Mth 7/8: Honours Clculus John C. Bowmn University of Albert Edmonton, Cnd November 9, 5
c 5 John C. Bowmn ALL RIGHTS RESERVED Reproduction of these lecture notes in ny form, in whole or in prt, is permitted only for nonprofit, eductionl use.
Contents Rel Numbers 7.A Elementry Concepts from Set Theory................. 7.B Hierrchy of Sets of Numbers...................... 8.C Algebric Properties of the Rel Numbers................D Absolute Vlue.............................. 5.E Induction................................. 6.F Binomil Theorem.............................G Open nd Closed Intervls........................ 6.H Lower nd Upper Bounds........................ 6.I Supremum nd Infimum......................... 7.J Completeness Axiom........................... 8 Sequences 9.A Limit of Sequence............................ 9.B Monotone Sequences........................... 36.C Subsequences............................... 39.D Bolzno Weierstrss Theorem...................... 43.E Cuchy Criterion............................. 44 3 Functions 49 3.A Exmples of Functions.......................... 49 3.B Trigonometric Functions......................... 5 3.C Limit of Function............................ 59 3.D Properties of Limits............................ 63 3.E Continuity................................. 66 3.F One-Sided Limits............................. 69 3.G Properties of Continuous Functions................... 7 4 Differentition 74 4.A The Derivtive nd Its Properties.................... 74 4.B Mxim nd Minim........................... 83 4.C Monotonic Functions........................... 87 4.D First Derivtive Test........................... 89 3
4 CONTENTS 4.E Second Derivtive Test.......................... 9 4.F L Hôpitl s Rule.............................. 9 4.G Tylor s Theorem............................. 94 4.H Inverse Functions nd Their Derivtives................ 96 4.I Convex nd Concve Functions..................... 4.J Implicit Differentition.......................... 6 5 Integrtion 7 5.A The Riemnn Integrl.......................... 7 5.B Cuchy Criterion............................. 4 5.C Riemnn Sums.............................. 6 5.D Properties of Integrls.......................... 8 5.E Fundmentl Theorem of Clculus................... 5.F Averge Vlue of Function....................... 7 6 Logrithmic nd Exponentil Functions 8 6.A Exponentils nd Logrithms...................... 8 6.B Logrithmic Differentition........................ 36 6.C Hyperbolic Functions........................... 37 7 Techniques of Integrtion 4 7.A Chnge of Vribles............................ 4 7.B Integrtion by Prts........................... 43 7.C Integrls of Trigonometric Functions.................. 48 7.D Prtil Frction Decomposition..................... 5 7.E Integrtion of Certin Irrtionl Expressions.............. 6 7.F Trigonometric & Hyperbolic Substitution................ 63 7.G Strtegy for Integrtion.......................... 65 7.H Numericl Approximtion of Integrls.................. 66 8 Applictions of Integrtion 7 8.A Ares between Curves.......................... 7 8.B Arc Length................................ 73 8.C Volumes by Cross Sections........................ 76 8.D Volume by Shells............................. 8 8.E Work.................................... 84 8.F Hydrosttic Force............................. 86 8.G Surfces of Revolution.......................... 87 8.H Centroids nd Pppus Theorems.................... 89 8.I Polr Coordintes............................. 93
CONTENTS 5 9 Improper Integrls nd Infinite Series 96 9.A Improper Integrls............................ 96 9.B Infinite Series............................... 9.C Power Series................................ 7 A Complex Numbers 9
Prefce These notes were developed for first-yer honours-level mthemtics course on differentil nd integrl clculus t the University of Albert. The uthor would like to thnk the mny students who took Mth 7/8 from September April 3 for their help in developing these notes. Prticulr thnks goes to Mnde Leung for typesetting the originl version of these notes, to Dniel Hrrison for his creful proofreding, nd to Andy Hmmerlindl nd Tom Prince for couthoring the high-level grphics lnguge Asymptote (freely vilble t http://symptote.sourceforge.net) tht ws used to drw the mthemticl figures in this text. 6
Chpter Rel Numbers.A Elementry Concepts from Set Theory Definition: A set is collection of distinct objects. Here re some exmples of sets: {,, 3}, {, }, {}, {book, pen}, N = {,, 3,...}, the set of nturl (counting) numbers, = {}, the empty set. Remrk: Not ll sets cn be enumerted like this, s (finite or infinite) list of elements. The set of rel numbers is one such exmple. Remrk: If we cn write the elements of set in list, the order in which we list them is not importnt. Definition: We sy tht set A is subset of set B if every element of A is lso n element of B. We write A B. Definition: We sy tht set A contins set B if every element of B is lso n element of A. We write A B. Note tht this definition implies tht B A. Definition: We sy tht two sets A nd B re equl if A B nd B A, tht is, if every element in A is lso in B nd vice-vers, so tht A nd B contin exctly the sme elements. We write A = B. See the excellent rticle on countbility, How do I love thee? Let me count the wys! by L. Mrcoux, http://www.pims.mth.c/pi/issue/pge-4.pdf,. Some uthors write this s A B nd reserve the nottion A B for the cse where A is subset of B but is not identiclly the sme set s B, tht is, where A is proper subset of B. In our nottion, if we wnt to emphsize tht A must be proper subset of B, we explicitly write A B. 7
8 CHAPTER. REAL NUMBERS {, } = {, }. Definition: The set contining ll elements of A nd ll elements of B (but no dditionl elements) is clled the union of A nd B nd is denoted A B. Definition: The set contining exctly those elements common to both A nd B is clled the intersection of A nd B nd is denoted A B. These definitions re illustrted in Figure.. {} {} = {, }. {,, 3} {, 4} = {}. {, } {} = {, }. A B A B A B Figure.: Venn Digrm.B Hierrchy of Sets of Numbers We will find it useful to consider the following sets ( mens is n element of): = {} the empty set, N = {,, 3,...}, the set of nturl (counting) numbers, Z = { n : n N} {} N, the set of integers, Q = { p : p, q Z, q }, the set of rtionl numbers, q R, the set of ll rel numbers. Notice tht N Z Q R.
.B. HIERARCHY OF SETS OF NUMBERS 9 Q. Why do we need the set R of rel numbers to develop clculus? Why cn t we just use the set Q of rtionl numbers? One might try to rgue, for exmple, tht every number representble on (finite-precision) digitl computer is rtionl. If subset of Q is good enough for computers, shouldn t it be good enough for mthemticins, too? To nswer this question, it will be helpful to recll Pythgors Theorem, which sttes tht the squre of the length c of the hypotenuse of right-ngle tringle equls the sum of the squres of the lengths nd b of the other two sides. A simple geometric proof of this importnt result is illustrted in Figure.. Four identicl b c b Figure.: Pythgors Theorem copies of the tringle, ech with re b/, re plced round squre of side c, so s to form lrger squre with side + b. The re c of the inner squre is then just the re ( + b) = + b + b of the lrge squre minus the totl re b of the four tringles. Tht is, c = + b. Consider now the following problem. Suppose you drw right-ngle tringle hving two sides of length one. x The Greek mthemticins of ntiquity noticed tht the length of the hypotenuse of such tringle cnnot possibly be rtionl number; tht is, it cnnot be expressed s the rtio of two integers. Let us denote the length of the hypotenuse by x. From Pythgors Theorem, we know tht x = + =. Suppose tht we could indeed write x = P/Q, where P nd Q re integers (with Q ). By cncelling out ny common integer fctors greter thn one, it would then lwys be possible to find
CHAPTER. REAL NUMBERS new integers p nd q tht re reltively prime (hve no common fctors) such tht x = p/q. Then = x = p q p = q p is even. If p were n odd integer, sy n +, then p = (n + ) = 4n + 4n + could not be even. Thus, p must be even: tht is, p = n for some integer n. Then (n) = q 4n = q n = q. This lst result sys tht q nd hence q re lso even, so now we know tht both p nd q re divisible by. But this contrdicts the fct tht p nd q re reltively prime! Hence our originl ssumption tht x = P/Q must be flse; tht is, x cnnot be represented s rtionl number. Remrk: This style of mthemticl proof is known s proof by contrdiction. Remrk: If A nd B re two sttements, the nottion A B sys tht if A holds, then B must lso hold; tht is, A only if B. The nottion A B sys tht if B holds, then A must lso hold; tht is, A if B. If A nd B re equivlent to ech other, we write A B, which mens A if nd only if B. Thus, the length of the hypotenuse of right-ngle tringle with unit sides cnnot be expressed s rtionl number. Mthemticins hve invented new number system, the rel numbers, precisely to circumvent this kind of deficiency with the rtionl numbers Q. The rel numbers, denoted by R, include ll rtionl numbers plus the curious missing irrtionl numbers (like ). In prticulr, the length of ny line segment is contined in the set of rel numbers. This mens tht there re no holes in the rel line. Mthemticins express this fct by sying tht the rel numbers re complete. Another importnt property of rel numbers is tht they cn be written in prescribed order on horizontl number line, in such wy tht every nonzero number is either to the right of the position occupied by the rel number (so tht its negtive is to the left of ), or to the left of (so tht its negtive is to the right of ). Moreover, the sum nd product of two numbers to the right of zero will lso pper to the right of zero. Mthemticins express this prticulr property of the set of rel numbers by sying tht it cn be ordered. Remrk: It is esy to see tht the deciml expnsion of rtionl number must end in repeting pttern (which could be ll zeros, in which cse the rtionl number cn be represented exctly s deciml number with finite number of digits). When we divide the integer p by the integer q, the reminder cn only tke on one of q different vlues, nmely,,...(q ). If the number cn be represented exctly with finitely mny digits, then the deciml expnsion will end with the repeting pttern... (which we represent using the nottion ). Otherwise, we
.C. ALGEBRAIC PROPERTIES OF THE REAL NUMBERS cn never obtin the reminder, so in fct only q vlues of the reminder re possible. Upon doing q steps of long division, we will therefore encounter repeted reminder, by the Pigeon-Hole Principle. 3 At the second occurrence of the repeted reminder, the pttern of digits in the quotient will then begin to repet itself. For q, there will never be more thn q digits in this pttern. For exmple, when computing /7 by long division, the pttern of quotient digits will strt repeting t the second occurrence of the reminder. In this exmple, the mximum possible number of digits in the pttern, q = 6, is ctully chieved. Problem.: Show tht the converse of the bove remrk holds; tht is, if the deciml expnsion of number eventully ends in repeting pttern of digits, the number must be rtionl. Problem.: Show tht every rel number my be pproximted by rtionl number s ccurtely s desired. This shows tht the rtionls densely cover the rel line. We sy tht the rtionls re dense in R. Problem.3: Prove tht 3 is n irrtionl number..c Algebric Properties of the Rel Numbers [Spivk 994, pp. 3 ] We now list the lgebric properties of the rel numbers tht we will use in our development of clculus. (P) If, b, nd c re ny rel numbers, then + (b + c) = ( + b) + c. (ssocitive) (P) There is rel number (the dditive identity) such tht for ny rel number, + = + =. (identity) Problem.4: Show tht the dditive identity is unique, tht is, if +θ = θ + = for ll, then θ =. Hint: set = θ in one set of equtions, set = in the other. (P3) Every rel number hs n dditive inverse such tht + ( ) = ( ) + =. (inverse) 3 The Pigeon-Hole Principle [Fomin et l. 996, pp. 3 37] (lso known s Dirichlet s Box Principle) sttes tht if you try to stuff more thn n pigeons into n holes, t lest one hole must contin two (or more) pigeons!
CHAPTER. REAL NUMBERS Problem.5: Show tht postultes (P P3) imply tht every number hs unique dditive inverse. Tht is, if + b =, show tht b =. Definition: We define b =.. + ( b). (We use the symbol = to emphsize definition, lthough the nottion := is more common.) Problem.6: If b =, show tht = b. (P4) If nd b re rel numbers, then + b = b +. (commuttive) Remrk: Not ll opertions hve this property. Cn you give n exmple of n noncommuttive opertion? (P5) If, b, nd c re ny rel numbers, then (b c) = ( b) c. (ssocitive) (P6) There is rel number (the multiplictive identity) such tht if is ny rel number, = =. (identity) (P7) If, b, nd c re ny rel numbers, then (b + c) = b + c. (distributive) Remrk: = for ll rel. Proof: + = + = ( + ) = =. =. Note: the symbol mens therefore. (P8) For ny rel number, there is rel number such tht = =. (inverse) Q. Why do we restrict here?
.C. ALGEBRAIC PROPERTIES OF THE REAL NUMBERS 3 Problem.7: Show tht both the multiplictive identity nd the multiplictive inverse of ny rel number is unique. (P9) If nd b re rel numbers, then b = b. (commuttive) Definition: If b >, we write > b. Similrly, if b <, we write < b. (P) Given two rel numbers nd b, exctly one of the following reltionships holds: < b, = b, > b. (Trichotomy Lw) (P) (P) > nd b > + b >. (closure under +) > nd b > b >. (closure under ) Definition: If < b or = b we write b. If > b or = b we write b. Q. Is it correct to write? Why or why not? Remrk: All the elementry rules of lgebr nd inequlities follow from these twelve properties. To see tht ( )( b) = b, use the distributive property: ( )( b) b = ( b + b) = () =, so ( )( b) = b. If < nd b <, then > nd b > ( )( b) = b >. Remrk: By setting = b in the bove exmple nd in (P), we see tht the squre of ny nonzero number is positive. If > b nd b > c, then b > nd b c > c > by (P) > c. (trnsitive) If > b nd c >, then
4 CHAPTER. REAL NUMBERS b > nd c > c bc > i.e. c > bc. by (P) If > b nd c <, then b > nd c > c + bc > i.e. c < bc. by (P) If > b, c R ( + c) (b + c) = b > + c > b + c. { > nd b >, b > or < nd b <. Since = or b = b = contrdicts b > nd >, b < >, b > b > contrdicts b > nd <, b > contrdicts b >. Problem.8: If < b nd c < d, show tht + c < b + d. Definition: If < x nd x < b, we write < x < b nd sy x is between nd b. Lemm. (Midpoint Lemm): < b < + b < b. Proof: < b + < + b < b + b = + < + b < b + b = b. Remrk: This lemm (smll theorem) estblishes tht there is no lest positive number. Moreover, between ny two distinct numbers there exists nother one. Q. Wht bout.9?
.D. ABSOLUTE VALUE 5.D Absolute Vlue [Muldowney 99, pp. 3] The fct tht for ny nonzero rel number either x > or x > mkes it convenient to define n bsolute vlue function: { x if x, x = x if x <. Properties: Let x nd y be ny rel numbers. (A) x. (A) x = x =. (A3) x = x. (A4) xy = x y. (A5) If c, then x c c x c. Proof: x c x c or < x c c x c. (A6) x x x. Proof: Apply (A5) with c = x. (A7) x y x ± y x + y. (Tringle Inequlity) Proof: RHS: (A6) { x x x y y y ( x + y ) x + y x + y = c (A5) x + y x + y. Let y y : x y x + y. Thus x ± y x + y.
6 CHAPTER. REAL NUMBERS LHS: x = (x + y) y x + y + y.e Induction x y x + y x y : y x y + x = x + y x y x + y. Let y y : x y x y. [Muldowney 99, pp. 7] Suppose tht in certin city locted on the west cost of Cnd, the wether office mkes long-term forecst consisting of two sttements: (A) If it rins on ny given dy, then it will lso rin on the following dy. (B) It will rin tody. Wht would we conclude from these two sttements? We would conclude tht it will rin every single dy from now on! Or, consider secret pssed long n infinite line of people, P P...P n P n+..., ech of whom enjoys gossiping. If we know for every n N tht P n will lwys pss on secret to P n+, then the mere ct of telling secret to the first person in line will result in everyone in the line eventully knowing the secret! These musing exmples encpsulte the xiom of Mthemticl Induction: If subset S N stisfies (i) S, (ii) k S k + S, then S = N. For exmple, suppose we wish to find the sum of the first n nturl numbers. For smll vlues of n, we could just compute the totl of these n numbers directly. But for lrge vlues of n, this tsk could become quite time consuming! The gret mthemticin nd physicist Crl Friedrich Guss (777 855) t ge noticed tht the rte of increse of the terms in the sum + +... + n could be exctly compensted by first writing the sum bckwrds, s n + (n ) +... +,
.E. INDUCTION 7 nd then verging the two equl expressions term-by-term to obtin sum of n identicl terms: n + + n + +... + n + ( ) n + = n. } {{ } n terms We will use mthemticl induction to verify Guss clim tht n n(n + ) + +... + n i =. (.) Let S be the set of numbers n for which Eq. (.) holds. Step : Check S: Step : Suppose k S, i.e. Then = k i = i= i= ( + ) =. k(k + ). ( k+ k ) i = i + (k + ) i= Hence k + S. Tht is, k S k + S. i= k(k + ) = + (k + ) ( ) k = (k + ) + (k + )(k + ) =. By the Axiom of Mthemticl Induction, we know tht S = N. In other words, n n(n + ) i =, n N. i= Here, the symbol mens for ll. Prove tht for ll nturl numbers n, ( n n i 3 = i). i= i=
8 CHAPTER. REAL NUMBERS We hve just seen tht n i = i= Hence wht we relly wnt to show is tht n i= n(n + ). Step : We see for n = tht = ( + ) /4. Step : Suppose Then n i= i 3 = n (n + ). (.) 4 i 3 = n (n + ) 4 ( n+ n ) i 3 = i 3 + (n + ) 3 i= i= =. = S n. = n (n + ) + (n + ) 3 (n + ) = (n + 4n + 4) 4 4 = (n + ) (n + ) = S n+. 4 Hence by induction, Eq. (.) holds. If < < b, show tht for ll n N. < n < b n (.3) Step : For n = we know tht < < b. Step : Assume < k < b k. On multiplying this inequlity by > nd the inequlity < b by b k >, we obtin < k+ < b k < b k b = b k+, from which we see tht the cse n = k + lso holds. by induction, Eq. (.3) holds for ll n. (Bernoulli Inequlity) If then ( + ) n + n n N.
.E. INDUCTION 9 Step : When n =, we see tht + +. Step : Assume tht the cse n = k holds: ( + ) k + k. Then ( + ) k+ ( + k)( + ) = + k + + k }{{} + k + = + (k + ), so tht the cse n = k + lso holds. All students re geniuses! We clim tht ll students in ny group of n students must be geniuses, for ech n =,,.... Assume tht the cse n = k holds. Given group of k + students, remove one of the students from the group. We know tht ech of the remining k students re geniuses. Now swp one of these geniuses with the removed student. Since every student in this new group of k students re lso geniuses, we deduce tht ll k + students re geniuses. By induction, the clim holds. Problem.9: Is there n error in the bove proof? If so, where is the flw? All girls hve the sme hir colour. We clim tht ll girls in ny group of n girls hve the sme hir colour, for ech n =,,.... Step : When n =, there is only one girl in the group, so ll girls in the group clerly hve the sme hir colour. Step : Assume tht the cse n = k holds. Given group of k + girls, remove one of them from the group. By ssumption, ech of the remining k girls hve the sme hir colour. Now swp one of these girls with the girl we removed. Since every girl in this new group of k girls lso hve the sme hir colour, we know now tht ll k + girls hve the sme hir colour! By induction, the clim holds. Problem.: Is there n error in the bove proof? If so, where is the flw? Use induction to show tht one cn extrct n distinct subsets from every set of n elements. For exmple:
CHAPTER. REAL NUMBERS set subsets { } { } {} { } {} {, b} { } {} {b} {, b} {, b, c} { } {} {b} {c} {, b} {, c} {b, c} {, b, c} Step : Let n =. There re = subsets of ny such singleton set, nmely the empty set nd the set itself. Step : Suppose the clim holds for n = k. Given ny set S k+ with k + elements, denote the collection of its subsets by S k+. Now remove one element from S k+, leving set S k contining k elements. Denote the collection of subsets of S k by S k. Notice tht S k S k+. The remining members of S k+ re obtined by dding the element we removed from S k+ to ech of the sets in S k. Hence the number of sets in S k+ is exctly twice the number of sets in S k. Given tht there re k members in S k, we deduce tht S k+ hs k+ members. By induction, we see tht exctly n distinct subsets cn be formed from every set contining n elements. Q. Cn you think of more direct wy to estblish this result? Summtion Nottion Recll k=n k = + +... + n = k= n(n + ). k=n Q. Wht is k? k= A. k=n k=n k = + k = + k= k= n(n + ) = n(n + ). Q. How bout A. k=n+ k= k=n+ k= k = k? ( k=n ) k + (n + ) = k= n(n + ) + n + = (n + )(n + ).
.E. INDUCTION k=n Q. How bout (k + )? k= A. Method : k=n k=n k=n (k + ) = k + = k= k= k= Method : First, let k = k + : n(n + ) + n = n(n + 3). k=n (k + ) = k= k =n+ k = Next, it is convenient to replce the symbol k with k (since it is only dummy index nywy): k =n+ k = In generl, k = k=n+ k= k = ( k=n+ k= k ) = k=u k+m = k=l k. (n + )(n + ) = k=u+m k=l+m Verify this by writing out both sides explicitly. k. n(n + 3). Problem.: For ny rel numbers,,... n, b, b,...b n, nd c prove tht Telescoping Sum: n c( k + b k ) = c k= n ( k+ k ) = k= n k + c k= n k+ k= n+ = k = k= n k= n k= k k n b k. k= / ( n k + n+ + k= = n+. / ) n k k=
CHAPTER. REAL NUMBERS.F Binomil Theorem Definition: n! =... (n ) n if n N,! =. Equivlently,! = nd (k + )! = (k + )k! for k =,,,....! =,! =,! =, 3! = 6, 4! = 4. [Muldowney 99, pp. 8 ] Definition: We introduce the binomil coefficient ( ) n n! if k =, = k k!(n k)! = n(n )...(n k + ) if k n.... k We ( find ) 3 =, ( ) 3 = 3 = 3, ( ) 3 ( ) 3 3 = 3 = 3, = 3 3 =. Also, ( ) 7 = 7 6 5 3 3 = 35. ( ) ( ) n n Remrk: =. k n k ( ) ( ) n n Remrk: = =. n ( ) ( ) n n Remrk: = = n. n Remrk: If n, k N, with k n, then ( ) ( ) ( ) n n n + + =. (Pscl s Tringle Lw) k k k
.F. BINOMIAL THEOREM 3 Proof: ( ) n + k ( ) n = k n! (k )!(n k + )! + n! k!(n k)! [ n! = (k )!(n k)! n k + + ] k ] n! [ k + (n k + ) = (k )!(n k)! (n k + )k ( ) (n + )! n + = k!(n k + )! =. k n\k 3 4 sum or 4 3 3 3 3 3 8 4 4 6 4 4 6 4 6 Problem.: Show tht ( n k) is n integer for ll integers k nd n stisfying k n. Hint: use induction on n nd Pscl s Tringle Lw. Alterntively, one cn mke use of the fct tht ( n k) is the number of distinct wys of choosing k objects from set of n objects. Clim: n i= ( ) n = n, n N {}. i Proof (by induction): Step : Cse n = : ( ) = =. Step : Suppose the clim holds for n = k: k+ ( ) k = k. i i= Then k+ ( ) ( ) k + k + = + i i= k ( ) k + + i i= ( ) k + k +
4 CHAPTER. REAL NUMBERS Now Thus k ( ) k = i i= k ( ) k = i i= k [( ) ( )] k k = + + + i i i= k ( ) k k ( ) k = + + +. i i i= i= i= i= k ( ) ( ) k k = k, i k k ( ) ( ) k k = k. i k+ ( ) k + = + ( k ) + ( k ) + i i= = k = k+. Cse n = k + holds. Theorem. (Binomil Theorem): For ll n N, ( ) ( ) ( n n n ( + b) n = n + n b + n = k= Proof (by induction): ) n b +... + ( ) n b n + n ( ) n b n n ( ) n n k b k. (.4) k Step : Cse n = : ( + b) = ( ) + ( ) b = + b. Step : Suppose Eq. (.4) holds for some n. Then ( + b) n+ = ( + b)( + b) n n ( n = ( + b) k = n k= ( n k k= ) n k b k ) n+ k b k + n k= ( ) n n k b k+ k
.F. BINOMIAL THEOREM 5 = = n ( n k k= ( n = n+ + = n+ + ) n+ k b k + ) n+ + k= k= n+ ( n + = k k= n+ k= ( ) n n (k ) b k k n ( ) n n k+ b k + k n ( ) n n+ k b k + k k= k= n [( ) ( )] n n + n+ k b k + b n+ k k n ( ) n + n+ k b k + b n+ k ) n+ k b k. Thus, by induction, Eq. (.4) holds for ll n N. Remrk: Alterntive form of Binomil Theorem: ( + b) n = n + n n b + Remrk: When = nd b = x, we find n(n ) n b +... + nb n + b n. ( + x) n n(n ) = + nx + x +... + nx n + x n + nx if x nd n. In fct, Bernoulli s Inequlity shows this is true even for x. ( ) n b n+ n Remrk: Let =. Then Set b = : n k= n k= ( ) n b k = ( + b) n. k ( ) n = ( + ) n = n. k Set b = : Set b = : n k= n k= ( ) n ( ) k = ( ) n =. k ( ) ( n k = + ) n = 3n k n.
6 CHAPTER. REAL NUMBERS Set b = : Set b = x : n k= n k= ( ) ( n k ( = k ) ) n = n. ( ) n (x ) k = ( + x ) n = x n. k.g Open nd Closed Intervls [Muldowney 99, pp. 3 4] Let, b R nd < b. There re 4 types of intervls: [,b] = {x : x b}, contins both end points (, b) = {x : < x < b}, [,b) = {x : x < b}, (,b] = {x : < x b}. It will be convenient to define lso: However, these re not (finite) intervls. (, ) = R, [, ) = {x : x }, (, ) = {x : x > }, (,] = {x : x }, (,) = {x : x < }..H Lower nd Upper Bounds [Muldowney 99, pp. 4 5] Definition: Given S R, we sy tht rel number b is n upper bound of S if x b for ech x S. Q. Do ll sets S hve n upper bound? Definition: If S hs n upper bound we sy S is bounded bove. Otherwise we sy S is unbounded bove. Remrk: An upper bound of S my, or my not, be n element of S.
.I. SUPREMUM AND INFIMUM 7 Definition: Given S R we sy tht rel number is lower bound of S if x for ech x S. Definition: If S hs lower bound we sy tht S is bounded below. Otherwise we sy tht S is unbounded below. Definition: If S is bounded bove nd below we sy S is bounded. Otherwise, we sy tht S is unbounded. Tht is, S R is bounded S [, b] for some, b R. Q. Consider the intervls [, ] nd [, ). Do these sets hve (i) n upper bound, (ii) lower bound, (ii) n upper bound in the set itself, (iv) lower bound in the set itself?.i Supremum nd Infimum [Muldowney 99, pp. 5] Definition: Let S R. Suppose there exists rel number b such tht (i) x b for ech x S (b is n upper bound for S), (ii) If u is n upper bound of S, then b u. Then b is clled the lest upper bound, or supremum, of S. We write b = l.u.b. S or b = sup S. Definition: If b = sup S nd b S, we sy b is the mximum of S. We write b = mxs. Remrk: A finite set of elements {,,..., n } lwys hs mximum element mx(,,..., n ). Note tht mx(,,..., n ) i for i =,,..., n. Note tht [, ] hs mximum element, but [, ) hs no mximum element. Definition: Let S R. Suppose there exists rel number such tht (i) x for ech x S ( is lower bound for S), (ii) If l is lower bound of S, then l. Then is clled the gretest lower bound, or infimum, of S. We write = g.l.b. S or b = inf S.
8 CHAPTER. REAL NUMBERS Definition: If = inf S nd S, we sy is the minimum of S. We write = min S. Remrk: A finite set of elements {,,..., n } lwys hs minimum element min(,,..., n ). Note tht min(,,..., n ) i for i =,,..., n. { } p Consider S = q : p q, p Z, q N. The lest upper bound of S is the rel number, so S hs supremum in R. However, / Q, so the supremum of S is not itself in S; tht is, S hs no mximum element..j Completeness Axiom [Muldowney 99, pp. 6] The completeness xiom sttes tht every nonempty subset of R with n upper bound hs lest upper bound in R. { } p q : p q, p Z, q N hs lest upper bound in R. [, ] hs the supremum. [, ) hs the supremum. Remrk: = {} hs no supremum. Any rel number is n upper bound of the empty set, so the empty set cnnot hve lest upper bound. Lemm. (Archimeden Property): No rel number is n upper bound for N. Note: here N is the subset of R defined inductively by (A) N, (B) k N k + N. Proof (by contrdiction): Suppose tht N hd n upper bound. Then N R, N b = sup N, where b is some rel number. Here the symbol mens there exists. By the definition, b = sup N mens (i) b k k N, (ii) b k k N. Tht is, b < k for some k N b < k +. But k N k + N, so (ii) contrdicts (i)!
Chpter Sequences [Muldowney 99, Chpter ] [Spivk 994, Chpter ].A Limit of Sequence Definition: A (rel-vlued) function f is rule tht ssocites rel number f(x) to every number x in some subset D R. The set D is clled the domin of f. Definition: The rnge f(d) of f is the set {f(x) : x D}. f(x) = x on domin D = [, ): f(d) = {x : x [, )} = [, 4). Definition: A sequence is function on the domin N. The vlue of function f t n N is often denoted by n, n = f(n). The consecutive function vlues re often written in list: { n } n= = {,,...} Repeted vlues re llowed. n = f(n) = n, { n } n= = {, 4, 9, 6...}. { } ( ) n = {, n, 3, 4 },.... n= Notice tht s n gets lrge, the terms of this sequence get closer nd closer to zero. We sy tht they converge to. However, n is not equl to for ny n N. We cn formlize this observtion with the following concept: 9
3 CHAPTER. SEQUENCES Definition: The sequence { n } n= exist number N such tht is convergent with limit L if, for ech ǫ >, there We bbrevite this s: lim n n = L. n > N n L < ǫ. If no such number L exists, we sy { n } n= diverges. Remrk: lim n n = L mens tht n L cn be mde s smll s we plese, simply by choosing n lrge enough. Remrk: Equivlently, s illustrted in Fig.., lim n n = L mens tht ll but finite number of terms of { n } n= re contined in ny open intervl bout L, tht is, in ny intervl (L ǫ, L + ǫ), where ǫ >. Remrk: If sequence { n } n= converges to L, the previous remrk implies tht every open intervl (L ǫ, L + ǫ) will contin n infinite number of terms of the sequence (there cnnot be only finite number of terms inside the intervl since sequence hs infinitely mny terms nd only finitely mny of them re llowed to lie outside the intervl). n 3 L = n = +, ǫ = n 4 L + ǫ L ǫ N = ǫ n Figure.: Limit of sequence Let n =, n N i.e. {,,,...}.
.A. LIMIT OF A SEQUENCE 3 Let ǫ >. Choose N =. Tht is, L =. Write lim n n =. n > n = = < ǫ. Remrk: Here N does not depend on ǫ, but normlly it will. n = ( )n n. lim n = since n = n ( ) n n = n < if n > N. N So, given ǫ >, we my force n < ǫ for n > N simply by picking N ǫ : n > N n < N ǫ. Proposition. (Uniqueness of Limits): If lim n n = L nd lim n n = L, then L = L. Proof: Given ǫ >, N, N N such tht n > N n L < ǫ, Let N = mx(n, N ). Then n > N n L < ǫ. n > N n > N nd n > N n L < ǫ nd n L < ǫ L L = L n + n L L n + n L < ǫ + ǫ = ǫ. Tht is, given ny number = ǫ >, then L L <, i.e. L L is smller thn ny positive number! But we hve lredy estblished from Lemm. tht there is no smllest positive number, nd since n bsolute vlue cn never be negtive, the only choice left is L L = L = L. Problem.: Suppose n nd { n } n= is convergent with limit. Show tht { n } n= converges with limit. Hints: Do the cse = seprtely. When >, note tht n = n n +.
3 CHAPTER. SEQUENCES Problem.: Suppose tht { n } nd {b n } re convergent sequences such tht n < b n for ll n N. Use proof by contrdiction to show tht lim n lim b n. Cn n n we conclude lim n < lim b n? n n Problem.3 (Squeeze Principle): Suppose x n z n y n for ll n N. If the sequences {x n } nd {y n } both converge to the sme number c, show tht {z n } is lso convergent nd hs limit c. Definition: A sequence is bounded if there exists number B such tht n B n N. Theorem. (Convergent Bounded): A convergent sequence is bounded. Proof: Suppose lim n = L. n Let ǫ =. Then N (the symbol mens such tht ) n > N n L < n L n L n L < n < + L, n > N. Hence n mx{,,..., N, + L }. = B for ll n N. Remrk: A bounded sequence need not be convergent. {( ) n } is bounded since ( ) n = B, if we tke B =. However, the sequence is not convergent: Suppose lim n n = L where n = ( ) n. Given ǫ =, then for n sufficiently lrge, n L = L < nd n+ L = L = + L < = L + L + L + L + < +, i.e. <, contrdiction. ( n ) lim + n =. n Given ǫ >, we cn mke n + n = n + n n + + n n + + n if n > N, s long s N ǫ, i.e. N ǫ. n + n = n + + n < ǫ n
.A. LIMIT OF A SEQUENCE 33 Remrk: By Theorem., we see tht { n + n} n= is bounded. Corollry.. (Unbounded Divergent): An unbounded sequence is divergent. Remrk: We sy tht Corollry.. is the contrpositive of Theorem.. Consider {n} n=. Theorem. (Properties of Limits): Let { n } nd {b n } be convergent sequences. Let L = lim n nd M = lim b n. Then n n () lim n ( n + b n ) = L + M; (b) lim n n b n = LM; n (c) lim = L n b n M if M. Proof of (): We wnt to show, given ǫ >, tht n + b n L M < ǫ (.) for ll sufficiently lrge n. Since the Tringle Inequlity tells us tht n + b n L M n L + b n M, it is enough to show tht for ll sufficiently lrge n. We know N, N n L + b n M < ǫ (.) n > N n L < ǫ, n > N b n M < ǫ. When n > mx(n, N ) then Eq. (.) will hold, which in turn implies Eq. (.). Hence lim n ( n + b n ) = L + M. Proof of (b): Note tht n b n LM = n b n Lb n + Lb n LM n b n Lb n + Lb n LM n L b n + b n M L.
34 CHAPTER. SEQUENCES Theorem. {b n } is bounded B > b n B n. Given ǫ >, let ǫ = n b n LM n L B + b n M L. ǫ B + L >. There exists N, N such tht n > N n L < ǫ, So n > mx(n, N ) n > N b n M < ǫ. Tht is, n b n LM ǫ B + ǫ L = ǫ (B + L ) = ǫ. lim nb n = LM. n Proof of (c): We only need to prove the specil cse lim b n = M lim = n n b n M for (c) would then follow from (b): ( ) n lim = lim n n b n n b n = lim n lim = L n n b n M To prove Eq. (.3), consider b n M = M b n b n M = b n b n M. M We know tht there exists number N such tht Thus n > N b n M < M M b n b n M < M < M < b n b n < M. n > N b n M < M b n M. Now given ǫ >, there exists number N n > N b n M < M ǫ since M. if M, (.3) if M.
.A. LIMIT OF A SEQUENCE 35 Hence n > mx(n, N ) Thus b n M < M M ǫ = ǫ. lim n b n = M. ( n + lim = lim + ) = lim + lim n n n n n n n. lim n n + 3n + n + = lim n + n 3 + n + n lim n ( + ) n = lim n (3 + + ) = 3. n n m n m + m n m +... + lim = n b m n m + b m n m +... + b m + m lim + n m +... + n n m n b m + b m + b n m +... + b n n m = m b m. n + lim n n 3 + 3 = lim + n n 3 n + 3 = lim n ( + ) n n 3 lim n 3 n ( + 3 ) = =. n 3 n 3 + 3 lim n n + = lim + 3 n 3 n + lim 3 n ( + ) n 3 lim n n 3 n ( + ). n n 3 Corollry.. (Cse L, M = ): Let { n } nd {b n } be convergent sequences. n If lim n nd lim b n =, then lim does not exist. n n n b n n Proof: Suppose lim = K. n b n Then Theorem. (b) lim n = lim b n lim n n This contrdicts the fct tht lim n n. Returning to the previous exmple, we see tht n n + 3 n lim 3 n + /. n n 3 b n = K =.
36 CHAPTER. SEQUENCES n 4 + 8n 3 + 6n + lim n n 3 + 3n + = lim + 8 + 6 + n n 3 n 4 n + 3 + /. n n n 4 In generl, if m nd b k, then m n m +... + lim = n b k n k +... + b Remrk: We cnnot use Theorem. to sy if m < k, m bm if m = k, / if m > k. n + = lim = lim n n n n lim (n + ) = lim (n + ) n n becuse lim n (n + ) does not exist. Whenever we use theorem, we must be very creful to check tht the conditions of the theorem pply!.b Monotone Sequences Definition: A sequence {} is incresing if nd decresing if 3..., i.e. n n+ n N 3..., i.e. n n+ n N. Q. By the bove definition, is it possible for sequence to be both incresing nd decresing? Definition: A sequence is monotone if it is either (i) n incresing sequence or (ii) decresing sequence. Definition: A sequence { n } is strictly incresing (strictly decresing) if < < 3 <... ( > > 3 >...). {n}, the sequence of ll nturl numbers, is strictly incresing. {n}, {n + }, {n }, nd {,,...} re ll strictly incresing sequences. Recll convergent bounded. When does bounded convergent? The next theorem ddresses this question in the specil cse of monotone sequences.
.B. MONOTONE SEQUENCES 37 Theorem.3 (Monotone Sequences: Convergent Bounded): Let { n } be monotone sequence. Then { n } is convergent { n } is bounded. Proof: Let { n } be convergent. Then { n } is bounded by Theorem.. Suppose { n } is incresing nd bounded. Let L = sup{ n : n =,,...} (why does this lwys exist?) We show tht lim n = L. Given n ǫ >, then L ǫ is not n upper bound of { n : n =,,...} by the definition of supremum. Tht is, element N L ǫ < N. Now n > N n N (why?). Thus n > N L ǫ < N n L L ǫ < n < L + ǫ. n L < ǫ. Hence lim n n = L. The proof for the cse where { n } decresing nd bounded is similr. Problem.4: Complete the bove proof for the cse of decresing bounded sequence { n }. Suggestion: consider the sequence { n }. { n } : n =,,... is incresing nd bounded. ( lim ). n n Of course, we lredy knew from Theorem. () tht lim n ( n ) = lim lim n n n = =. If n = ( + n) n, show tht { n } n= is convergent sequence. In the Binomil Theorem (x + y) n = n ( n k= k) x n k y k, set x = nd y = /n. Then ( n = + ) n = + n ( ) ( ) ( ) 3 n(n ) n(n )(n ) + + n! n! n 3! n
38 CHAPTER. SEQUENCES ( ) n n(n )...3 +... + + n! ( ) n (n )! n n! n = + + ( ) + ( ) ( )! n 3! n n ( +... + ) ( ) (... n (n )! n n n + ( ) ( ) (... n ). n! n n n This expression for n cn be used to estblish two key properties. Clim: { n } is (strictly) incresing. n+ = + + ( ) + ( ) ( )! n + 3! n + n + +... + ( ) ( ) (... n ) n! n + n + n + ( + ) ( ) (... n ) > n. (n + )! n + n + n + }{{} positive ) Clim: { n } is bounded. ) n + +! + ( 3! + 4! +... + n! + + + ( + +... + ) since n! n for n 4 (induction) 4 n + + + + ( + 3 +... + ) 4 n + ( )n+ < + = 3. Thus, by Theorem.3, { n } converges. We define e =. ( lim + ) n n n The bove rgument shows tht e 3. In fct e.78888459.... Problem.5 (Sequence Limit Rtio Test)): Let { n } n= be sequence such tht lim n+ n n = r, where r [, ). Let s = ( + r)/, so tht r < s <.
.C. SUBSEQUENCES 39 () Show tht there exists number N such tht n N Consider ǫ = s r >. (b) Use prt () nd induction to show for n N tht n s N N s n. (c) Prove tht lim n n =. (d) Apply this result to prove for ny x R tht x n lim n n! =, n+ n s. Hint: (e) If r > show tht { n } is divergent. Hint: consider the sequence {/ n }. (f) If r = give exmples to illustrte tht { n } my be either convergent or divergent..c Subsequences Definition: Given sequence { n } n= nd strictly incresing sequence of nturl numbers {n k } k=, we cn form the subsequence { n k } k= of { n} n=. {(k ) } k= = {, 9, 5,...} is subsequence of {n } n= = {, 4, 9, 6, 5,...}. Remrk: Note tht n k k for ll k. This is esily proven by induction: n nd if n k k then n k+ > n k k, so tht n k+ k +. {,...} is subsequence of { n } n=. So is {,,...}. So is { n } n= itself (here n k = k)., Theorem.4 (Convergent All Subsequences Convergent): A sequence { n } n= is convergent with limit L ech subsequence { nk } k= of { n} n= is convergent with limit L. Proof: Suppose { n } is convergent with limit L. Given ǫ >, N Let { nk } be subsequence of { n }. Then n > N n L < ǫ. (.4) k > N n k k > N.
4 CHAPTER. SEQUENCES Considering only the indices n = n k in Eq. (.4) nk L < ǫ. lim k nk = L. Suppose ech subsequence of { n } is convergent with limit L. Note tht { n } is subsequence of itself. Hence { n } is convergent with limit L. {( ) n } is not convergent since {( )n } = {,,,...} {( ) n+ } = {,,,...} nd. In the exmples to follow, we will mke use of lemm. Lemm.: For ll n N, () c < c n <, (b) c > c n >. Proof: () Holds for n = ; c n < c n+ = cc n c <. (b) Holds for n = ; c n > c n+ = cc n > c >. lim n c n = if c <, if c =, / if c > (divergent; in fct, unbounded (exercise)). Cse c < : We hve c n+ c n c <, n. {c n } n= is decresing, bounded sequence. L = lim n c n nd L c <. {c n+ } n= is subsequence of {cn } n=, so lim n cn+ = L. But c n+ = c c n, so Theorem. (b) implies. Cse c = : lim n c n = lim n =. L = lim n c n+ = c lim n c n = c L L( c) = L = (why?)
.C. SUBSEQUENCES 4 Cse c > : Now < c n < c n+ {c n } is (strictly) incresing. If {c n } were convergent, we could let L = lim n c n. Then L = lim n c n+ = c lim n c n = c L L( c) = L =. This would contrdict c n > for ll n N, which requires tht L. Hence {c n } is divergent. ( ) n lim n = lim =. n n Definition: If c >, let c /n denote the n th positive root of c (which we will soon see lwys exists). This rel root is unique: consider x n y n = (x y)(x n + x n y + x n 3 y +... + xy n + y n ) nd observe tht, if x nd y re both positive, x n y n = x = y. The bove fctoriztion my be estblished using Telescoping Sum: n n n n (x y) x n k y k = x n k y k x n k y k+ = x n k y k k= k= = x n y n. k= An immedite corollry of Lemm. is Lemm.: For ll n N, () c < c /n <, (b) c > c /n >. We now estblish tht lim n c /n = for ny c >. Cse c > : Note tht < c /n c < cc /n = c (n+)/n. Let n = c /n. Then k= n+ n+ = ( c /(n+)) n+ = c < c (n+)/n = n+ n ( ) n+ n+ < n+ < n n { n } decresing nd bounded: < n+ < n = c. Hence lim n c /n = L. But then lim n c /(n) = L. n x n k y k k=
4 CHAPTER. SEQUENCES Note tht c /n > for ll n N implies tht L. But lim c /(n) lim c /(n) = lim c /n n n n Hence L L =. Cse c = : lim n c /n = lim n =. Cse < c < : Let k = c c = k Then k > lim n k /n =. Theorem. (c) lim n c /n = The sequence n = n n converges. n+ = n + n+ = L L = L L(L ) =. c /n = k /n = k /n. lim n k /n = =. ( n + n ) ( n n ) = ( n + n ) n < n+ n since n + n n N. { n } is convergent. Its limit cn be found from the observtion tht ( ) n + lim n = lim n+ = lim n = n n n n lim n. n Hence lim n n =. Consider the sequence x n+ = (x n + Axn ), A >, x >. Suppose p = lim n x n. Then p = p + A p p = A. This sequence cn therefore be used s n lgorithm for computing squre roots. To show tht it ctully converges, consider x n > x n+ > n. Consider x > x n > n. x n+ x n = x n + A x n = A x n x n.
.D. BOLZANO WEIERSTRASS THEOREM 43 Now, for ll n N, x n+ A = ) n A (x + A + A = 4 x n 4 x n A, n =, 3,... x n+ x n, n =, 3,... ( x n A x n ). Since {x n+ } n= is decresing nd bounded below by A >, we know tht {x n+ } n=, nd hence {x n } n=, is convergent. Problem.6: Let r be rel number. Consider the sequence {S n } n=, where S n is the prtil sum of the geometric series S n = n r k. k= For wht vlues of r does S = lim n S n exist? Compute S (when the limit exists) in terms of r. When the limit exists, we sy tht the infinite series converges nd hs limit S. Hint: Consider the telescoping sum rs n S n. Problem.7: Compute the sum k= r k 9 k. k=.d Bolzno Weierstrss Theorem The converse of Theorem. (convergent bounded) does not necessrily hold for nonmonotonic sequence. Nevertheless, Theorem. does hve the following prtil converse: Theorem.5 (Bolzno Weierstrss Theorem): A bounded sequence hs convergent subsequence. Proof: Let { n } n= be bounded sequence. Then there exists rel numbers A nd B such tht n [A, B] n =,,... Split [A, B] into subintervls [A, A+B ] nd [ A+B, B]. At lest one of the subintervls must contin infinitely mny members of the sequence { n } n= (why?). Denote this intervl by [A, B ] nd let n [A, B ] be one such member.
44 CHAPTER. SEQUENCES Set i =. Hving constructed the intervl [A i, B i ], consider the two subintervls [A i, A i+b i ] nd [ A i+b i, B i ]. Agin t lest one of the intervls, cll it [A i+, B i+ ], contins infinitely mny elements of { n } n=. Let n i+ [A i+, B i+ ] be one such element. Proceeding inductively, we define sequence of nested intervls [A i, B i ] such tht for ech i =,,..., ni+ [A i+, B i+ ] [A i, B i ], i.e. A i A i+ ni+ B i+ B i, i =,,.... Note tht {A i } i= is bounded incresing sequence lim i A i = L. Likewise, {B i } i= is bounded decresing sequence lim i B i. Moreover, lim B B A i lim A i = lim(b i A i ) = lim = (B A) lim i i i i i lim B i = lim A i = L. i i i i = Note tht A i ni B i i N. Since {A i } i= nd {B i} i= hve the sme limit L, the Squeeze Principle tells us lim n i = L. i We hve thus constructed convergent subsequence { ni } i= of { n } n=..e Cuchy Criterion [Muldowney 99, pp. 38] Q. To prove convergence by the ǫ, N definition, we need to first know the limit L. For monotonic sequence we sw we could prove convergence without knowing L: ll we hve to estblish is tht the sequence is bounded. But wht cn we do in the cse of nonmonotonic sequences when we don t know which vlue of L we should use in the limit definition? A. Use the Cuchy Criterion. Definition: The sequence { n } is Cuchy Sequence if for ll ǫ >, there exists number N such tht m, n > N m n < ǫ. N.B. This must hold not just for m = n+ nd n, but lso for m = n+, m = n+3,..., deep into the til of the sequence.
.E. CAUCHY CRITERION 45 n =. Given ǫ >, the Tringle Inequlity implies n m n = m n m + n < N = ǫ whenever m, n > N provided we choose N = ǫ. Hence { } is Cuchy Sequence. Notice tht { } is convergent. n n Remrk: Without loss of generlity we cn lwys tke m n > N. Remrk: If for every N we cn find even single n, m pir tht violtes n m < ǫ, the sequence is not Cuchy Sequence. {( ) n } is not Cuchy Sequence. Notice {( ) n } is divergent. Pick m = n + : ( ) n+ ( ) n = ǫ, hd we been given ǫ = (sy). Wht we hve observed in these two cses ws formulted by Cuchy into the next theorem. { n } is Cuchy se- Theorem.6 (Cuchy Criterion): { n } is convergent quence. Proof: Suppose { n } is convergent. Let L = lim n n. Given ǫ >, N n > N n L < ǫ. Also, m > N m L < ǫ. Therefore m, n > N m n m L + L n < ǫ + ǫ = ǫ. Thus { n } is Cuchy sequence. Let { n } be Cuchy sequence. Step : { n } is bounded. There exists number N such tht m, n > N n m < (in prticulr).
46 CHAPTER. SEQUENCES E.g. Let m = N +. Then n > N n N+ n N+ < n + N+, so n B = mx{,..., N, N+ + } for ll n N. Step : { n } n= hs convergent subsequence { nk } k=. This follows from Step nd the Bolzno Weierstrss Theorem. Step 3: { n } converges to the limit L. = lim k nk. We re given tht { n } is Cuchy sequence. Given ǫ >, N m, n > N n m < ǫ. Also, from Step we know tht there exists number K, which we my tke to be t lest s lrge s N, such tht k > K nk L < ǫ. We then tke m = n k nd use the fct tht subsequence indices stisfy n k k: Thus n k n > N, k > K N n nk < ǫ. n > N n L n nk + nk L < ǫ + ǫ = ǫ. Hence { n } lso converges (to the sme limit L). {n} is not Cuchy sequence. To see this, just pick m = n + nd ǫ =. From Theorem.6, we then know tht {n} diverges. } { + ( )n is Cuchy sequence. n Given ǫ >, we cn mke ) ) ( + ( )n ( + ( )m = ( ) n n m ( )m n m n + m < N + N = N = ǫ, provided we mke the nturl numbers n nd m both greter thn N = /ǫ. Hence this is Cuchy sequence nd it therefore converges.
.E. CAUCHY CRITERION 47 Wht bout the sequence defined by n = + + 3 +... + n? Pick m = n: n n = n + + n + +... + n n + n +... + n } {{ } n terms = n n =. Then, if we should be given ǫ =, we won t be ble to stisfy m n < ǫ. So { n } is not Cuchy sequence. Hence, by Theorem.6 we see tht { n } diverges, i.e. k= k. = lim n n k= k /. This is known s the hrmonic series. It diverges, but only very slowly. Q. Using the bove estimte, how mny terms would you wnt to tke to be sure tht the sum is greter thn? Problem.8 (Infinite limits): Let { n } n= be sequence of rel numbers. If, given ny nturl number M, we cn find number N such tht n > N n > M, we sy tht lim n =. n () Suppose { n } n= nd {b n} n= re sequences of rel numbers with lim n = n nd lim b n =. Show tht lim ( n + b n ) = nd lim n b n =. n n n (b) Under the conditions of prt (), find exmples tht demonstrte lim ( n b n ) n nd lim n /b n my exist s rel number, my hve n infinite limit, or my fil n to exist t ll. (c) Show tht lim n = lim =. n n n (d) Does the converse to (c) hold? Tht is does, lim n n = lim n n =? Problem.9 (Limit Superior nd Limit Inferior): Let { n } n= be bounded sequence. Consider the sequence {s n } n= defined by s n = sup{ n, n+, n+,...} for n N.
48 CHAPTER. SEQUENCES () Prove tht {s n } n= is bounded sequence. (b) Prove tht {s n } n= is monotone sequence. Is {s n} n= n incresing or decresing sequence? (c) Prove tht {s n } n= is convergent. Note: The limit of the sequence {s n } n= is known s the limit superior of the sequence { n } n= nd is written lim sup n. This is just the supremum of the vlues in the til of the sequence. In similr mnner, one cn define the limit n inferior: lim inf n = lim i n, where i n = inf{ n, n+, n+,...}. A bounded n n sequence { n } n= converges lim inf n = lim sup n. For exmple, the n n bounded sequence {sin n} n= does not converge becuse lim inf n lim sup sin n =. n sin n = nd
Chpter 3 Functions [Spivk 994, Chpter 3] 3.A Exmples of Functions In the previous chpter, we defined function f s rule tht ssocites number y to ech number x in its domin. An equivlent definition is [Spivk 994, p. 47]: Definition: A function is collection of pirs of numbers (x, y) such tht if (x, y ) nd (x, y ) re in the collection, then y = y. Tht is, x = x f(x ) = f(x ). This cn be restted s the verticl line test: n set of ordered pirs (x, y) is function if every verticl line intersects their grph t most once. Definition: If function f hs domin A nd rnge B, we write f : A B. Definition: Constnt functions re functions of the form f(x) = c, where c is constnt. Definition: Polynomils re functions of the form f(x) = n x n + n x n +... + x + x +. When n, we sy tht the degree of f is n nd write deg f = n. While nonzero constnt function hs degree, it turns out to be convenient to define the degree of the zero function f(x) = to be. Note tht polynomil f(x) with only even-degree terms (ll the odd-degree coefficients re zero) stisfies the property f( x) = f(x), while polynomil f(x) with only odd-degree terms stisfies f( x) = f(x). We generlize this notion with the following definition. 49
5 CHAPTER 3. FUNCTIONS Definition: A function f is sid to be even if f( x) = f(x) for every x in the domin of f. Definition: A function f is sid to be odd if f( x) = f(x) for every x in the domin of f. Problem 3.: Show tht every function f : R R cn be written s sum of n even function nd n odd function. Definition: Rtionl functions re functions of the form f(x) = P(x), where P(x) Q(x) nd Q(x) re polynomils. They re defined on the set of ll x for which Q(x). x3 + 3x + x + nd x re both rtionl functions. Composition Once we hve defined few elementry functions, we cn crete new functions by combining them together using +,,,, or by introducing the composition opertor. Definition: If f : A B nd g : B C then we define g f : A C to be the function tht tkes x A to g(f(x)) C. f(x) = x + g(x) = x g(f(x)) = x + f : R [, ), g : [, ) [, ), g f : R [, ). Note however tht f(g(x)) = 4x +, so tht f g : [, ) [, ). f(x) = x + f : R [, ), g(x) = g : [, ) (, ], x g(f(x)) = g f : R (, ]. x + One cn lso build new functions from old ones using cses, or piecewise definitions. x <, f(x) = x =, x >. f(x) = x = { x x, x x <. Notice tht cses cn sometimes introduce jumps in function.
3.B. TRIGONOMETRIC FUNCTIONS 5 3.B Trigonometric Functions Trigonometric functions re functions relting the shpe of right-ngle tringle to one of its other ngles. Definition: If we lbel one of the non-right ngles by θ, the length of the hypotenuse by hyp, nd the lengths of the sides opposite nd djcent to x by opp nd dj, respectively, then sin θ = opp hyp, cos θ = dj hyp, tn θ = opp dj. Note here tht since x is one of the nonright ngles of right-ngle tringle, these definitions pply only when < x < 9. Note lso tht tnθ = sin θ/ cosθ. Sometimes it is convenient to work with the reciprocls of these functions: csc θ = sin θ, sec θ = cosθ, cot θ = tn θ. Pythgoren Identities: Recll from Pythgors Theorem tht (opp) + (dj) = (hyp). Dividing by the squre of the length of the hypotenuse, we see tht sin θ + cos θ =. (3.) Other useful identities result from dividing both sides of this eqution either by sin θ: + cot θ = csc θ, or by cos θ: tn θ + = sec θ. Note tht Eq. (3.) implies both tht sin θ nd cosθ. Definition: We define the number π to be the re of unit circle ( circle with rdius ).