P, b L(c f, P) = c L( f, P) c f c U( f, P) = U(c f, P).

Integrtion theory 149 Since there is only one number l such tht L(c f, P) l U(c f, P) for ll prtitions P, nd this number is by definition b (c f ), then b (c f ) = c b f. Consider then the cse where c < 0; in fct, it is sufficient to consider the cse of c = 1, since for rbitrry c < 0 we write c f = ( 1) c f. For every prtition P, L( f, P) = U( f, P) nd U( f, P) = L( f, P). Since f is integrble on [, b], then there exists for every ɛ > 0 prtition P, such tht U( f, P) L( f, P) < ɛ, hence U( f, P) L( f, P) = (L( f, P) U( f, P)) < ɛ, which proves tht ( f ) is integrble on [, b]. Then, we note tht for every prtition P, b L( f, P) = U( f, P) f L( f, P) = U( f, P), nd the rest is s bove. (45 hrs, 2010) This lst theorem is in fct specil cse of more generl theorem: Theorem 4.8 Suppose tht both f nd g re integrble on [, b], then so is their product f g. Proof : We will prove this theorem for the cse where f, g 0. The generliztion to rbitrry functions is not hrd (for exmple, since f nd g re bounded, we cn dd to ech constnt to mke it positive nd then use the rithmetic of integrbility ).

150 Chpter 4 Let P be prtition. It is useful to introduce the following nottions: m f j = inf{ f (x) : x j 1 x x j } m g j = inf{g(x) : x j 1 x x j } m f g j = inf{ f (x)g(x) : x j 1 x x j } M f j = sup{ f (x) : x j 1 x x j } M g j = sup{g(x) : x j 1 x x j } M f g j = sup{ f (x)g(x) : x j 1 x x j }. We lso define M f = sup{ f (x) : x b} M g = sup{ g(x) : x b}. For every x j 1 x x j, hence It further follows tht m f j mg j f (x)g(x) M f j Mg j, m f j mg j m f g j nd M f g j M f j Mg j. M f g j m f g j M f j Mg j m f j mg j (M f j m f j )Mg j +m f j (Mg j mg j ) Mg (M f j m f j )+M f (M g j mg j ). This inequlity implies tht U( f g, P) L( f g, P) M g [U( f, P) L( f, P)] + M f [U(g, P) L(g, P)]. It is then simple tsk to show tht the integrbility of f nd g implies the integrbility of f g. Theorem 4.9 If f is integrble on [, b] nd m f (x) M for ll x, then m(b ) b f M(b ).

Integrtion theory 151 Proof : Let P 0 = {x 0, x n }, then L( f, P 0 ) m(b ) nd U( f, P 0 ) M(b ), nd the proof follows from the fct tht L( f, P 0 ) b f U( f, P 0 ). Suppose now tht f is integrble on [, b]. We proved tht it is integrble on every sub-segment. Define for every x, F(x) = x f. Theorem 4.10 The function F defined bove is continuous on [, b]. Proof : Since f is integrble it is bounded. Define M = sup{ f (x) : x b}. Given ɛ > 0 tke δ = ɛ/m. Then for x y < δ (suppose wlog tht y > x), y x y F(y) F(x) = f f = x f. By the previous theorem, since M f (x) M for ll x, M(y x) F(y) F(x) M(y x), i.e., F(y) F(x) M(y x) < ɛ. Tht is, given ɛ > 0, we tke δ = ɛ/m, nd then F(y) F(x) < ɛ whenever x y < δ, which proves tht F is (uniformly) continuous on [, b]. (41 hrs, 2009)

152 Chpter 4 Exercise 4.6 Let f be bounded nd integrble in [, b] nd denote M = sup{ f (x) : x b}. Show tht for every p > 0, ( b ) 1/p f p M b 1/p, nd in prticulr, tht the integrl on the left hnd side exists. Exercise 4.7 Show tht if f is bounded nd integrble on [, b], then f is integrble on [, b] nd b b f f. Exercise 4.8 Let < b nd c > 0 nd let f be integrble on [c, cb]. Show tht where g(x) = f (cx). cb c f = c Exercise 4.9 Let f be integrble on [, b] nd g(x) = f (x c). Show tht g is integrble on [ + c, b + c] nd b g, b f = b+c +c g. Exercise 4.10 Let f : [, b] R be bounded, nd define the vrition of f in [, b] s ω f = sup{ f (x) : x b} inf{ f (x) : x b}. ➀ Prove tht ω f = sup{ f (x 1 ) f (x 2 ) : x 1, x 2 b} = sup{ f (x 1 ) f (x 2 ) : x 1, x 2 b}. ➁ Show tht f is integrble on [, b] if nd only if there exists for every ɛ > 0 prtition P = {x 0,..., x n }, such tht n i=1 ω i < ɛ, where ω i is the vrition of f in the segment [x i 1, x i ]. ➂ Prove tht if f is integrble on [, b], nd there exists n m > 0 such tht f (x) m for ll x [, b], then 1/ f is integrble on [, b].

Integrtion theory 153 פונקצית) Exercise 4.11 A function s : [, b] R is clled step function (מדרגות if there exists prtition p such tht s is constnt on every open intervl (we do not cre for the vlues t the nodes), ➀ Show tht if f is integrble on [, b] then there exist for every ɛ > 0 two step functions, s 1 f nd s 2 f, such tht b f b s 1 < ɛ nd b s 2 b f < ɛ. ➁ Show tht if f is integrble on [, b] then there exists for every ɛ > 0 continuous function g, g f, such tht b f b g < ɛ. Discussion: A question tht occupied mny mthemticins in the second hlf of the 19th century is wht chrcterizes integrble functions. Hermnn Hnkel (1839-1873) proved tht if function is integrble, then it is necessrily continuous on dense set of points (he lso proved the opposite, but hd n error in his proof). The ultimte nswer is due to Henri Lebesgue (1875-1941) who proved tht bounded function is integrble if nd only if its points of discontinuity hve mesure zero ( notion not defined in this course). (44 hrs, 2011) 4.3 The fundmentl theorem of clculus Theorem 4.11 היסודי של החשבון הדיפרנציאלי )) ((המשפט Let f be integrble on [, b] nd define for every x [, b], F(x) = x If f is continuous t c (, b), then F is differentible t c nd F (c) = f (c). f.

154 Chpter 4 Comment: The theorem sttes tht in certin wy the integrl is n opertion inverse to the derivtive. Note the condition on f being continuous. For exmple, if f (x) = 0 everywhere except for the point x = 1 where it is equl to one, then F(x) = 0, nd it is not true tht F (1) = f (1). Proof : We need to show tht lim c F,c = f (c). We re going to show tht this holds for the right derivtive; the sme rguments cn then be pplied for the left derivtive. For x > c, If we define then, or F(x) F(c) = c x f. m(c, x) = inf{ f (y) : x < y < c} M(c, x) = sup{ f (y) : x < y < c}, m(c, x)(x c) F(x) F(c) M(c, x)(x c), m(c, x) F,c (x) M(c, x). Then, F,c(x) f (c) mx { m(c, x) f (y), M(c, x) f (y) }. It remins tht the right-hnd side s limit zero t c. Since f is continuous t c, For c < x < c + δ, hence ( ɛ > 0)( δ > 0) : ( c < y < c + δ)( f (y) f (c) < ɛ/2). ( ɛ > 0)( δ > 0) : ( c < y < x)( f (y) f (c) < ɛ/2), m(c, x) f (c) ɛ/2 < ɛ nd M(c, x) f (c) ɛ/2 < ɛ. To conclude, for every ɛ > 0 there exists δ > 0 such tht for ll c < x < c + δ, F,c (x) f (c) < ɛ, which concludes the proof.

Integrtion theory 155 Corollry 4.2 (Newton-Leibniz formul) If f is continuous on [, b] nd f = g for some function g, then b f = g(b) g(). Proof : We hve seen tht F(x) = stisfies the identity F (x) = f (x) for ll x (since f is continuous on [, b]). We hve lso seen tht it implies tht F = g + c, for some constnt c, hence x x f = F(x) = g(x) + c, but since F() = 0 it follows tht c = g(). Setting x = b we get the desired result. This is very useful result. It mens tht whenever we know pir of functions f, g, such tht g = f, we hve n esy wy to clculte the integrl of f. On the other hnd, it is totlly useless if we wnt to clculte the integrl of f, but no primitive function קדומה ) (פונקציה g is known. (43 hrs, 2009) (47 hrs, 2010) Exercise 4.12 Let f : R R be continuous nd g, h : (, b) R differentible. Define F : (, b) R by f F(x) = h(x) g(x) f. ➀ Show tht F is well-defined for ll x (, b). ➁ Show tht F is differentible, nd clculte its derivtive. Exercise 4.13 Differentite the following functions: ➀ 1 0 cos(t2 ) dt. ➁ e x cos(t) dt. 0

156 Chpter 4 ➂ x 2 x e sin(t) dt. Exercise 4.14 Let f : R R be continuous. For every δ > 0 we define Show tht for every x R, F δ (x) = 1 2δ x+δ x δ f. f (x) = lim δ 0 + F δ(x). Exercise 4.15 Let f, g : [, b] R be integrble, such tht g(x) 0 for ll x [, b]. ➀ Show tht there exists λ R, such tht nd inf{ f (x) : x b} λ sup{ f (x) : x b}, b ( f g) = λ Hint: distinguish between the cses where b g > 0 nd b g = 0. ➁ Show tht if f is continuous, then there exists c [, b], such tht b ( f g) = f (c) b b (This is known s the integrl men vlue theorem.) Exercise 4.16 Let f : [, b] R be monotonic nd let g : [, b] R be non-negtive nd integrble. Show tht there exists ξ [, b], such tht b ( f g) = f () ξ g + f (b) Exercise 4.17 Let f : [, b] R be integrble, c (, b), nd F(x) = x f. Prove or disprove: ➀ If f is differentible t c then F is differentible t c. ➁ If f is differentible t c then F is continuous t c. ➂ If f is differentible in neighborhood of c nd f is continuous t c then F is twice differentible t c. g. g. b ξ g.

Integrtion theory 157 4.4 Riemnn sums Let s sk prcticl question. Let f be given integrble function on [, b], nd suppose we wnt to clculte the integrl of f on [, b], however we do not know ny primitive function of f. Suppose tht we re content with n pproximtion of this integrl within mximum error of ɛ. Tke prtition P: since L( f, P) b f U( f, P), it follows tht if the difference between the upper nd lower sum is less thn ɛ, then their verge differs from the integrl by less tht ɛ/2. If the difference is greter thn ɛ, then it cn be reduced by refining the prtition. Thus, up to the fct tht we hve to compute lower nd upper sums, we cn keep refining until they differ by less thn ɛ, nd so we obtin n pproximtion to the integrl within the desired ccurcy. There is one drwbck in this pln. Infimums nd supremums my be hrd to clculte. On the other hnd, since m i f (x) M i for ll x i 1 x x i, it follows tht if in every intervl in the prtition we tke some point t i ( smple point), then n L( f, P) f (t i )(x i 1 x i ) U( f, P). i=1 The sum in the middle is clled Riemnn sum. Its evlution only requires function evlutions, without ny inf or sup. If the prtition is sufficiently fine, such tht U( f, P) L( f, P) < ɛ, then n b f (t i )(x i 1 x i ) f < ɛ. i=1 Note tht we don t know priori whether Riemnn is greter or smller thn the integrl. Wht we re going to show is tht Riemnn sums converge to the integrl s we refine the prtition, irrespectively on how we smple the points t i. Definition 4.5 The dimeter of prtition P is (45 hrs, 2011) dim P = mx{x i x i 1 : 1 i n}.

158 Chpter 4 Theorem 4.12 Let f be n integrble function on [, b]. For every ɛ > 0 there exists δ > 0, such tht n f (t i )(x i 1 x i ) i=1 b f < ɛ for every prtition P such tht dim P < δ nd ny choice of points t i. Proof : It is enough to show tht given ɛ > 0 there exists δ > 0 such tht U( f, P) L( f, P) < ɛ whenever dim P < δ, for ny Riemnn sum with the sme prtition will differ from the integrl by less thn ɛ. Let M = sup{ f (x) : x b}, nd tke some prtition P = {y 0,..., y k } for which U( f, P ) L( f, P ) < ɛ 2. We set δ < ɛ 4Mk. For every prtition P of dimeter less thn δ, n U( f, P) L( f, P) = (M i m i )(x i x i 1 ) i=1 cn be divided into two sums. One for which ech of the intervls [x i 1, x i ] is contined in n intervl [y j 1, y j ]. The contribution of these terms is t most ɛ/2. The second sum includes those intervls for which x i 1 < y j < x i. There re t most k such terms nd ech contributes t most Mδ. This concludes the proof. 4.5 The trigonometric functions Discussion: Describe the wy the trigonometric functions re defined in geometry, nd explin tht we wnt definitions tht only rely on the 13 xioms of rel numbers.

Integrtion theory 159 Definition 4.6 The number π is defined s 1 π = 2 1 x2 dx. 1 Comment: We hve used here the stndrd nottion of integrls. In our nottion, we hve function ϕ : [ 1, 1] R defined s ϕ : x 1 x 2, nd we define π = 2 1 This is well defined becuse ϕ is continuous, hence integrble. Consider now unit circle. For 1 x 1 we look t the point (x, 1 x 2 ), connect it to the origin, nd express the re of the sector bounded by this segment nd the positive x-xis. This geometric construction is just the motivtion; formlly, we define function A : [ 1, 1] R, 1 ϕ. A(x) = x 1 x 2 2 + 1 x ϕ. If θ is the ngle of this sector, then we know tht x = cos θ, nd tht A(x) = θ/2. Thus, A(cos θ) = θ 2, or, ( θ ) cos θ = A 1. 2 This observtion motivtes the wy we re going to define the cosine function. A!x" x

160 Chpter 4 We now exmine some properties of the function A. First, A( 1) = π/2 nd A(1) = 0. Second, A 1 (x) = 2 1 x, 2 i.e., the function A is monotoniclly decresing. It follows tht it is invertible, with its inverse defined on [0, π/2]. Definition 4.7 We define the functions cos : [0, π] R nd sin : [0, π] R by cos x = A 1 (x/2) nd sin x = 1 cos 2 x. With these definitions in hnd we define the other trigonometric functions, tn, cot, etc. We now need to show tht the sine nd cosine functions stisfy ll known differentil nd lgebric properties. Theorem 4.13 For 0 < x < π, sin x = cos x nd cos x = sin x. Proof : Since A is differentible nd A (x) 0 for 0 < x < π, we hve by the chin rule nd the derivtive of the inverse function formul, cos x = 1 2 (A 1 ) (x/2) = 1 2 1 A (A 1 (x/2)) = 1 2 2 1 = = sin x. 1 1 (A 1 (x/2)) 2 The derivtive of the sine function is then obtined by the composition rule. (45 hrs, 2009) We re now in mesure to strt picturing the grphs of these two functions. First, by definition, sin x > 0 0 < x < π. It thus follows tht cos is decresing in this domin. We lso hve cos π = A 1 (π/2) = 1 nd cos 0 = A 1 (0) = 1,

Integrtion theory 161 from which follows tht sin π = 0 nd sin 0 = 0. By the intermedite vlue theorem, cos must vnish for some 0 < y < π. Since A(cos y) = y 2, it follows tht y = 2A(0) = 2 1 where we hve used the fct tht by symmetry, 0 ϕ = 0 1 1 0 ϕ = π 2, ϕ. Thus, cos π 2 = 0 nd sin π 2 = 1. We cn then extend the definition of the sine nd cosine to ll of R by setting nd sin x = sin(2π x) nd cos x = cos(2π x) for π x 2π, sin(x + 2πk) = sin x nd cos(x + 2πk) = cos x for ll k Z, It is esy to verify tht the expressions for the derivtives of sine nd cosine remin the sme for ll x R. Since we defined the trigonometric functions (more precisely their inverses) vi integrls, it is not surprising tht it ws strightforwrd to differentite them. A little more subtle is to derive their lgebric properties. Lemm 4.3 Let f : R R be twice differentible, nd stisfy the differentil eqution f + f = 0, with f (0) = f (0) = 0. Then, f is identiclly zero.

162 Chpter 4 Proof : Multiply the eqution by 2 f, then 2 f f + 2 f f = [( f ) 2 + f 2 ] = 0, from which we deduce tht ( f ) 2 + f 2 = c, for some c R. The conditions t zero imply tht c = 0, which implies tht f is identiclly zero. Theorem 4.14 Let f : R R be twice differentible, nd stisfy the differentil eqution f + f = 0, with f (0) = nd f (0) = b. Then, f = cos +b sin. Proof : Define then, g = f cos b sin, g + g = 0, nd g(0) = g (0) = 0. It follows from the lemm tht g is identiclly zero, hence f = cos +b sin. Theorem 4.15 For every x, y R, sin(x + y) = sin x cos y + cos x sin y cos(x + y) = cos x cos y sin x sin y. Proof : Fix y nd define f : x sin(x + y),

Integrtion theory 163 from which follows tht Thus, f + f = 0 nd f : x cos(x + y) f : x sin(x + y). f (0) = sin y nd f (0) = cos y. It follows from the previous theorem tht f (x) = sin y cos x + cos y sin x. We proceed similrly for the cosine function. 4.6 The logrithm nd the exponentil Our (strting) gol is to define the function f : x 10 x for x R, long with its inverse, which we denote by log 10. For n N we hve n inductive definition, which in prticulr, stisfies the identity 10 n 10 m = 10 n+m. We then extend the definition of the exponentil (with bse 10) to integers, by requiring the bove formul to remin correct. Thus, we must set 10 0 = 1 nd 10 n = 1 10 n. The next extension is to frctionl powers of the form 10 1/n. Since we wnt } 10 1/n 1/n {{ 10 1/n } = 10 1 = 10, n times this forces the definition 10 1/n = n 10. Then, we extend the definition to rtionl powers, by 10 } 1/n 10 1/n {{ 10 1/n } = 10 m/n. m times

164 Chpter 4 We hve proved in the beginning of this course tht the definition of rtionl power is independent of the representtion of tht frction. The question is how to extend the definition of 10 x when x is rel number? This cn be done with lots of sups nd infs, but here we shll dopt nother pproch. We re looking for function f with the property tht f (x + y) = f (x) f (y), nd f (1) = 10. Cn we find differentible function tht stisfies these properties? If there ws such function, then ( ) f f (x + h) f (x) f (x) f (h) f (x) f (h) 1 (x) = lim = lim = lim f (x). h 0 h h 0 h h 0 h Suppose tht the limit on the right hnd side exists, nd denote it by α. Then, f (x) = α f (x) nd f (1) = 10. This is not very helpful, s the derivtive of f is expressed in terms of the (unknown) f. It is cler tht f should be monotoniclly incresing (it is so for rtionl powers), hence invertible. Then, ( f 1 ) (x) = 1 f ( f 1 (x)) = 1 α f ( f 1 (x)) = 1 αx. From this we deduce tht f 1 (x) = x 1 dt αt = log 10 (x). We seem to hve succeeded in defining (using integrtion) the logrithm-bse-10 function, but there is cvet. We do not know the vlue of α. Since, in ny cse, the so fr distinguished role of the number 10 is unnturl, we define n lterntive logrithm function: Definition 4.8 For ll x > 0, log(x) = x 1 dt t.

Integrtion theory 165 Comment: Since 1/x is continuous on (0, ), this integrl is well-defined for ll 0 < x <. Theorem 4.16 For ll x, y > 0, log(xy) = log(x) + log(y). Proof : Fix y nd consider the function f : x log(xy). Then, f (x) = 1 xy y = 1 x. Since log (x) = 1/x, it follows tht log(xy) = log(x) + c. The vlue of the constnt is found by setting x = 1. Corollry 4.3 For integer n, log(x n ) = n log(x). Proof : By induction on n. Corollry 4.4 For ll x, y > 0, log ( ) x = log(x) log(y). y

166 Chpter 4 Proof : ( ) ( ) x x log(x) = log y y = log + log(y). y Definition 4.9 The exponentil function is defined s exp(x) = log 1 (x). Comment: The logrithm is invertible becuse it is monotoniclly incresing, s its derivtive is strictly positive. Wht is less esy to see is tht the domin of the exponentil is the whole of R. Indeed, since log(2 n ) = n log(2), it follows tht the logrithm is not bounded bove. Similrly, since log(1/2 n ) = n log(2), it is neither bounded below. Thus, its rnge is R. Theorem 4.17 exp = exp. Proof : We proceed s usul, exp (x) = (log 1 ) (x) = 1 log (log 1 (x)) = log 1 (x) = exp(x). Theorem 4.18 For every x, y, exp(x + y) = exp(x) exp(y).

Integrtion theory 167 Proof : Set t = exp(x) nd s = exp(y). Thus, x = log(t) nd y = log(s), nd x + y = log(t) + log(s) = log(ts). Exponentiting both sides, we get the desired result. Definition 4.10 e = exp(1). There is plenty more to be done with exponentils nd logrithms, like for exmple, show tht for rtionl numbers r, e r = exp(r), nd the pssge between bses. Due to the lck of time, we will stop t this point. (47 hrs, 2011) 4.7 Integrtion methods Given function f, question of prcticl interest is how to clculte integrls, b If we know function F, which is primitive of f, in the sense tht F = f, then the nswer is known, b f. f = F(b) F() def = F b. To effectively use this formul, we must know mny pirs of functions f, F. Here is tble of pirs f, F, which we lredy know: f (x) F(x) 1 x x n x n+1 /(n + 1) cos x sin x sin x cos x 1/x log(x) exp(x) exp(x) sec 2 (x) tn(x) 1/(1 + x 2 ) rctn(x) 1/ 1 x 2 rcsin(x)

168 Chpter 4 Sums of such functions nd multipliction by constnts re esily treted with our integrtion theorems, for b ( f + g) = b f + b g nd b (c f ) = c Functions tht cn be expressed in terms of powers, trigonometric, exponentil nd logrithmic functions nd clled elementry. In this section we study techniques tht in some cses llow to express primitive function in terms of elementry functions. There re bsiclly two techniques integrtion by prts, nd substitution both we will study. Comment: Every continuous function f hs primitive, F(x) = where is rbitrry. It is however not lwys the cse tht the primitive of n elementry function is itself nd elementry function. Exmple: Consider the function It is esily checked tht As result of this observtion, x F(x) = x rctn(x) 1 2 log(1 + x2 ). f, f (x) = F (x) = rctn(x). b rctn(x) dx = F b. The question is how could hve we guessed this nswer if the function F hd not been given to us. b f. Theorem 4.19 (Integrtion by prts בחלקים ) ((אינטגרציה Let f, g hve continuous first derivtives, then b f g = ( f g) b b f g.

Integrtion theory 169 Comment: When we use this formul, f nd g re given. Then, g on the righthnd side my be ny primitive of g. Also, setting b = x nd noting tht primitives re defined up to constnts, we hve tht Primitive( f g ) = f Primitive(g ) Primitive( f Primitive(g)) + const. Exmple: Exmple: The 1-trick: b b x e x dx. log(x) dx. Exmple: The bck-to-originl-problem-trick b (1/x) log(x) dx. Once we hve used this method to get new f, F pirs, we use this knowledge to obtin new such pirs: Exmple: Using the fct tht primitive of log(x) is x log(x) x, we solve b log 2 (x) dx. Theorem 4.20 (Substitution formul ההצבה ) ((שיטת For continuously differentible f nd g, b (g f ) f = f (b) f () g.

170 Chpter 4 Proof : Let G be primitive of g, then hence b (g f ) f = b (G f ) = (g f ) f, f (G f ) = (G f ) b = G f (b) (b) f () = g. The simplest use of this formul is by recognizing tht function is of the form (g f ) f. Exmple: Tke the integrl b If f (x) = sin(x) nd g(x) = x 5, then sin 5 (x) cos(x) dx. sin 5 (x) cos(x) = g( f (x)) f (x), f () hence (46 hrs, 2009) (48 hrs, 2011) b sin 5 (x) cos(x) dx = sin(b) sin() t 5 dt = t6 6 sin(b) sin().

Chpter 5 Sequences 5.1 Bsic definitions An (infinite) sequence (סדרה ) is n infinite ordered set of rel numbers. We denote sequences by letter which lbels the sequence, nd subscript which lbels the position in the sequence, for exmple, 1, 2, 3,.... The essentil fct is tht to ech nturl number corresponds rel number. Hving sid tht, we cn define: Definition 5.1 An infinite sequence is function N R. Comment: Let : N R. Rther thn writing (3), (17), we write 3, 17. Comment: As for functions, where we often refer to the function x 2, we often refer to the sequence 1/n. Nottion: We usully denote sequence : N R by ( n ). Another ubiquitous nottion is ( n ) n=1. Exmples: ➀ : n n or n = n.

172 Chpter 5 ➁ b : n ( 1) n or b n = ( 1) n. ➂ c : n 1/n or c n = 1/n. ➃ (d n ) = {2, 3, 5, 7, 11,... }, the sequence of primes. In the first nd fourth cses we would sy tht the sequence goes (שואפת ) to infinity. In the second cse we would sy tht it jumps between 1 nd 1. In the third cse we would sy tht it goes to zero. We will mke such sttements precise momentrily. Like functions, sequences cn be grphed. More useful often is to mrk the elements of the sequence on the number xis. 1 5 3 2 4 5.2 Limits of sequences Definition 5.2 A sequence ( n ) converges (מתכנסת ) to l, denoted lim n = l, n if ( ɛ > 0)( N N) : ( n > N)( n l < ɛ). Definition 5.3 A sequence is clled convergent (מתכנסת ) if it converges to some (rel) number; otherwise it is clled divergent (מתבדרת ) (there is no such thing s convergence to infinity). Comment: There is close connection between the limit of sequence nd the limit of function t infinity. We will elborte on tht lter on. Nottion: Like for functions, clener nottion for the limit of sequence ( n ) would be lim.

Sequences 173 Exmple: Let : n 1/n, or ( n ) = 1, 1 2, 1 3, 1 4,... It is esy to show tht lim = 0, since, ( ɛ > 0)(set N = 1/ɛ ) then ( n > N) ( n = 1n < 1N ) < ɛ. Exmple: Let : n n + 1 n, or ( n ) = 2 1, 3 2, 4 3,.... It seems resonble tht lim = 0, however showing it requires some mnipultions. One wy is by n lgebric trick, n = ( n + 1 n)( n + 1 + n) n + 1 + n = 1 < 1 n + 1 + n 2 n, from which it is esy to proceed. The sme inequlity cn be obtined by defining function f : x x, nd using the men-vlue theorem, y x = y x 2 ξ, for some x < ξ < y. Substituting x = n nd y = n + 1, we obtin the desired inequlity. Exmple: Let now : n 3n3 + 7n 2 + 1 4n 3 8n + 63.

174 Chpter 5 It seems resonble tht lim = 3 4, but once gin, proving it is not totlly strightforwrd. One wy to proceed is to note tht 3 + 7/n + 1/n3 n = 4 8/n 2 + 63/n, 3 nd use limit rithmetic s stted below. Exercise 5.1 For ech of the following sequence determine whether it is converging or diverging. Prove your clim using the definition of the limit: ➀ n = 7n2 + n n 2 +3n. ➁ n = n4 +1 n 3 3. ➂ n = n 2 + 1 n. Theorem 5.1 Let nd b be convergent sequences. Then the sequences ( + b) n = n + b n nd ( b) n = n b n re lso convergent, nd lim ( + b) = lim + lim b lim ( b) = lim lim b. If furthermore lim b 0, then there exists n N N such tht b n 0 for ll n > N, the sequence (/b) converges nd lim b = lim lim b. Comment: Since the limit of sequence is only determined by the til of the sequence, we do not cre if finite number of elements is not defined. Proof : Esy once you ve done it for functions.

Sequences 175 Exercise 5.2 Show tht if nd b re converging sequences, such tht lim = L nd lim b = M, then lim ( + b) = L + M. We now estblish the nlogy between limits of sequences nd limits of functions. Let be sequence, nd define piecewise-constnt function f : [1, ) R by f (x) = n n x n + 1. Proposition 5.1 lim = l if nd only if lim f = l. Proof : Suppose first tht the sequence converges. Thus, ( ɛ > 0)( N N) : ( n > N)( n l < ɛ). Since for every x, f (x) = f ( x ), it follows tht for ll x > N + 1, f (x) l x l < ɛ, hence ( ɛ > 0)( N N) : ( x > N)( f (x) l < ɛ). Conversely, if f converges to l s x, then, ( ɛ > 0)( N N) : ( x > N)( f (x) l < ɛ), nd this is in prticulr true for x N. Conversely, given function f : R R, it is often useful to define sequence n = f (n). If f (x) converges s x, then so does the sequence ( n ).

176 Chpter 5 Exmple: Let 0 < < 1, then We define f (x) = x. Then lim n n =? f (x) = e x log, nd n = f (n). Since log < 0 it follows tht lim x f (x) = 0, hence the sequence converges to zero. Comment: Like for functions, we my define convergence of sequences to ±, only tht we don t use the term convergence, but rther sy tht sequence pproches (שואפת ) infinity. Theorem 5.2 Let f be defined on (punctured) neighborhood of point c. Suppose tht lim c f = l. If is sequence whose entries belong to the domin of f, such tht n c, nd lim = c, then lim f () = l, where the composition of function nd sequence is defined by [ f ()] n = f ( n ). Conversely, if lim f () = l for every sequence tht converges to c, then lim c f = l. Comment: This theorem provides chrcteriztion of the limit of function t point in terms of sequences. This is known s Heine s chrcteriztion of the limit. Proof : Suppose first tht lim c f = l.

Sequences 177 Thus, ( ɛ > 0)( δ > 0) : ( x : 0 < x c < δ)( f (x) l < ɛ). Let be sequence s specified in the theorem, nmely, n c nd lim = c. Then ( δ > 0)( N N) : ( n > N)(0 < n c < δ). Combining the two, i.e., lim f () = l. ( ɛ > 0)( N N) : ( n > N)( f ( n ) l < ɛ), Conversely, suppose tht for ny sequence s bove lim f () = l. Suppose, by contrdiction tht lim f l c (i.e., either the limit does not exist or it is not equl to l). This mens tht ( ɛ > 0) : ( δ > 0)( x : 0 < x c < δ) : ( f (x) l ɛ). In prticulr, setting δ n = 1/n, ( ɛ > 0) : ( n N)( n : 0 < n c < δ n ) : ( f ( n ) l ɛ). The sequence converges to c but f () does not converge to l, which contrdicts our ssumptions. Exmple: Consider the sequence : n sin (13 + 1n 2 ). This sequence cn be written s = sin(b), where b n = 13 + 1/n 2. Since it follows tht (48 hrs, 2009) lim 13 sin = sin(13) nd lim b = 13, lim = sin(13).

178 Chpter 5 Theorem 5.3 A convergent sequence is bounded. Proof : Let be sequence tht converges to limit α. We need to show tht there exists L 1, L 2 such tht L 1 n L 2 n N. By definition, setting ɛ = 1, ( N N) : ( n > N)( n α < 1), or, Set then for ll n, ( N N) : ( n > N)(α 1 < n < α + 1). M = mx{ n : 1 n N} m = min{ n : 1 n N}. min(m, α 1) n mx(m, α + 1). Theorem 5.4 Let be bounded sequence nd let b be sequence tht converges to zero. Then lim(b) = 0. n Proof : Let M be bound on, nmely, Since b converges to zero, n M n N. ( ɛ > 0)( N N) : ( n > N) ( b n < ɛ ). M Thus, ( ɛ > 0)( N N) : ( n > N) ( n b n M b n < ɛ), which implies tht the sequence b converges to zero.

Sequences 179 Exmple: The sequence converges to zero. : n sin n n Theorem 5.5 Let be sequence with non-zero elements, such tht lim = 0. Then lim 1 =. Proof : By definition, ( M > 0)( N N) : ( n > N)(0 < n < 1/M), hence which concludes the proof. ( M > 0)( N N) : ( n > N)(1/ n > M), Theorem 5.6 Let be sequence such tht lim =. Then lim 1 = 0. Proof : By definition, ( ɛ > 0)( N N) : ( n > N)( n > 1/ɛ), hence which concludes the proof. ( ɛ > 0)( N N) : ( n > N)(0 < 1/ n < ɛ),

180 Chpter 5 Theorem 5.7 Suppose tht nd b re convergent sequences, lim = α nd lim b = β, nd β > α. Then there exists n N N, such tht b n > n for ll n > N, i.e., the sequence b is eventully greter (term-by-term) thn the sequence. Proof : Since (β α)/2 > 0, ( N 1 N) : ( n > N 1 ) ( n < α + 1 2 (β α)), nd ( N 2 N) : ( n > N 2 ) ( b n > β 1(β α)). 2 Thus, ( N 1, N 2 N) : ( n > mx(n 1, N 2 )) (b n n > 0). (50 hrs, 2011) Corollry 5.1 Let be sequence nd α, β R. If the sequence converges to α nd α > β then eventully n > β. Theorem 5.8 Suppose tht nd b re convergent sequences, lim = α nd lim b = β, nd there exists n N N, such tht n b n for ll n > N. Then α β. Proof : By contrdiction, using the previous theorem.

Sequences 181 Comment: If insted n < b n for ll n > N, then we still only hve α β. Tke for exmple the sequences n = 1/n nd b n = 2/n. Even though n < b n for ll n, both converge to the sme limit. Theorem 5.9 (Sndwich) Suppose tht nd b re sequences tht converge to the sme limit l. Let c be sequence for which there exists n N N such tht n c n b n for ll n > N. Then lim c = l. Proof : It is given tht ( ɛ > 0)( N 1 N) : ( n > N 1 )( n l < ɛ), nd Thus, ( ɛ > 0)( N 2 N) : ( n > N 2 )( b n l < ɛ). ( ɛ > 0)( N 1, N 2 N) : ( n > mx(n 1, N 2 ))( ɛ < n l c n l b n l < ɛ), or ( ɛ > 0)( N N) : ( n > N)( c n l < ɛ). Exmple: Since for ll n, 1 < 1 + 1/n < 1 + 2/n + 1/n 2 = 1 + 1/n, it follows tht lim 1 + 1/n = 1. n We cn lso deduce this from the fct tht lim x = 1 nd lim (1 + 1/n) = 1. x 1 n

182 Chpter 5 Exercise 5.3 ➀ Find sequences nd b, such tht converges to zero, b is unbounded, nd b converges to zero. ➁ Find sequences nd b, such tht converges to zero, b is unbounded, nd b converges to limit other then zero. ➂ Find sequences nd b, such tht converges to zero, b is unbounded, nd b is bounded but does not converge. Definition 5.4 A sequence is clled incresing (עולה ) if n+1 > n for ll n. It is clled non-decresing יורדת ) (לא if n+1 n for ll n. We define similrly decresing nd non-incresing sequence. Theorem 5.10 (Bounded + monotonic = convergent) Let be non-decresing sequence bounded from bove. Then it is convergent. Proof : Set α = sup{ n : n N}. (Note tht supremum is property of set, i.e., the order in the set does not mtter.) By the definition of the supremum, Since the sequence is non-decresing, nd in prticulr, ( ɛ > 0)( N N) : (α ɛ < N α). ( ɛ > 0)( N N) : ( n > N)(α ɛ < N n α), ( ɛ > 0)( N N) : ( n > N)( n α < ɛ). Exmple: Consider the sequence : n ( 1 + 1 n) n.

Sequences 183 We will first show tht n < 3 for ll n. Indeed 1, ( 1 + 1 n ( ) ( ) ( ) n 1 n 1 n 1 = 1 + n) 1 n + 2 n + + 2 n n n n(n 1) = 1 + 1 + + + n! 2! n 2 n! n n 1 + 1 + 1 2! + + 1 n! 1 + 1 + 1 2 + + 1 < 3. 2n 1 Second, we clim tht this sequence is incresing, s n = 1 + 1 + 1 ( 1 1 ) + 1 ( 1 1 ) ( 1 2 ) +. 2! n 3! n n When n grows, both the number of elements grows s well s their vlues. It follows tht ( 1 + n) 1 n exists (nd equls to 2.718 e). lim n Definition 5.5 Let be sequence. A subsequence סדרה ) (תת of is ny sequence n1, n2,..., such tht n 1 < n 2 <. More formlly, b is subsequence of if there exists monotoniclly incresing sequence of nturl numbers (n k ), such tht b k = nk for ll k N. Exmple: Let by the sequence of nturl numbers, nmely n = n. The subsequence b of ll even numbers is b k = 2k, i.e., n k = 2k. 1 The binomil formul is ( + b) n = n k=0 ( ) n k b n k. k

184 Chpter 5 Lemm 5.1 Any sequence contins subsequence which is either non-decresing or non-incresing. Proof : Let be sequence. Let s cll number n pek point שיא ) (נקודת of the sequence if m < n for ll m > n. pek poin! pek poin! There re now two possibilities. There re infinitely mny pek points: If n 1 < n 2 < re sequence of pek points, then the subsequence nk is decresing. There re finitely mny pek points: Then let n 1 be greter thn ll the pek points. Since it is not pek point, there exists n n 2, such tht n2 n1. Continuing this wy, we obtin non-decresing subsequence. Corollry 5.2 (Bolzno-Weierstrß) Every bounded sequence hs convergent subsequence. (50 hrs, 2009) In mny cses, we would like to know whether sequence is convergent even if we do not know wht the limit is. We will now provide such convergence criterion. Definition 5.6 A sequence is clled Cuchy sequence if ( ɛ > 0)( N N) : ( m, n > N)( n m < ɛ).

Sequences 185 Comment: A common nottion for the condition stisfied by Cuchy sequence is lim n,m n m = 0. (52 hrs, 2011) Theorem 5.11 A sequence converges if nd only if it is Cuchy sequence. Proof : One direction is esy 2. If sequence converges to limit α, then hence ( ɛ > 0)( N N) : ( n > N)( n α < ɛ/2), ( ɛ > 0)( N N) : ( m, n > N)( n m n α + m α < ɛ), i.e., the sequence is Cuchy sequence. Suppose now tht is Cuchy sequence. We first show tht the sequence is bounded. Tking ɛ = 1, ( N N) : ( n > N)( n N+1 < 1), which proves tht the sequence is bounded. By the Bolzno-Weierstrß theorem, it follows tht hs converging subsequence. Denote this subsequence by b with b k = nk nd its limit by β. We will show tht the whole sequence converges to β. Indeed, by the Cuchy property ( ɛ > 0)( N N) : ( m, n > N)( n m < ɛ/2), nd wheres by the convergence of the sequence b, ( N N)( K N : n K > N) : ( k > K)( b k β < ɛ/2). 2 There is something musing bout clling sequences stisfying this property Cuchy sequence. Cuchy ssumed tht sequences tht get eventully rbitrrily close converge, without being wre tht this is something tht ought to be proved.

186 Chpter 5 Combining the two, ( ɛ > 0)( N, K N : n K > N) : ( n > N)( n β n b k + b k β < ɛ). This concludes the proof. (51 hrs, 2009) (53 hrs, 2011) 5.3 Infinite series Let be sequence. We define the sequence of prtil sums חלקיים ) (סכומים of by n S n = k. k=1 Note tht S does not depend on the order of summtion ( finite summtion). If the sequence S converges, we will interpret its limit s the infinite sum of the sequence. Definition 5.7 A sequence is clled summble (סכימה ) if the sequence of prtil sums S converges. In this cse we write k=1 k def = lim S. The sequence of prtil sum is clled series.(טור ) Often, the term series refers lso to the limit of the sequence of prtil sums. Exmple: The cnonicl exmple of series is the geometric series. Let be geometric sequence n = q n q < 1. The geometric series is the sequence of prtil sums, for which we hve n explicit expresion: S n = q 1 qn 1 q

Sequences 187 This series converges, hence this geometric sequence is summble nd q n = n=1 q 1 q. Comment: Note the terminology: sequence is summble = series is convergent. Theorem 5.12 Let nd b be summble sequences nd γ R. Then the sequences + b nd γ re summble nd ( n + b n ) = n=1 n + n=1 b n, n=1 nd (γ n ) = γ n. n=1 n=1 Proof : Very esy. For exmple, S +b n = n ( k + b k ) = S n + S n, b k=1 nd the rest follows from limits rithmetic. An importnt question is how to identify summble sequences, even in cses where the sum is not known. This brings us to discuss convergence criteri. Theorem 5.13 (Cuchy s criterion) The sequence is summble if nd only if for every ɛ > 0 there exists n N N, such tht n+1 + n+2 + + m < ɛ for ll m > n > N.

188 Chpter 5 Proof : This is just resttement of Cuchy s convergence criterion for sequences. We know tht the series converges if nd only if ( ɛ > 0)( N N) : ( m > n > N)( S m S n < ɛ), but S m S n = n+1 + n+2 + + m. Corollry 5.3 If is summble then lim = 0. Proof : We pply Cuchy s criterion with m = n + 1. Tht is, ( ɛ > 0)( N N) : ( n > N)( n+1 < ɛ). The vnishing of the sequence is only necessry condition for its summbility. It is not sufficient s proves the hrmonic series, n = 1/n, S n = 1 + 1 2 + 1 3 + 1 + 1 }{{} 4 5 + 1 6 + 1 7 + 1 +. } {{ } 8 1/2 1/2 This grouping of terms shows tht the sequence of prtil sums is unbounded. For while we will del with non-negtive sequences. The reson will be mde cler lter on. Theorem 5.14 A non-negtive sequence is summble if nd only if the series (i.e., the sequence of prtil sums) is bounded. Proof : Consider the sequence of prtil sums, which is monotoniclly incresing. Convergence implies boundedness. Monotonicity nd boundedness imply convergence.

Sequences 189 Theorem 5.15 (Comprison test) If 0 n b n for every n nd the sequence b is summble, then the sequence is summble. Proof : Since b is summble then the corresponding series S b is bounded, nmely. ( M > 0) : ( n N) ( 0 S b n M ). The bound M bounds the series S, which therefore converges. Exmple: Consider the following sequence: Since n = 2 + sin3 (n + 1) 2 n + n 2. 0 < n 3 2 n def = b n, nd the series ssocited with b converges, it follows tht the series of converges. Exmple: Consider now the sequence n = 1 2 n 1 + sin 3 (n 2 ). Here we hve 0 < 1 2 1 n n 2 n 2 2 2, n 2, n so gin, bsed on the comprison test nd the convergence of the geometric series we conclude tht the series of converges. Exmple: The next exmple is n = n + 1 n 2 + 1. It is cler tht the corresponding series diverges, since n roughly behves like 1/n. This cn be shown by the following inequlity 1 n = n + 1 n 2 + n n,

190 Chpter 5 which proves tht the series S is unbounded. The following theorem provides n even esier wy to show tht. (53 hrs, 2009) Theorem 5.16 (Limit comprison) Let nd b be positive sequences, such tht lim b = γ 0. Then is summble if nd only if b is summble. Exmple: Bck to the previous exmple, setting b n = 1/n we find hence the series of diverges. lim b = lim n 2 + n n n 2 + 1 = 1, Proof : Suppose tht b is summble. Since lim γ b = 1, it follows tht or Now, S n = S N + ( ) n ( N N) : ( n > N) < 2, γ b n ( N N) : ( n > N) ( n < 2γb n ). n k=n+1 k < S N + 2γ n k=n+1 b k S N + 2γS b n. Since the right hnd sides is uniformly bounded, then is summble. The roles of nd b cn then be interchnged. Theorem 5.17 (Rtio test המנה ) ((מבחן Let n > 0 nd suppose tht n+1 lim = r. n n (i) If r < 1 then ( n ) is summble. (ii) If r > 1 then ( n ) is not summble.

Sequences 191 Proof : Suppose first tht r < 1. Then, ( s : r < s < 1)( N N) : ( n > N) ( n+1 In prticulr, N+1 s N, N+2 s 2 N, nd so on. Thus, n N n k = k + N s k N. k=1 k=1 k=n+1 n ) s. The second term on the right hnd side is prtil sum of convergent geometric series, hence the prtil sums of ( n ) re bounded, which proves their convergence. Conversely, if r > 1, then ( s : 1 < s < r)( N N) : ( n > N) ( n+1 n ) s, from which follows tht n s n N N for ll n > N, i.e., the sequence ( n ) is not even bounded (nd certinly not summble). Exmples: Apply the rtio test for n = 1 n! b n = rn n! c n = n r n. Comment: Wht bout the cse r = 1 in the bove theorem? This does not provide ny informtion bout convergence, s show the two exmples 3, n=1 1 n < nd 1 2 n =. These exmples motivte, however, nother comprison test. Note tht x lim x 1 dt t n=1 x dt = wheres lim x 1 t <. 2 Indeed, there is close reltion between the convergence of the series nd the convergence of the corresponding integrls. is not. 3 This nottion mens tht the sequence n = 1/n 2 is summble, wheres the sequence b n = 1/n

192 Chpter 5 Definition 5.8 Let f be integrble on ny segment [, b] for some nd b >. Then x f def = lim f. x Theorem 5.18 (Integrl test) Let f be positive decresing function on [1, ) nd set n = f (n). Then n=1 n converges if nd only if 1 f exists. Proof : The proof hinges on showing tht x k x k=2 1 x f k. k=1 TO COMPLETE. Exmple: Consider the series for p > 0. We pply the integrl text, nd consider the integrl x dt t = log x p = 1 p (1 p) 1 (x p+1 1) p 1. 1 n=1 This integrl converges s x if nd only if p > 1, hence so does the series. (55 hrs, 2009) 1 n p Exmple: One cn tke this further nd prove tht 4 4 Use the fct tht n=1 1 n(log n) p nd F(x) = F(x) = (log x) p+1 p + 1 (log log x) p+1 p + 1 implies F (x) = implies F (x) = 1 x (log x) p, 1 x log x(log log x) p.

Sequences 193 converges if nd only if p > 1, nd so does n=1 1 n log n(log log n) p, nd so on. This is known s Bertrnd s ldder. We hve delt with series of non-negtive sequences. The cse of non-positive sequences is the sme s we cn define b n = ( n ). Sequences of non-fixed sign re different story. Definition 5.9 A series מתכנסת) n=1 n is sid to be bsolutely convergent (בהחלט if n=1 n converges. A series tht converges but does not converge.(מתכנסת בתנאי ) bsolutely is clled conditionlly convergent Exmple: The series 1 1 2 + 1 3 1 4 +, converges conditionlly. Exmple: The series 1 1 3 + 1 5 1 7 +, converges conditionlly s well. It is well-known lternting series, which Leibniz showed (without ny rigor) to be equl to π/4. It is known s the Leibniz series. Theorem 5.19 Every bsolutely convergent series is convergent. Also, series is bsolutely convergent, if nd only if the two series formed by its positive elements nd its negtive elements both converge. Proof : The fct tht n bsolutely convergent series converges follows from Cuchy s criterion. Indeed, for every ɛ > 0 there exists n N N such tht whenever m, n > N. n+1 + + m n+1 + + m < ɛ

194 Chpter 5 Define now 5 + n n > 0 n = 0 n 0 nd n n < 0 n = 0 n 0. Then, n = + n + n, n = + n n, nd conversely, Thus, + n = 1 2 ( n + n ) nd n = 1 2 ( n n ). n k = k=1 n + n k=1 n n, hence if the two series of positive nd negtive terms converge so does the series of bsolute vlues. On the other hnd, n + n = 1 2 k=1 n n = 1 2 k=1 n n + 1 2 k=1 n n 1 2 k=1 k=1 n n k=1 n n k=1 so tht the other direction holds s well. (56 hrs, 2009) Theorem 5.20 (Leibniz) Suppose tht n is non-incresing sequence of nonnegtive numbers, i.e., 1 2 0, nd lim n = 0. n 5 For exmple, if n = ( 1) n+1 /n, then + n = 1, 0, 1 3, 0, 1 5, 0, n = 0, 1 2, 0, 1 4, 0, 1 6,.

Sequences 195 Then the series converges. ( 1) n+1 n = 1 2 + 3 4 + n=1 Proof : Simple considertions show tht Indeed, s 2 s 4 s 6 s 5 s 3 s 1. s 2n s 2n + ( 2n+1 2n+2 ) = s 2n+2 = s 2n+1 2n+2 s 2n+1. Thus, the sequence b n = s 2n is non-decresing nd bounded from bove, wheres the sequence c n = s 2n+1 is non-incresing nd bounded from below. It follows tht both subsequences converge, By limit rithmetic lim b n = b nd lim c n = c. n n lim (c n b n ) = lim(s 2n+1 s 2n ) = lim 2n+1 = c b, n n n however 2n+1 converges to zero, hence b = c. It remins to show tht if both the odd nd even elements of sequence converge to the sme limit then the whole sequence converges to this limit. Indeed, nd i.e., ( ɛ > 0)( N 1 N) : ( n > N 1 )( s 2n+1 b < ɛ) ( ɛ > 0)( N 2 N) : ( n > N 2 )( s 2n b < ɛ), ( ɛ > 0)( N 1, N 2 N) : ( n > mx(2n 1 + 1, 2N 2 ))( s n b < ɛ).

196 Chpter 5 Definition 5.10 A rerrngement מחדש ) (סידור of sequence n is sequence b n = f (n), where f : N N is one-to-one nd onto. Theorem 5.21 (Riemnn) If n=1 n is conditionlly convergent then for every α R there exists rerrngement of the series tht converges to α. Proof : I will not give forml proof, but only sketch. Exmple: Consider the series S = ( 1) n+1 n=1 n = 1 1 2 + 1 3 1 4 + 1 5 +. We proved tht this series does indeed converge to some 1 2 rithmetic for sequences (or series), < S < 1. By limit 1 2 S = n=1 ( 1) n+1 2n = 1 2 1 4 + 1 6 1 8 + 1 10 +, which we cn lso pd with zeros, 1 2 S = 0 + 1 2 + 0 1 4 + 0 + 1 6 + 0 1 8 + 0 + 1 10 +. Using once more sequence rithmetic, 3 2 S = 1 + 1 3 1 2 + 1 5 + 1 7 1 4 +..., which directly proves tht rerrngement of the lternting hrmonic series converges to three hlves of its vlue. In this exmple the rerrngement is b 3n+1 = 4n+1 b 3n+2 = 4n+3 b 3n = 2n.

Sequences 197 Comment: Here is n importnt fct bout Cuchy sequences: let ( n ) be Cuchy sequence with limit. Then, ( ɛ > 0)( N N) : ( m, n > N)( n m < ɛ). Think now of the left hnd side s sequence with index m with n fixed. As m this sequence converges to n. This implies tht n ɛ for ll n > N. Theorem 5.22 If n=1 n converges bsolutely then ny rerrngement n=1 b n converges bsolutely, nd n = n=1 b n. In other word, the order of summtion does not mtter in n bsolutely convergent series. n=1 Proof : Let b n = f (n) nd set n s n = k nd t n = k=1 n b k. k=1 Let ɛ > 0 be given. Since n=1 n converges, then there exists n N such tht s N k < ɛ 2. k=1 Moreover, since the sequence is bsolutely convergent, we cn choose N lrge enough such tht N k k = k < ɛ 2. k=1 k=1 k=n+1 Tke now M be sufficiently lrge, so tht ech of the 1,..., N ppers mong the b 1,..., b M. For ll m > M, t m s N = k k < ɛ 2, k>n, f (k) m k=n+1

198 Chpter 5 i.e., This proves tht i.e., k t m k s N + s N t m ɛ. k=1 k=1 lim t m = m b n = n=1 k, k=1 n. n=1 Tht n=1 b n converges bsolutely follows form the fct tht b n is rerrngement of n nd the first prt of this proof. Absolute convergence is lso necessry in order to multiply two series. Consider the product n b k = ( 1 + 2 + )(b 1 + b 2 + ). n=1 k=1 This product seems to be sum of ll products n b k, i.e., ( n b k ), k,n=1 except tht these numbers to be dded form mtrix, i.e., set tht cn be ordered in different wys. Theorem 5.23 Suppose tht n=1 n nd n=1 b n converge bsolutely, nd tht c n is ny sequence tht contins ll terms n b k. Then c n = n b k. n=1 n=1 k=1

Sequences 199 Proof : Consider the sequence s n = k b l. k=1 This sequence converges s n (by limit rithmetic), which implies tht it is Cuchy sequence. For every ɛ > 0 there exists n N N, such tht n n m m k b l k b l < ɛ 2 k=1 k=1 l=1 for ll m, n > N. By the comment bove, m m k b l k b l ɛ 2 for ll m > N. Thus, l=1 k or l>n k=1 k=1 l=1 l=1 l=1 k b l ɛ 2 < ɛ. Let now (c n ) be sequence comprising of ll pirs n b l. There exists n M sufficiently lrge, such tht m > M implies tht either k or l is greter thn N, where c m = k b l. I.e., M N N c n k b l n=1 k=1 only consists of terms k b l with either k or l greter thn N, hence for ll m > M, m N N c n k b l k b l < ɛ. n=1 k=1 l=1 l=1 k or l>n Letting N, m c n k b l ɛ, n=1 k=1 for ll m > M, which proves the desires result. l=1

200 Chpter 5