One Variable Advanced Calculus. Kenneth Kuttler

Transcription

1 One Vrible Advnced Clculus Kenneth Kuttler August 5, 2009

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3 Contents 1 Introduction 7 2 The Rel And Complex Numbers The Number Line And Algebr Of The Rel Numbers Exercises Set Nottion Order Exercises The Binomil Theorem Well Ordering Principle And Archimedin Property Exercises Completeness of R Exercises The Complex Numbers Exercises Set Theory Bsic Definitions The Schroder Bernstein Theorem Equivlence Reltions Exercises Functions And Sequences Generl Considertions Sequences Exercises Worked Exercises The Limit Of A Sequence The Nested Intervl Lemm Exercises Worked Exercises Sequentil Compctness Sequentil Compctness Closed And Open Sets Exercises Worked Exercises Cuchy Sequences And Completeness Decimls lim sup nd lim inf Shrinking Dimeters Exercises

4 4 CONTENTS 4.14 Worked Exercises Infinite Series Of Numbers Bsic Considertions Exercises Worked Exercises More Tests For Convergence Convergence Becuse Of Cncelltion Rtio And Root Tests Double Series Exercises Worked Exercises Continuous Functions Continuity And The Limit Of A Sequence Exercises Worked Exercises The Extreme Vlues Theorem The Intermedite Vlue Theorem Exercises Worked Exercises Uniform Continuity Exercises Worked Exercises Sequences And Series Of Functions Sequences Of Polynomils, Weierstrss Approximtion Exercises Worked Exercises The Derivtive Limit Of A Function Exercises Worked Exercises The Definition Of The Derivtive Continuous And Nowhere Differentible Finding The Derivtive Men Vlue Theorem And Locl Extreme Points Exercises Worked Exercises Men Vlue Theorem Exercises Worked Exercises Derivtives Of Inverse Functions Derivtives And Limits Of Sequences Exercises Worked Exercises Power Series Functions Defined In Terms Of Series Opertions On Power Series The Specil Functions Of Elementry Clculus The Functions, sin, cos, exp ln And log b

5 CONTENTS The Binomil Theorem Exercises Worked Exercises L Hôpitl s Rule Interest Compounded Continuously Exercises Multipliction Of Power Series Exercises Worked Exercises Worked Exercises The Fundmentl Theorem Of Algebr Some Other Theorems The Riemnn And Riemnn Stieltjes Integrls The Drboux Stieltjes Integrl Upper And Lower Drboux Stieltjes Sums Exercises Functions Of Drboux Integrble Functions Properties Of The Integrl Fundmentl Theorem Of Clculus Exercises The Riemnn Stieltjes Integrl Chnge Of Vribles A Simple Procedure For Finding Integrls Generl Riemnn Stieltjes Integrls Exercises Fourier Series The Complex Exponentil Definition And Bsic Properties The Riemnn Lebesgue Lemm Dini s Criterion For Convergence Integrting And Differentiting Fourier Series Wys Of Approximting Functions Uniform Approximtion With Trig. Polynomils Men Squre Approximtion Exercises The Generlized Riemnn Integrl Definitions And Bsic Properties Integrls Of Derivtives Exercises Copyright c 2007,

6 6 CONTENTS

7 Introduction The difference between dvnced clculus nd clculus is tht ll the theorems re proved completely nd the role of plne geometry is minimized. Insted, the notion of completeness is of preeminent importnce. Silly gimmicks re of no significnce t ll. Routine skills involving elementry functions nd integrtion techniques re supposed to be mstered nd hve no plce in dvnced clculus which dels with the fundmentl issues relted to existence nd mening. This is subject which plces clculus s prt of mthemtics nd involves proofs nd definitions, not lgorithms nd busy work. An orderly development of the elementry functions is included but it is ssumed the reder is fmilir enough with these functions to use them in problems which illustrte some of the ides presented. 7

8 8 INTRODUCTION

9 The Rel And Complex Numbers 2.1 The Number Line And Algebr Of The Rel Numbers To begin with, consider the rel numbers, denoted by R, s line extending infinitely fr in both directions. In this book, the nottion, indictes something is being defined. Thus the integers re defined s the nturl numbers, Z { 1, 0, 1, }, N {1, 2, } nd the rtionl numbers, defined s the numbers which re the quotient of two integers. { m } Q n such tht m, n Z, n 0 re ech subsets of R s indicted in the following picture / As shown in the picture, 1 2 is hlf wy between the number 0 nd the number, 1. By nlogy, you cn see where to plce ll the other rtionl numbers. It is ssumed tht R hs the following lgebr properties, listed here s collection of ssertions clled xioms. These properties will not be proved which is why they re clled xioms rther thn theorems. In generl, xioms re sttements which re regrded s true. Often these re things which re self evident either from experience or from some sort of intuition but this does not hve to be the cse. Axiom x + y = y + x, (commuttive lw for ddition) Axiom x + 0 = x, (dditive identity). Axiom For ech x R, there exists x R such tht x + ( x) = 0, (existence of dditive inverse). 9

10 10 THE REAL AND COMPLEX NUMBERS Axiom (x + y) + z = x + (y + z), (ssocitive lw for ddition). Axiom xy = yx, (commuttive lw for multipliction). Axiom (xy) z = x (yz), (ssocitive lw for multipliction). Axiom x = x, (multiplictive identity). Axiom For ech x 0, there exists x 1 such tht xx 1 = 1.(existence of multiplictive inverse). Axiom x (y + z) = xy + xz.(distributive lw). These xioms re known s the field xioms nd ny set (there re mny others besides R) which hs two such opertions stisfying the bove xioms is clled field. Division nd subtrction re defined in the usul wy by x y x + ( y) nd x/y x ( y 1). It is ssumed tht the reder is completely fmilir with these xioms in the sense tht he or she cn do the usul lgebric mnipultions tught in high school nd junior high lgebr courses. The xioms listed bove re just creful sttement of exctly wht is necessry to mke the usul lgebric mnipultions vlid. A word of dvice regrding division nd subtrction is in order here. Whenever you feel little confused bout n lgebric expression which involves division or subtrction, think of division s multipliction by the multiplictive inverse s just indicted nd think of subtrction s ddition of the dditive inverse. Thus, when you see x/y, think x ( y 1) nd when you see x y, think x + ( y). In mny cses the source of confusion will dispper lmost mgiclly. The reson for this is tht subtrction nd division do not stisfy the ssocitive lw. This mens there is nturl mbiguity in n expression like Do you men (6 3) 4 = 1 or 6 (3 4) = 6 ( 1) = 7? It mkes difference doesn t it? However, the so clled binry opertions of ddition nd multipliction re ssocitive nd so no such confusion will occur. It is conventionl to simply do the opertions in order of ppernce reding from left to right. Thus, if you see 6 3 4, you would normlly interpret it s the first of the bove lterntives. In doing lgebr, the following theorem is importnt nd follows from the bove xioms. The resoning which demonstrtes this ssertion is clled proof. Proofs nd definitions re very importnt in mthemtics becuse they re the mens by which truth is determined. In mthemtics, something is true if it follows from xioms using correct logicl rgument. Truth is not determined on the bsis of experiment or opinions nd it is this which mkes mthemtics useful s lnguge for describing certin kinds of relity in precise mnner. 1 It is lso the definitions nd proofs which mke the subject of mthemtics intellectully worth while. Tke these wy nd it becomes gry wstelnd filled with endless tedium nd meningless mnipultions. In the first prt of the following theorem, the clim is mde tht the dditive inverse nd the multiplictive inverse re unique. This mens tht for given number, only one number hs the property tht it is n dditive inverse nd tht, given nonzero number, only one number hs the property tht it is multiplictive inverse. The significnce of this is tht if you re wondering if given number is the dditive inverse of given number, ll you hve to do is to check nd see if it cts like one. Theorem The bove xioms imply the following. 1. The multiplictive inverse nd dditive inverses re unique. 1 There re certinly rel nd importnt things which should not be described using mthemtics becuse it hs nothing to do with these things. For exmple, feelings nd emotions hve nothing to do with mth.

11 2.1. THE NUMBER LINE AND ALGEBRA OF THE REAL NUMBERS x = 0, ( x) = x, 3. ( 1) ( 1) = 1, ( 1) x = x 4. If xy = 0 then either x = 0 or y = 0. Proof:Suppose then tht x is rel number nd tht x + y = 0 = x + z. It is necessry to verify y = z. From the bove xioms, there exists n dditive inverse, x for x. Therefore, x + 0 = ( x) + (x + y) = ( x) + (x + z) nd so by the ssocitive lw for ddition, (( x) + x) + y = (( x) + x) + z which implies 0 + y = 0 + z. Now by the definition of the dditive identity, this implies y = z. You should prove the multiplictive inverse is unique. Consider 2. It is desired to verify 0x = 0. From the definition of the dditive identity nd the distributive lw it follows tht 0x = (0 + 0) x = 0x + 0x. From the existence of the dditive inverse nd the ssocitive lw it follows 0 = ( 0x) + 0x = ( 0x) + (0x + 0x) = (( 0x) + 0x) + 0x = 0 + 0x = 0x To verify the second clim in 2., it suffices to show x cts like the dditive inverse of x in order to conclude tht ( x) = x. This is becuse it hs just been shown tht dditive inverses re unique. By the definition of dditive inverse, x + ( x) = 0 nd so x = ( x) s climed. To demonstrte 3., ( 1) (1 + ( 1)) = ( 1) 0 = 0 nd so using the definition of the multiplictive identity, nd the distributive lw, ( 1) + ( 1) ( 1) = 0. It follows from 1. nd 2. tht 1 = ( 1) = ( 1) ( 1). To verify ( 1) x = x, use 2. nd the distributive lw to write x + ( 1) x = x (1 + ( 1)) = x0 = 0. Therefore, by the uniqueness of the dditive inverse proved in 1., it follows ( 1) x = x s climed. To verify 4., suppose x 0. Then x 1 exists by the xiom bout the existence of multiplictive inverses. Therefore, by 2. nd the ssocitive lw for multipliction, y = ( x 1 x ) y = x 1 (xy) = x 1 0 = 0. This proves 4. nd completes the proof of this theorem.

12 12 THE REAL AND COMPLEX NUMBERS Recll the notion of something rised to n integer power. Thus y 2 = y y nd b 3 = 1 b etc. 3 Also, there re few conventions relted to the order in which opertions re performed. Exponents re lwys done before multipliction. Thus xy 2 = x ( y 2) nd is not equl to (xy) 2. Division or multipliction is lwys done before ddition or subtrction. Thus x y (z + w) = x [y (z + w)] nd is not equl to (x y) (z + w). Prentheses re done before nything else. Be very creful of such things since they re source of mistkes. When you hve doubts, insert prentheses to resolve the mbiguities. Also recll summtion nottion. If you hve not seen this, the following is short review of this topic. Definition Let x 1, x 2,, x m be numbers. Then m x j x 1 + x x m. j=1 Thus this symbol, m j=1 x j mens to tke ll the numbers, x 1, x 2,, x m nd dd them ll up. Note the use of the j s generic vrible which tkes vlues from 1 up to m. This nottion will be used whenever there re things which cn be dded, not just numbers. As n exmple of the use of this nottion, you should verify the following. Exmple k=1 (2k + 1) = 48. Be sure you understnd why m+1 k=1 x k = As slight generliztion of this nottion, m x k + x m+1. k=1 m x j x k + + x m. j=k It is lso possible to chnge the vrible of summtion. m x j = x 1 + x x m j=1 while if r is n integer, the nottion requires m+r j=1+r x j r = x 1 + x x m nd so m j=1 x j = m+r j=1+r x j r. Summtion nottion will be used throughout the book whenever it is convenient to do so. Another thing to keep in mind is tht you often use letters to represent numbers. Since they represent numbers, you mnipulte expressions involving letters in the sme mnner s you would if they were specific numbers. Exmple Add the frctions, x x 2 +y + y x 1.

13 2.2. EXERCISES 13 You dd these just like they were numbers. Write the first expression s nd the second s you dd them s follows. x(x 1) (x 2 +y)(x 1) y(x 2 +y) (x 1)(x 2 +y). Then since these hve the sme common denomintor, x x 2 + y + y x Exercises = x (x 1) (x 2 + y) (x 1) + y ( x 2 + y ) (x 1) (x 2 + y) = x2 x + yx 2 + y 2 (x 2 + y) (x 1). 1. Consider the expression x + y (x + y) x (y x) f (x, y). Find f ( 1, 2). 2. Show (b) = ( ) b. 3. Show on the number line the effect of dding two positive numbers, x nd y. 4. Show on the number line the effect of subtrcting positive number from nother positive number. 5. Show on the number line the effect of multiplying number by Add the frctions x x x 1 x Find formul for (x + y) 2, (x + y) 3, nd (x + y) 4. Bsed on wht you observe for these, give formul for (x + y) When is it true tht (x + y) n = x n + y n? 9. Find the error in the following rgument. Let x = y = 1. Then xy = y 2 nd so xy x 2 = y 2 x 2. Therefore, x (y x) = (y x) (y + x). Dividing both sides by (y x) yields x = x + y. Now substituting in wht these vribles equl yields 1 = Find the error in the following rgument. x = x + 1 nd so letting x = 2, 5 = 3. Therefore, 5 = Find the error in the following. Let x = 1 nd y = 2. Then 1 3 = 1 x+y = 1 x + 1 y = = 3 2. Then cross multiplying, yields 2 = Simplify x2 y 4 z 6 x 2 y 1 z. 13. Simplify the following expressions using correct lgebr. In these expressions the vribles represent rel numbers. () x2 y+xy 2 +x x (b) x2 y+xy 2 +x xy (c) x3 +2x 2 x 2 x Find the error in the following rgument. Let x = 3 nd y = 1. Then 1 = 3 2 = 3 (3 1) = x y (x y) = (x y) (x y) = 2 2 = Verify the following formuls. () (x y) (x + y) = x 2 y 2

14 14 THE REAL AND COMPLEX NUMBERS (b) (x y) ( x 2 + xy + y 2) = x 3 y 3 (c) (x + y) ( x 2 xy + y 2) = x 3 + y Find the error in the following. xy + y x = y + y = 2y. Now let x = 2 nd y = 2 to obtin 3 = Show the rtionl numbers stisfy the field xioms. You my ssume the ssocitive, commuttive, nd distributive lws hold for the integers. 2.3 Set Nottion A set is just collection of things clled elements. Often these re lso referred to s points in clculus. For exmple {1, 2, 3, 8} would be set consisting of the elements 1,2,3, nd 8. To indicte tht 3 is n element of {1, 2, 3, 8}, it is customry to write 3 {1, 2, 3, 8}. 9 / {1, 2, 3, 8} mens 9 is not n element of {1, 2, 3, 8}. Sometimes rule specifies set. For exmple you could specify set s ll integers lrger thn 2. This would be written s S = {x Z : x > 2}. This nottion sys: the set of ll integers, x, such tht x > 2. If A nd B re sets with the property tht every element of A is n element of B, then A is subset of B. For exmple, {1, 2, 3, 8} is subset of {1, 2, 3, 4, 5, 8}, in symbols, {1, 2, 3, 8} {1, 2, 3, 4, 5, 8}. The sme sttement bout the two sets my lso be written s {1, 2, 3, 4, 5, 8} {1, 2, 3, 8}. The union of two sets is the set consisting of everything which is contined in t lest one of the sets, A or B. As n exmple of the union of two sets, {1, 2, 3, 8} {3, 4, 7, 8} = {1, 2, 3, 4, 7, 8} becuse these numbers re those which re in t lest one of the two sets. In generl A B {x : x A or x B}. Be sure you understnd tht something which is in both A nd B is in the union. It is not n exclusive or. The intersection of two sets, A nd B consists of everything which is in both of the sets. Thus {1, 2, 3, 8} {3, 4, 7, 8} = {3, 8} becuse 3 nd 8 re those elements the two sets hve in common. In generl, A B {x : x A nd x B}. When with rel numbers, [, b] denotes the set of rel numbers, x, such tht x b nd [, b) denotes the set of rel numbers such tht x < b. (, b) consists of the set of rel numbers, x such tht < x < b nd (, b] indictes the set of numbers, x such tht < x b. [, ) mens the set of ll numbers, x such tht x nd (, ] mens the set of ll rel numbers which re less thn or equl to. These sorts of sets of rel numbers re clled intervls. The two points, nd b re clled endpoints of the intervl. Other intervls such s (, b) re defined by nlogy to wht ws just explined. In generl, the curved prenthesis indictes the end point it sits next to is not included while the squre prenthesis indictes this end point is included. The reson tht there will lwys be curved prenthesis next to or is tht these re not rel numbers. Therefore, they cnnot be included in ny set of rel numbers.

15 2.4. ORDER 15 A specil set which needs to be given nme is the empty set lso clled the null set, denoted by. Thus is defined s the set which hs no elements in it. Mthemticins like to sy the empty set is subset of every set. The reson they sy this is tht if it were not so, there would hve to exist set, A, such tht hs something in it which is not in A. However, hs nothing in it nd so the lest intellectul discomfort is chieved by sying A. If A nd B re two sets, A \ B denotes the set of things which re in A but not in B. Thus A \ B {x A : x / B}. Set nottion is used whenever convenient. 2.4 Order The rel numbers lso hve n order defined on them. This order my be defined by reference to the positive rel numbers, those to the right of 0 on the number line, denoted by R + which is ssumed to stisfy the following xioms. The sum of two positive rel numbers is positive. (2.1) The product of two positive rel numbers is positive. (2.2) For given rel number, x, one nd only one of the following lterntives holds. Either x is positive, x equls 0 or x is positive. (2.3) Definition x < y exctly when y + ( x) y x R +. In the usul wy, x < y is the sme s y > x nd x y mens either x < y or x = y. The symbol is defined similrly. Theorem The following hold for the order defined s bove. 1. If x < y nd y < z then x < z (Trnsitive lw). 2. If x < y then x + z < y + z (ddition to n inequlity). 3. If x 0 nd y 0, then xy If x > 0 then x 1 > If x < 0 then x 1 < If x < y then xz < yz if z > 0, (multipliction of n inequlity). 7. If x < y nd z < 0, then xz > zy (multipliction of n inequlity). 8. Ech of the bove holds with > nd < replced by nd respectively except for 4 nd 5 in which we must lso stipulte tht x For ny x nd y, exctly one of the following must hold. Either x = y, x < y, or x > y (trichotomy).

16 16 THE REAL AND COMPLEX NUMBERS Proof: First consider 1, the trnsitive lw. Suppose x < y nd y < z. Why is x < z? In other words, why is z x R +? It is becuse z x = (z y) + (y x) nd both z y, y x R +. Thus by 2.1 bove, z x R + nd so z > x. Next consider 2, ddition to n inequlity. If x < y why is x + z < y + z? it is becuse (y + z) + (x + z) = (y + z) + ( 1) (x + z) = y + ( 1) x + z + ( 1) z = y x R +. Next consider 3. If x 0 nd y 0, why is xy 0? First note there is nothing to show if either x or y equl 0 so ssume this is not the cse. By 2.3 x > 0 nd y > 0. Therefore, by 2.2 nd wht ws proved bout x = ( 1) x, ( x) ( y) = ( 1) 2 xy R +. Is ( 1) 2 = 1? If so the clim is proved. But ( 1) = ( 1) 2 nd ( 1) = 1 becuse = 0. Next consider 4. If x > 0 why is x 1 > 0? By 2.3 either x 1 = 0 or x 1 R +. It cn t hppen tht x 1 = 0 becuse then you would hve to hve 1 = 0x nd s ws shown erlier, 0x = 0. Therefore, consider the possibility tht x 1 R +. This cn t work either becuse then you would hve ( 1) x 1 x = ( 1) (1) = 1 nd it would follow from 2.2 tht 1 R +. But this is impossible becuse if x R +, then ( 1) x = x R + nd contrdicts 2.3 which sttes tht either x or x is in R + but not both. Next consider 5. If x < 0, why is x 1 < 0? As before, x 1 0. If x 1 > 0, then s before, x ( x 1) = 1 R + which ws just shown not to occur. Next consider 6. If x < y why is xz < yz if z > 0? This follows becuse yz xz = z (y x) R + since both z nd y x R +. Next consider 7. If x < y nd z < 0, why is xz > zy? This follows becuse zx zy = z (x y) R + by wht ws proved in 3. The lst two clims re obvious nd left for you. This proves the theorem. Note tht trichotomy could be stted by sying x y or y x. { Definition x if x 0, x x if x < 0. Note tht x cn be thought of s the distnce between x nd 0. Theorem xy = x y. Proof: You cn verify this by checking ll vilble cses. Do so.

17 2.4. ORDER 17 Theorem The following inequlities hold. x + y x + y, x y x y. Either of these inequlities my be clled the tringle inequlity. Proof: First note tht if, b R + {0} then b if nd only if 2 b 2. Here is why. Suppose b. Then by the properties of order proved bove, 2 b b 2 becuse b 2 b = b (b ) R + {0}. Next suppose 2 b 2. If both, b = 0 there is nothing to show. Assume then they re not both 0. Then b 2 2 = (b + ) (b ) R +. By the bove theorem on order, ( + b) 1 R + nd so using the ssocitive lw, Now ( + b) 1 ((b + ) (b )) = (b ) R + x + y 2 = (x + y) 2 = x 2 + 2xy + y 2 x 2 + y x y = ( x + y ) 2 nd so the first of the inequlities follows. Note I used xy xy = x y which follows from the definition. To verify the other form of the tringle inequlity, so nd so x = x y + y x x y + y x y x y = y x Now repet the rgument replcing the roles of x nd y to conclude Therefore, This proves the tringle inequlity. y x y x. y x y x. Exmple Solve the inequlity 2x + 4 x 8 Subtrct 2x from both sides to yield 4 x 8. Next dd 8 to both sides to get 12 x. Then multiply both sides by ( 1) to obtin x 12. Alterntively, subtrct x from both sides to get x+4 8. Then subtrct 4 from both sides to obtin x 12. Exmple Solve the inequlity (x + 1) (2x 3) 0. If this is to hold, either both of the fctors, x + 1 nd 2x 3 re nonnegtive or they re both non-positive. The first cse yields x nd 2x 3 0 so x 1 nd x 3 2 yielding x 3 2. The second cse yields x nd 2x 3 0 which implies x 1 nd x 3 2. Therefore, the solution to this inequlity is x 1 or x 3 2.

18 18 THE REAL AND COMPLEX NUMBERS Exmple Solve the inequlity (x) (x + 2) 4 Here the problem is to find x such tht x 2 + 2x However, x 2 + 2x + 4 = (x + 1) for ll x. Therefore, the solution to this problem is ll x R. Exmple Solve the inequlity 2x + 4 x 8 This is written s (, 12]. Exmple Solve the inequlity (x + 1) (2x 3) 0. This ws worked erlier nd x 1 or x 3 2 nottion this is denoted by (, 1] [ 3 2, ). Exmple Solve the eqution x 1 = 2 ws the nswer. In terms of set This will be true when x 1 = 2 or when x 1 = 2. Therefore, there re two solutions to this problem, x = 3 or x = 1. Exmple Solve the inequlity 2x 1 < 2 From the number line, it is necessry to hve 2x 1 between 2 nd 2 becuse the inequlity sys tht the distnce from 2x 1 to 0 is less thn 2. Therefore, 2 < 2x 1 < 2 nd so 1/2 < x < 3/2. In other words, 1/2 < x nd x < 3/2. Exmple Solve the inequlity 2x 1 > 2. This hppens if 2x 1 > 2 or if 2x 1 < 2. Thus the solution is x > 3/2 or x < 1/2. Written in terms of intervls this is ( 3 2, ) (, 1 2). Exmple Solve x + 1 = 2x 2 There re two wys this cn hppen. It could be the cse tht x + 1 = 2x 2 in which cse x = 3 or lterntively, x + 1 = 2 2x in which cse x = 1/3. Exmple Solve x + 1 2x 2 In order to keep trck of wht is hppening, it is very good ide to grph the two reltions, y = x + 1 nd y = 2x 2 on the sme set of coordinte xes. This is not hrd job. x + 1 = x + 1 when x > 1 nd x + 1 = 1 x when x 1. Therefore, it is not hrd to drw its grph. Similr considertions pply to the other reltion. The result is y = x + 1 1/3 3 Equlity holds exctly when x = 3 or x = 1 3 s in the preceding exmple. Consider x between 1 3 nd 3. You cn see these vlues of x do not solve the inequlity. For exmple x = 1 does not work. Therefore, ( 1 3, 3) must be excluded. The vlues of x lrger thn 3 do not produce equlity so either x + 1 < 2x 2 for these points or 2x 2 < x + 1 for these points. Checking exmples, you see the first of the two cses is the one which holds. Therefore, [3, ) is included. Similr resoning obtins (, 1 3 ]. It follows the solution set to this inequlity is (, 1 3 ] [3, ).

19 2.5. EXERCISES 19 Exmple Suppose ε > 0 is given positive number. Obtin number, δ > 0, such tht if x 1 < δ, then x 2 1 < ε. First of ll, note x 2 1 = x 1 x + 1 ( x + 1) x 1. Now if x 1 < 1, it follows x < 2 nd so for x 1 < 1, x 2 1 < 3 x 1. Now let δ = min ( 1, ε 3). This nottion mens to tke the minimum of the two numbers, 1 nd ε 3. Then if x 1 < δ, x 2 1 < 3 x 1 < 3 ε 3 = ε. 2.5 Exercises 1. Solve (3x + 2) (x 3) Solve (3x + 2) (x 3) > Solve x+2 3x 2 < Solve x+1 x+3 < Solve (x 1) (2x + 1) Solve (x 1) (2x + 1) > Solve x 2 2x Solve (x + 2) (x 2) Solve 3x 4 x 2 +2x Solve 3x+9 x 2 +2x Solve x2 +2x+1 3x+7 < Solve x + 1 = 2x Solve 3x + 1 < 8. Give your nswer in terms of intervls on the rel line. 14. Sketch on the number line the solution to the inequlity x 3 > Sketch on the number line the solution to the inequlity x 3 < Show x = x Solve x + 2 < 3x Tell when equlity holds in the tringle inequlity. 19. Solve x x Solve (x + 1) (2x 2) x Solve x+3 2x+1 > Solve x+2 3x+1 > 2.

20 20 THE REAL AND COMPLEX NUMBERS 23. Describe the set of numbers, such tht there is no solution to x + 1 = 4 x Suppose 0 < < b. Show 1 > b Show tht if x 6 < 1, then x < Suppose x 8 < 2. How lrge cn x 5 be? 27. Obtin number, δ > 0, such tht if x 1 < δ, then x 2 1 < 1/ Obtin number, δ > 0, such tht if x 4 < δ, then x 2 < 1/ Suppose ε > 0 is given positive number. Obtin number, δ > 0, such tht if x 1 < δ, then x 1 < ε. Hint: This δ will depend in some wy on ε. You need to tell how. 2.6 The Binomil Theorem Consider the following problem: You hve the integers S n = {1, 2,, n} nd k is n integer no lrger thn n. How mny wys re there to fill k slots with these integers strting from left to right if whenever n integer from S n hs been used, it cnnot be re used in ny succeeding slot? k of these slots {}}{,,,,, This number is known s permuttions of n things tken k t time nd is denoted by P (n, k). It is esy to figure it out. There re n choices for the first slot. For ech choice for the fist slot, there remin n 1 choices for the second slot. Thus there re n (n 1) wys to fill the first two slots. Now there remin n 2 wys to fill the third. Thus there re n (n 1) (n 2) wys to fill the first three slots. Continuing this wy, you see there re P (n, k) = n (n 1) (n 2) (n k + 1) wys to do this. Now define for k positive integer, k! k (k 1) (k 2) 1, 0! 1. This is clled k fctoril. Thus P (k, k) = k! nd you should verify tht P (n, k) = n! (n k)! Now consider the number of wys of selecting set of k different numbers from S n. For ech set of k numbers there ( ) re P (k, k) = k! wys of listing these numbers in order. n Therefore, denoting by the number of wys of selecting set of k numbers from k S n, it must be the cse tht ( n k ) n! k! = P (n, k) = (n k)! Therefore, ( n k ) = n! k! (n k)!.

21 2.7. WELL ORDERING PRINCIPLE AND ARCHIMEDIAN PROPERTY 21 How mny wys re there to select no numbers from S n? Obviously one wy. Note the bove formul gives the right nswer in this cse s well s in ll other cses due to the definition which sys 0! = 1. Now consider the problem of writing formul for (x + y) n where n is positive integer. Imgine writing it like this: n times {}}{ (x + y) (x + y) (x + y) Then you know the result will be sums of terms of the form k x k y n k. Wht is k? In other words, how mny wys cn you pick x from k of the fctors bove nd y from the other n k. There re n fctors so the number of wys to do it is ( ) n. k Therefore, k is the bove formul nd so this proves the following importnt theorem known s the binomil theorem. Theorem The following formul holds for ny n positive integer. (x + y) n = k=0 ( n k ) x k y n k. 2.7 Well Ordering Principle And Archimedin Property Definition A set is well ordered if every nonempty subset S, contins smllest element z hving the property tht z x for ll x S. Axiom Any set of integers lrger thn given number is well ordered. In prticulr, the nturl numbers defined s N {1, 2, } is well ordered. The bove xiom implies the principle of mthemticl induction. Theorem (Mthemticl induction) A set S Z, hving the property tht S nd n + 1 S whenever n S contins ll integers x Z such tht x. Proof: Let T ([, ) Z) \ S. Thus T consists of ll integers lrger thn or equl to which re not in S. The theorem will be proved if T =. If T then by the well ordering principle, there would hve to exist smllest element of T, denoted s b. It must be the cse tht b > since by definition, / T. Then the integer, b 1 nd b 1 / S becuse if b 1 S, then b = b S by the ssumed property of S. Therefore, b 1 ([, ) Z) \ S = T which contrdicts the choice of b s the smllest element of T. (b 1 is smller.) Since contrdiction is obtined by ssuming T, it must be the cse tht T = nd this sys tht everything in [, ) Z is lso in S. Mthemticl induction is very useful device for proving theorems bout the integers. Exmple Prove by induction tht n k=1 k2 = n(n+1)(2n+1) 6.

22 22 THE REAL AND COMPLEX NUMBERS By inspection, if n = 1 then the formul is true. The sum yields 1 nd so does the formul on the right. Suppose this formul is vlid for some n 1 where n is n integer. Then n+1 k 2 = k 2 + (n + 1) 2 k=1 = k=1 n (n + 1) (2n + 1) 6 + (n + 1) 2. The step going from the first to the second line is bsed on the ssumption tht the formul is true for n. This is clled the induction hypothesis. Now simplify the expression in the second line, n (n + 1) (2n + 1) + (n + 1) 2. 6 This equls nd Therefore, n (2n + 1) 6 n+1 k 2 = k=1 ( ) n (2n + 1) (n + 1) + (n + 1) 6 = + (n + 1) = 6 (n + 1) + 2n2 + n 6 (n + 2) (2n + 3) = 6 (n + 1) (n + 2) (2n + 3) 6 (n + 1) ((n + 1) + 1) (2 (n + 1) + 1), 6 showing the formul holds for n + 1 whenever it holds for n. This proves the formul by mthemticl induction. Exmple Show tht for ll n N, n 1 2n < 1 2n+1. If n = 1 this reduces to the sttement tht 1 2 < 1 3 which is obviously true. Suppose then tht the inequlity holds for n. Then n 1 2n + 1 2n 2n + 2 < = 1 2n + 1 2n + 1 2n + 2 2n + 1 2n The theorem will be proved if this lst expression is less thn 2n+3. This hppens if nd only if ( ) = 2n + 3 2n + 3 > 2n + 1 (2n + 2) 2 which occurs if nd only if (2n + 2) 2 > (2n + 3) (2n + 1) nd this is clerly true which my be seen from expnding both sides. This proves the inequlity. Lets review the process just used. If S is the set of integers t lest s lrge s 1 for which the formul holds, the first step ws to show 1 S nd then tht whenever n S,

23 2.7. WELL ORDERING PRINCIPLE AND ARCHIMEDIAN PROPERTY 23 it follows n + 1 S. Therefore, by the principle of mthemticl induction, S contins [1, ) Z, ll positive integers. In doing n inductive proof of this sort, the set, S is normlly not mentioned. One just verifies the steps bove. First show the thing is true for some Z nd then verify tht whenever it is true for m it follows it is lso true for m + 1. When this hs been done, the theorem hs been proved for ll m. Definition The Archimedin property sttes tht whenever x R, nd > 0, there exists n N such tht n > x. Axiom R hs the Archimedin property. This is not hrd to believe. Just look t the number line. This Archimedin property is quite importnt becuse it shows every rel number is smller thn some integer. It lso cn be used to verify very importnt property of the rtionl numbers. Theorem Suppose x < y nd y x > 1. Then there exists n integer, l Z, such tht x < l < y. If x is n integer, there is no integer y stisfying x < y < x + 1. Proof: Let x be the smllest positive integer. Not surprisingly, x = 1 but this cn be proved. If x < 1 then x 2 < x contrdicting the ssertion tht x is the smllest nturl number. Therefore, 1 is the smllest nturl number. This shows there is no integer, y, stisfying x < y < x + 1 since otherwise, you could subtrct x nd conclude 0 < y x < 1 for some integer y x. Now suppose y x > 1 nd let S {w N : w y}. The set S is nonempty by the Archimedin property. Let k be the smllest element of S. Therefore, k 1 < y. Either k 1 x or k 1 > x. If k 1 x, then y x y (k 1) = 0 {}}{ y k contrry to the ssumption tht y x > 1. Therefore, x < k 1 < y nd this proves the theorem with l = k 1. It is the next theorem which gives the density of the rtionl numbers. This mens tht for ny rel number, there exists rtionl number rbitrrily close to it. Theorem If x < y then there exists rtionl number r such tht x < r < y. Proof:Let n N be lrge enough tht n (y x) > 1. Thus (y x) dded to itself n times is lrger thn 1. Therefore, n (y x) = ny + n ( x) = ny nx > 1. It follows from Theorem there exists m Z such tht nd so tke r = m/n. nx < m < ny Definition A set, S R is dense in R if whenever < b, S (, b).

24 24 THE REAL AND COMPLEX NUMBERS Thus the bove theorem sys Q is dense in R. You probbly sw the process of division in elementry school. Even though you sw it t young ge it is very profound nd quite difficult to understnd. Suppose you wnt to do the following problem Wht did you do? You likely did process of long division which gve the following result. This ment 79 = 3 with reminder = 3 (22) You were given two numbers, 79 nd 22 nd you wrote the first s some multiple of the second dded to third number which ws smller thn the second number. Cn this lwys be done? The nswer is in the next theorem nd depends here on the Archimedin property of the rel numbers. Theorem Suppose 0 < nd let b 0. Then there exists unique integer p nd rel number r such tht 0 r < nd b = p + r. Proof: Let S {n N : n > b}. By the Archimedin property this set is nonempty. Let p + 1 be the smllest element of S. Then p b becuse p + 1 is the smllest in S. Therefore, r b p 0. If r then b p nd so b (p + 1) contrdicting p + 1 S. Therefore, r < s desired. To verify uniqueness of p nd r, suppose p i nd r i, i = 1, 2, both work nd r 2 > r 1. Then little lgebr shows p 1 p 2 = r 2 r 1 (0, 1). Thus p 1 p 2 is n integer between 0 nd 1, contrdicting Theorem The cse tht r 1 > r 2 cnnot occur either by similr resoning. Thus r 1 = r 2 nd it follows tht p 1 = p 2. This theorem is clled the Eucliden lgorithm when nd b re integers. 2.8 Exercises 1. By Theorem it follows tht for < b, there exists rtionl number between nd b. Show there exists n integer k such tht < k 2 m < b. 2. Show there is no smllest number in (0, 1). Recll (0, 1) mens the rel numbers which re strictly lrger thn 0 nd smller thn Show there is no smllest number in Q (0, 1). 4. Show tht if S R nd S is well ordered with respect to the usul order on R then S cnnot be dense in R. 5. Prove by induction tht n k=1 k3 = 1 4 n n n2.

25 2.8. EXERCISES It is fine thing to be ble to prove theorem by induction but it is even better to be ble to come up with theorem to prove in the first plce. Derive formul for n k=1 k4 in the following wy. Look for formul in the form An 5 + Bn 4 + Cn 3 + Dn 2 + En + F. Then try to find the constnts A, B, C, D, E, nd F such tht things work out right. In doing this, show (n + 1) 4 = ( ) A (n + 1) 5 + B (n + 1) 4 + C (n + 1) 3 + D (n + 1) 2 + E (n + 1) + F An 5 + Bn 4 + Cn 3 + Dn 2 + En + F nd so some progress cn be mde by mtching the coefficients. When you get your nswer, prove it is vlid by induction. 7. Prove by induction tht whenever n 2, n k=1 1 k > n. 8. If r 0, show by induction tht n k=1 rk = rn+1 r 1 r r Prove by induction tht n k=1 k = n(n+1) Let nd d be rel numbers. Find formul for n your result by induction. k=1 ( + kd) nd then prove 11. Consider the geometric series, n k=1 rk 1. Prove by induction tht if r 1, then k=1 r k 1 = rn 1 r. 12. This problem is continution of Problem 11. You put money in the bnk nd it ccrues interest t the rte of r per pyment period. These terms need little explntion. If the pyment period is one month, nd you strted with $100 then the mount t the end of one month would equl 100 (1 + r) = r. In this the second term is the interest nd the first is clled the principl. Now you hve 100 (1 + r) in the bnk. How much will you hve t the end of the second month? By nlogy to wht ws just done it would equl 100 (1 + r) (1 + r) r = 100 (1 + r) 2. In generl, the mount you would hve t the end of n months would be 100 (1 + r) n. (When bnk sys they offer 6% compounded monthly, this mens r, the rte per pyment period equls.06/12.) In generl, suppose you strt with P nd it sits in the bnk for n pyment periods. Then t the end of the n th pyment period, you would hve P (1 + r) n in the bnk. In n ordinry nnuity, you mke pyments, P t the end of ech pyment period, the first pyment t the end of the first pyment period. Thus there re n pyments in ll. Ech ccrue interest t the rte of r per pyment period. Using Problem 11, find formul for the mount you will hve in the bnk t the end of n pyment periods? This is clled the future vlue of n ordinry nnuity. Hint: The first pyment sits in the bnk for n 1 pyment periods nd so this pyment becomes P (1 + r) n 1. The second sits in the bnk for n 2 pyment periods so it grows to P (1 + r) n 2, etc. 13. Now suppose you wnt to buy house by mking n equl monthly pyments. Typiclly, n is pretty lrge, 360 for thirty yer lon. Clerly pyment mde 10 yers from now cn t be considered s vluble to the bnk s one mde tody.

26 26 THE REAL AND COMPLEX NUMBERS This is becuse the one mde tody could be invested by the bnk nd hving ccrued interest for 10 yers would be fr lrger. So wht is pyment mde t the end of k pyment periods worth tody ssuming money is worth r per pyment period? Shouldn t it be the mount, Q which when invested t rte of r per pyment period would yield P t the end of k pyment periods? Thus from Problem 12 Q (1 + r) k = P nd so Q = P (1 + r) k. Thus this pyment of P t the end of n pyment periods, is worth P (1 + r) k to the bnk right now. It follows the mount of the lon should equl the sum of these discounted pyments. Tht is, letting A be the mount of the lon, A = P (1 + r) k. k=1 Using Problem 11, find formul for the right side of the bove formul. This is clled the present vlue of n ordinry nnuity. 14. Suppose the vilble interest rte is 7% per yer nd you wnt to tke lon for $100,000 with the first monthly pyment t the end of the first month. If you wnt to py off the lon in 20 yers, wht should the monthly pyments be? Hint: The rte per pyment period is.07/12. See the formul you got in Problem 13 nd solve for P. 15. Consider the first five rows of Pscl s 2 tringle Wht is the sixth row? Now consider tht (x + y) 1 = 1x + 1y, (x + y) 2 = x 2 + 2xy + y 2, nd (x + y) 3 = x 3 + 3x 2 y + 3xy 2 + y 3. Give conjecture bout tht (x + y) Bsed on Problem 15 conjecture formul for (x + y) n nd prove your conjecture by induction. Hint: Letting the numbers of the n th row of Pscl s tringle be denoted by ( ( n 0), n ( 1),, n n) in reding from left to right, there is reltion between the numbers on the (n + 1) st row nd those on the n th row, the reltion being ( ) ( n+1 k = n ( k) + n k 1). This is used in the inductive step. 17. Let ( ) n k n! (n k)!k! where 0! 1 nd (n + 1)! (n + 1) n! for ll n 0. Prove tht whenever k 1 nd k n, then ( ) ( n+1 k = n ( k) + n ( k 1). Are these numbers, n k) the sme s those obtined in Pscl s tringle? Prove your ssertion. 18. The binomil theorem sttes ( + b) n = n k=0 ( n k) n k b k. Prove the binomil theorem by induction. Hint: You might try using the preceding problem. 19. Show tht for p (0, 1), n k=0 ( n k) kp k (1 p) n k = np. 2 Blise Pscl lived in the 1600 s nd is responsible for the beginnings of the study of probbility.

27 2.9. COMPLETENESS OF R Using the binomil theorem prove tht for ll n N, ( ) ( n n+1 n 1 + n+1) 1. Hint: Show first tht ( ) n k = n (n 1) (n k+1) k!. By the binomil theorem, ( n) n = k=0 ( ) ( ) k n 1 = k n k=0 k fctors {}}{ n (n 1) (n k + 1) k!n k. Now consider the term n (n 1) (n k+1) nd note tht similr term occurs in ( k!n k n+1 the binomil expnsion for 1 + n+1) 1 except tht n is replced with n + 1 whereever this occurs. Argue the term got bigger nd then note tht in the ( n+1 binomil expnsion for 1 + n+1) 1, there re more terms. 21. Prove by induction tht for ll k 4, 2 k k! 22. Use the Problems 21 nd 20 to verify for ll n N, ( n) n Prove by induction tht 1 + n i=1 i (i!) = (n + 1)!. 24. I cn jump off the top of the Empire Stte Building without suffering ny ill effects. Here is the proof by induction. If I jump from height of one inch, I m unhrmed. Furthermore, if I m unhrmed from jumping from height of n inches, then jumping from height of n + 1 inches will lso not hrm me. This is self evident nd provides the induction step. Therefore, I cn jump from height of n inches for ny n. Wht is the mtter with this resoning? 25. All horses re the sme color. Here is the proof by induction. A single horse is the sme color s himself. Now suppose the theorem tht ll horses re the sme color is true for n horses nd consider n + 1 horses. Remove one of the horses nd use the induction hypothesis to conclude the remining n horses re ll the sme color. Put the horse which ws removed bck in nd tke out nother horse. The remining n horses re the sme color by the induction hypothesis. Therefore, ll n + 1 horses re the sme color s the n 1 horses which didn t get moved. This proves the theorem. Is there something wrong with this rgument? ( ) n 26. Let denote the number of wys of selecting set of k k 1, k 2, k 1 things, set 3 of k 2 things, nd set of k 3 things from set of n things such tht 3 ( ) i=1 k i = n. n Find formul for. Now give formul for trinomil theorem, one k 1, k 2, k 3 which expnds (x + y + z) n. Could you continue this wy nd get multinomil formul? 2.9 Completeness of R By Theorem 2.7.9, between ny two rel numbers, points on the number line, there exists rtionl number. This suggests there re lot of rtionl numbers, but it is not cler from this Theorem whether the entire rel line consists of only rtionl numbers. Some people might wish this were the cse becuse then ech rel number could be described, not just s point on line but lso lgebriclly, s the quotient of integers. Before 500 B.C., group of mthemticins, led by Pythgors believed in this, but they discovered their beliefs were flse. It hppened roughly like this. They knew they

28 28 THE REAL AND COMPLEX NUMBERS could construct the squre root of two s the digonl of right tringle in which the two sides hve unit length; thus they could regrd 2 s number. Unfortuntely, they were lso ble to show 2 could not be written s the quotient of two integers. This discovery tht the rtionl numbers could not even ccount for the results of geometric constructions ws very upsetting to the Pythgorens, especilly when it becme cler there were n endless supply of such irrtionl numbers. This shows tht if it is desired to consider ll points on the number line, it is necessry to bndon the ttempt to describe rbitrry rel numbers in purely lgebric mnner using only the integers. Some might desire to throw out ll the irrtionl numbers, nd considering only the rtionl numbers, confine their ttention to lgebr, but this is not the pproch to be followed here becuse it will effectively eliminte every mjor theorem of clculus. In this book rel numbers will continue to be the points on the number line, line which hs no holes. This lck of holes is more precisely described in the following wy. Definition A non empty set, S R is bounded bove (below) if there exists x R such tht x ( ) s for ll s S. If S is nonempty set in R which is bounded bove, then number, l which hs the property tht l is n upper bound nd tht every other upper bound is no smller thn l is clled lest upper bound, l.u.b. (S) or often sup (S). If S is nonempty set bounded below, define the gretest lower bound, g.l.b. (S) or inf (S) similrly. Thus g is the g.l.b. (S) mens g is lower bound for S nd it is the lrgest of ll lower bounds. If S is nonempty subset of R which is not bounded bove, this informtion is expressed by sying sup (S) = + nd if S is not bounded below, inf (S) =. Every existence theorem in clculus depends on some form of the completeness xiom. Axiom (completeness) Every nonempty set of rel numbers which is bounded bove hs lest upper bound nd every nonempty set of rel numbers which is bounded below hs gretest lower bound. It is this xiom which distinguishes Clculus from Algebr. A fundmentl result bout sup nd inf is the following. Proposition Let S be nonempty set nd suppose sup (S) exists. Then for every δ > 0, S (sup (S) δ, sup (S)]. If inf (S) exists, then for every δ > 0, S [inf (S), inf (S) + δ). Proof:Consider the first clim. If the indicted set equls, then sup (S) δ is n upper bound for S which is smller thn sup (S), contrry to the definition of sup (S) s the lest upper bound. In the second clim, if the indicted set equls, then inf (S)+δ would be lower bound which is lrger thn inf (S) contrry to the definition of inf (S) Exercises 1. Let S = [2, 5]. Find sup S. Now let S = [2, 5). Find sup S. Is sup S lwys number in S? Give conditions under which sup S S nd then give conditions under which inf S S.

29 2.10. EXERCISES Show tht if S nd is bounded bove (below) then sup S (inf S) is unique. Tht is, there is only one lest upper bound nd only one gretest lower bound. If S = cn you conclude tht 7 is n upper bound? Cn you conclude 7 is lower bound? Wht bout 13.5? Wht bout ny other number? 3. Let S be set which is bounded bove nd let S denote the set { x : x S}. How re inf ( S) nd sup (S) relted? Hint: Drw some pictures on number line. Wht bout sup ( S) nd inf S where S is set which is bounded below? 4. Solve the following equtions which involve bsolute vlues. () x + 1 = 2x + 3 (b) x + 1 x + 4 = 6 5. Solve the following inequlities which involve bsolute vlues. () 2x 6 < 4 (b) x 2 < 2x Which of the field xioms is being bused in the following rgument tht 0 = 2? Let x = y = 1. Then 0 = x 2 y 2 = (x y) (x + y) nd so 0 = (x y) (x + y). Now divide both sides by x y to obtin 0 = x + y = = Give conditions under which equlity holds in the tringle inequlity. 8. Let k n where k nd n re nturl numbers. P (n, k), permuttions of n things tken k t time, is defined to be the number of different wys to form n ordered list of k of the numbers, {1, 2,, n}. Show P (n, k) = n (n 1) (n k + 1) = n! (n k)!. 9. Using the preceding problem, show the number of wys of selecting set of k things from set of n things is ( n k). 10. Prove the binomil theorem from Problem 9. Hint: When you tke (x + y) n, note tht the result will be sum of terms of the form, k x n k y k nd you need to determine wht k should be. Imgine writing (x + y) n = (x + y) (x + y) (x + y) where there re n fctors in the product. Now consider wht hppens when you multiply. Ech fctor contributes either n x or y to typicl term. 11. Prove by induction tht n < 2 n for ll nturl numbers, n Prove by the binomil theorem nd Problem 9 tht the number of subsets of given finite set contining n elements is 2 n. 13. Let n be nturl number nd let k 1 + k 2 + k r = n where k i is non negtive integer. The symbol ( n k 1 k 2 k r ) denotes the number of wys of selecting r subsets of {1,, n} which contin k 1, k 2 k r elements in them. Find formul for this number.