Advaced Aalysis Mi Ya Departmet of Mathematics Hog Kog Uiversity of Sciece ad Techology September 3, 009
Cotets Limit ad Cotiuity 7 Limit of Sequece 8 Defiitio 8 Property 3 3 Ifiity ad Ifiitesimal 8 4 Additioal Exercise Covergece of Sequece Limit Necessary Coditio Supremum ad Ifimum 4 3 Mootoe Sequece 8 4 Coverget Subsequece 3 5 Covergece of Cauchy Sequece 36 6 Ope Cover 37 7 Additioal Exercise 39 3 Limit of Fuctio 40 3 Defiitio 40 3 Variatio 43 33 Property 46 34 Limit of Trigometric Fuctio 50 35 Limit of Expoetial Fuctio 5 36 More Property 56 37 Order of Ifiity ad Ifiitesimal 59 38 Additioal Exercise 6 4 Cotiuous Fuctio 63 4 Defiitio 63 4 Uiformly Cotiuous Fuctio 67 43 Maximum ad Miimum 69 44 Itermediate Value Theorem 7 45 Ivertible Cotiuous Fuctio 73 46 Iverse Trigometric ad Logarithmic Fuctios 75 47 Additioal Exercise 79 Differetiatio 8 Approximatio ad Differetiatio 8 Approximatio 8 Differetiatio 84 3 Derivative 85 3
4 CONTENTS 4 Taget Lie ad Rate of Chage 88 5 Rules of Computatio 90 6 Basic Example 93 7 Derivative of Iverse Fuctio 96 8 Additioal Exercise 99 Applicatio of Differetiatio 00 Maximum ad Miimum 00 Mea Value Theorem 0 3 Mootoe Fuctio 05 4 L Ĥospital s Rule 08 5 Additioal Exercise 3 High Order Approximatio 4 3 Quadratic Approximatio 4 3 High Order Derivative 6 33 Taylor Expasio 34 Remaider 6 35 Maximum ad Miimum 9 36 Covex ad Cocave 30 37 Additioal Exercise 35 3 Itegratio 39 3 Riema Itegratio 40 3 Riema Sum 40 3 Itegrability Criterio 43 33 Itegrability of Cotiuous ad Mootoe Fuctios 46 34 Properties of Itegratio 48 35 Additioal Exercise 54 3 Atiderivative 58 3 Fudametal Theorem of Calculus 59 3 Atiderivative 64 33 Itegratio by Parts 70 34 Chage of Variable 7 35 Additioal Exercise 74 33 Topics o Itegratio 79 33 Itegratio of Ratioal Fuctios 79 33 Improper Itegratio 84 333 Riema-Stieltjes Itegratio 89 334 Bouded Variatio Fuctio 95 335 Additioal Exercise 0 4 Series 03 4 Series of Numbers 04 4 Sum of Series 04 4 Compariso Test 07 43 Coditioal Covergece 44 Rearragemet of Series 5 45 Additioal Exercise 9
CONTENTS 5 4 Series of Fuctios 4 Uiform Covergece 4 Properties of Uiform Covergece 8 43 Power Series 34 44 Fourier Series 38 45 Additioal Exercise 44 5 Multivariable Fuctio 49 5 Limit ad Cotiuity 50 5 Limit i Euclidea Space 50 5 Topology i Euclidea Space 53 53 Multivariable Fuctio 58 54 Cotiuous Fuctio 6 55 Multivariable Map 63 56 Additioal Exercise 67 5 Multivariable Algebra 69 5 Liear Trasform 69 5 Biliear ad Quadratic Form 7 53 Multiliear Map ad Polyomial 79 54 Additioal Exercises 87 6 Multivariable Differetiatio 89 6 Differetiatio 90 6 Differetiability ad Derivative 90 6 Partial Derivative 9 63 Rules of Differetiatio 95 64 Directioal Derivative 98 6 Iverse ad Implicit Fuctio 30 6 Iverse Differetiatio 30 6 Implicit Differetiatio 306 63 Hypersurface 309 64 Additioal Exercise 3 63 High Order Differetiatio 34 63 Quadratic Approximatio 34 63 High Order Partial Derivative 37 633 Taylor Expasio 30 634 Maximum ad Miimum 33 635 Costraied Extreme 38 636 Additioal Exercise 334 7 Multivariable Itegratio 337 7 Riema Itegratio 338 7 Volume i Euclidea Space 338 7 Riema Sum 344 73 Properties of Itegratio 348 74 Fubii Theorem 35 75 Volume i Vector Space 356
6 CONTENTS 76 Chage of Variable 360 77 Improper Itegratio 367 78 Additioal Exercise 37 7 Itegratio o Hypersurface 374 7 Rectifiable Curve 374 7 Itegratio of Fuctio o Curve 378 73 Itegratio of -Form o Curve 380 74 Surface Area 384 75 Itegratio of Fuctio o Surface 387 76 Itegratio of -Form o Surface 388 77 Volume ad Itegratio o Hypersurface 394 73 Stokes Theorem 399 73 Gree Theorem 399 73 Idepedece of Itegral o Path 404 733 Stokes Theorem 40 734 Gauss Theorem 45
Chapter Limit ad Cotiuity 7
8 CHAPTER LIMIT AND CONTINUITY Limit of Sequece A sequece is a ifiite list x, x, x 3,, x, x +, The sequece ca also be deoted as {x } The subscript is called the idex ad does ot have to start from For example, x 5, x 6, x 7,, x, x +,, is also a sequece, with the idex startig from 5 I this chapter, the terms x of a sequece are assumed to be real umbers ad ca be plotted o the real umber lie A sequece has a it l if the followig implicatio happes: is big = x is close to l Ituitively, this meas that the sequece accumulates aroud l However, to give a rigorous defiitio, the meaig of big, close ad implies has to be more precise The bigess of a umber is usually measured by > N for some big N For example, we say is i the thousads if N =, 000 The closeess betwee two umbers u ad v is usually measured by (the smalless of) the size of u v The implicatio meas that the predetermied smalless of x l may be achieved by the bigess of Defiitio Defiitio A sequece {x } of real umbers has it l (or coverges to l), ad deoted x = l, if for ay ɛ > 0, there is N, such that > N = x l < ɛ () A sequece is coverget if it has a (fiite) it Otherwise, the sequece is diverget l ɛ ɛ x x x 5 x N+ x x x 3 N+3 N+ x N+4 x 4 Figure : for ay ɛ, there is N Note the logical relatio betwee ɛ ad N The predetermied smalless ɛ for x l is arbitrarily give, while the size N for is to be foud after ɛ is give Thus the choice of N usually depeds o ɛ ad is ofte expressed as a fuctio of ɛ
LIMIT OF SEQUENCE 9 l ɛ ɛ 5 N+ N+4 Figure : aother plot of a covergig sequece Sice the it is about the log term behavior of a sequece gettig closer to a target, oly small ɛ ad big N eed to be cosidered establishig a it For example, the it of a sequece is ot chaged if the first oe hudred terms are replaced by other arbitrary umbers Exercise 9 cotais more examples Example Ituitively, the bigger is, the smaller gets This suggests = 0 Rigorously followig the defiitio, for ay ɛ > 0, choose N = ɛ The > N = 0 = < N = ɛ { } Thus the implicatio () is established for the sequece How did we fid the suitable N? Our goal is to achieve 0 < ɛ This is the same as > ɛ, which suggests us to take N = Note that our choice of N ɛ may ot be a atural umber { Example The sequece + ( ) 3, 3, 4 5, 5 4, 6 7, 7 6, Plottig the sequece suggests + ( ) = For the rigorous argumet, we observe that + ( ) = + ( ) }, with idex startig from =, is I order for the left side to be less tha ɛ, it is sufficiet to make is the same as > ɛ + < ɛ, which
0 CHAPTER LIMIT AND CONTINUITY Based o the aalysis, we have the followig formal ad rigorous argumet for the it: For ay ɛ > 0, choose N = + The ɛ > N = + ( ) = + ( ) < N = ɛ Example 3 Cosider the sequece 4, 4, 44, 44, 44, 443, 4435, 44356, of more ad more refied decimal approximatios of The ituitio suggests that x = The rigorous verificatio meas that for ay ɛ > 0, we eed to fid N, such that > N = x < ɛ Sice the -th term x is the decimal expasio up to the -th decimal poit, it satisfies x < 0 Therefore it suffices to fid N such that the followig implicatio holds > N = 0 < ɛ Assume > ɛ > 0 The ɛ has the decimal expasio ɛ = 000 0E N E N+ E N+, with E N is from {,,, 9} I other words, N is the locatio of the first ozero digit i the decimal expasio of ɛ The for > N, we have ɛ 000 0E N 000 0 = 0 N > 0 The argumet above assumes > ɛ > 0 This is ot a problem because if we ca achieve x l < 05 for > N, the we ca certaily achieve x l < ɛ for ay ɛ ad for the same > N I other words, we may add the assumptio that ɛ is less tha a certai fixed umber without hurtig the overall rigorous argumet for the it See Exercise 9 for more freedom we may have i choosig ɛ ad N Exercise Rigorously verify the its = = 0 3 ( + ) = 0 4 /3 = 0 / 5 cos = 0 6 cos + si = Exercise Let a positive real umber a > 0 have the decimal expasio a = XZ Z Z Z +, where X is a o-egative iteger, ad Z is a sigle digit iteger from {0,,,, 9} at the -th decimal poit Prove the sequece XZ, XZ Z, XZ Z Z 3, XZ Z Z 3 Z 4, of more ad more refied decimal approximatios coverges to a
LIMIT OF SEQUENCE Exercise 3 Suppose x l y ad (x y ) = 0 Prove x = y = l Exercise 4 Suppose x l y ad y = 0 Prove x = l Exercise 5 Suppose x = l Prove x = l Is the coverse true? Exercise 6 Suppose x = l Prove x +3 = l Is the coverse true? Exercise 7 Prove that the it is ot chaged if fiitely may terms are modified I other words, if there is N, such that x = y for > N, the x = l if ad oly if y = l Exercise 8 Prove the uiqueess of the it I other words, if x = l ad x = l, the l = l Exercise 9 Prove the followig are equivalet defiitios of x = l For ay c > ɛ > 0, where c is some fixed umber, there is N, such that x l < ɛ for all > N For ay ɛ > 0, there is a atural umber N, such that x l < ɛ for all > N 3 For ay ɛ > 0, there is N, such that x l ɛ for all > N 4 For ay ɛ > 0, there is N, such that x l < ɛ for all N 5 For ay ɛ > 0, there is N, such that x l ɛ for all > N Exercise 0 Which are equivalet to the defiitio of x = l? For ɛ = 000, we have N = 000, such that x l < ɛ for all > N For ay 000 ɛ > 0, there is N, such that x l < ɛ for all > N 3 For ay ɛ > 000, there is N, such that x l < ɛ for all N 4 For ay ɛ > 0, there is a atural umber N, such that x l ɛ for all N 5 For ay ɛ > 0, there is N, such that x l < ɛ for all > N 6 For ay ɛ > 0, there is N, such that x l < ɛ + for all > N 7 For ay ɛ > 0, we have N = 000, such that x l < ɛ for all > N 8 For ay ɛ > 0, there are ifiitely may, such that x l < ɛ 9 For ifiitely may ɛ > 0, there is N, such that x l < ɛ for all > N 0 For ay ɛ > 0, there is N, such that l ɛ < x < l + ɛ for all > N For ay atural umber K, there is N, such that x l < K for all > N The followig examples are the most importat basic its For ay give ɛ, the aalysis leadig to the suitable choice of N will be give It is left to the reader to write dow the rigorous formal argumet i lie with the defiitio of it
CHAPTER LIMIT AND CONTINUITY Example 4 We have = 0 for a > 0 () a The iequality a < ɛ is the same as < ɛ a Thus choosig N = ɛ a should make the implicatio () hold Example 5 We have a = 0 for a < (3) Let = + b The b > 0 ad a a = ( + b) = + b + ( ) b + > b This implies a < b I order to get a < ɛ, therefore, it suffices to make sure b < ɛ This suggests us to choose N = bɛ Example 6 We have Let x = The x > 0 ad = ( + x ) = + x + = (4) ( ) x + > ( ) x This implies x < I order to get = x < ɛ, therefore, it suffices to make sure < ɛ Thus we may choose N = ɛ + Example 7 For ay a, we have a! = 0 (5) Fix a iteger M > a The for > M, we have a! = a M a a M! M + M + a a a M a M! Thus i order to get a a M a! < ɛ, we oly eed to make sure < ɛ This leads } M! to the choice N = max {M, a M+ M!ɛ Exercise Prove! < (!) ad ()! < for > The use this to +! prove = (!) ()! = 0 Exercise Use the biary expasio of = (+) to prove > The prove = 0 Exercise 3 Prove 3 = ad + + 3 = Exercise 4 Prove! > The use this to prove ( )! = 0
LIMIT OF SEQUENCE 3 Property A sequece is bouded if there is a costat B, such that x B for all This is equivalet to the existece of costats B ad B, such that B x B for ay The costats B, B, B are respectively called a boud, a lower boud ad a upper boud Propositio Coverget sequeces are bouded Proof Suppose x = l For ɛ = > 0, there is N, such that > N = x l < Moreover, by takig a bigger atural umber if ecessary, we may further assume N is a atural umber The x N+, x N+,, have upper boud l + ad lower boud l, ad the whole sequece has upper boud max{x, x,, x N, l + } ad lower boud mi{x, x,, x N, l } Exercise 5 Prove that if x < B for > N, the the whole sequece {x } is bouded This implies that the boudedess is ot chaged by modifyig fiitely may terms i a sequece Exercise 6 Suppose x = 0 ad y is bouded Prove x y = 0 Propositio 3 (Arithmetic Rule) Suppose The x = l, (x + y ) = l + k, y = k x y = lk, where y 0 ad k 0 are assumed i the third equality Proof For ay ɛ > 0, there are N ad N, such that The for > max{n, N }, we have > N = x l < ɛ, > N = y k < ɛ x = l y k, (x + y ) (l + k) x l + y k < ɛ + ɛ = ɛ This completes the proof that (x + y ) = l + k By Propositio, we have y < B for a fixed umber B ad all For ay ɛ > 0, there are N ad N, such that > N = x l < ɛ B, > N = y k < ɛ l
4 CHAPTER LIMIT AND CONTINUITY The for > max{n, N }, we have x y lk = (x y ly ) + (ly lk) x l y + l y k < ɛ B B + l ɛ l = ɛ This completes the proof that x y = lk Assume y 0 ad k 0 We will prove = By the product y k property of the it, this implies x = x = l y y k = l k { ɛ k For ay ɛ > 0, we have ɛ = mi, k } > 0 The there is N, such that > N = y k < ɛ y k < ɛ k, y k < k = y k < ɛ k, y > k ɛ k = y k = y k < y k This completes the proof that y = k k k = ɛ Example 8 By the it () ad the arithmetic rule, we have = Here is a more complicated example = = 0 = 3 + + + 3 + 0 + = + 3 + 0 + 3 + = ( ( + 0 ) ( + ) ( + = + 0 + 0 3 + 0 0 + 0 3 = The idea ca be geeralized to obtai a p p + a p p + + a + a 0 0 if p < q ad b q 0 b q q + b q q = a p + + b + b 0 if p = q ad b q 0 (6) b q ) 3 ) 3
LIMIT OF SEQUENCE 5 Exercise 7 Suppose x = l ad y = k Prove max{x, y } = max{l, k} ad mi{x, y } = mi{l, k} You may use the formula max{x, y} = (x + y + x y ) ad the similar oe for mi{x, y} Propositio 4 (Order Rule) Suppose both {x } ad {y } coverge If x y for big, the x y If x > y, the x > y for big A special case of the property is that x l implies x l, ad x < l implies x < l for sufficietly big Proof We prove the secod statemet first Suppose x > y The by Propositio 3, (x y ) = x y > 0 For ɛ = (x y ) > 0, there is N, such that > N = (x y ) ɛ < ɛ = x y > ɛ ɛ = 0 x > y By exchagig x ad y i the secod statemet, we fid that x < y = x < y for big This further implies that we caot have x y for big The combied implicatio x < y = opposite of (x y for big ) is equivalet to the first statemet I the secod part of the proof above, we used the logical fact that A = B is the same as (ot B) = (ot A) Moreover, we ote that the followig two statemets are ot opposite of each other There is N, such that x < y for > N There is N, such that x y for > N Therefore although the proof above showed that the first statemet is a cosequece of the secod, the two statemets are ot logically equivalet Propositio 5 (Sadwich Rule) Suppose x y z, x = z = l The y = l
6 CHAPTER LIMIT AND CONTINUITY Proof For ay ɛ > 0, there are N ad N, such that > N = x l < ɛ, > N = z l < ɛ The > max{n, N } = ɛ < x l y l z l < ɛ = y l < ɛ { cos } Example 9 To fid the it of the sequece, we compare it with { } the sequece i Example Sice cos ad = = 0, we get cos = 0 si By similar reaso, we get = 0 ad ( ) = 0 The by the arithmetic rule, cos + ( ) si = ( = cos + si ( ) ( ) cos ) ( ) si ( ) + ( ) = 0 0 + 0 0 = 0 Example 0 Suppose {x } is a sequece satisfyig x l < The we have l < x < l + ( Sice l ) ( = l + ) = l, by the sadwich rule, we get x = l Example For ay a > ad > a, we have < a < Thus by the it (4) ad the sadwich rule, we have a = O the other had, for 0 < a <, we have b = a > ad a = b = b = Combiig all the cases, we get a = for ay a > 0 Furthermore, we have < ( + a) +b < (), < ( + a + b) +c < ( ) for sufficietly big By ( () = ( ( ) = ) ( ) = ( ) =, ) 4 = 4 =,
LIMIT OF SEQUENCE 7 ad the sadwich rule, we get ( + a) +b = ( + a + b) +c = The same idea leads to the it (a p p + a p p + + a + a 0 ) +a = (7) p Example Cosider for a > ad ay p The special case p = 0 a is the it (3), ad the special case p =, a = is Exercise Let a = + b Sice a >, we have b > 0 Fix a atural umber P p The for > P, a ( ) = + b + b + + ( ) ( P ) > b P + (P + )! ( ) ( P ) b P + + (P + )! Thus 0 < p a P a < (P + )! P ( ) ( P ) b P + = ( ) ( ) ( P ) (P + )! b P + By = 0, =, the fact that P ad b are fixed costats, k ad the arithmetic rule, the right side has it 0 as By the sadwich rule, we coclude that p = 0 for a > ad ay p (8) a Exercise 8 Redo Exercise 4 by usig the sadwich rule Exercise 9 Let a > 0 be a costat The the it (4) to prove a = Exercise 0 Compute the its < a < for big Use this ad 7/4 3 3/ (3 3/4 / + )( + ) ( + ) 3 3 8 7 + ( ) + 8 + ( + )( 5) 4 ( ) + 3 + ( ) 5! + 0 0 + Exercise Compute the its a a, where a + 6 ( + + 3)(! + 5 ) ( + )! + 5 7 + cos + si 8 + 5 + 5 9 5 ( + ) 0 + + 3 + 3 5 ( ) 4 5
8 CHAPTER LIMIT AND CONTINUITY ( + a) +b+c 3 a + b + c, where a, b, c > 0 ( ) 4 + + + + + + 3 Ifiity ad Ifiitesimal A chagig umerical quatity is a ifiity if it teds to get arbitrarily big For sequeces, this meas the followig Defiitio 6 A sequece {x } diverges to ifiity, deoted x =, if for ay b, there is N, such that > N = x > b (9) It diverges to positive ifiity, deoted x = +, if for ay b, there is N, such that > N = x > b (0) It diverges to egative ifiity, deoted x =, if for ay b, there is N, such that > N = x < b () Example 3 We rigorously verify choose N = b The = + For ay b > 0, + ( ) > N = + ( ) + > > N = b Exercise Rigorously verify the divergece to ifiity (00 + 0 ) = ( ) + si = 3 + ( ) + + + = + Exercise 3 Ifiities must be ubouded Is the coverse true? Exercise 4 Suppose x = + ad y = + Prove (x + y ) = + ad x y = + Exercise 5 Suppose x = ad x x + < c for some costat c Prove that either x = + or x = If we further kow x = +, prove that for ay a > x, some term x lies i the iterval (a, a + c)
LIMIT OF SEQUENCE 9 A chagig umerical quatity is a ifiitesimal if it teds to get arbitrarily small For sequeces, this meas that for ay ɛ > 0, there is N, such that > N = x < ɛ () This simply meas x = 0 Note that the implicatios (9) ad () are equivalet by chagig x to ad takig ɛ = Therefore we have x b { } {x } is a ifiity is a ifiitesimal For example, the ifiitesimals (),{ (3), } (5), { }(8) tell us that! a { a } (for a > 0), {a } (for a > ),, ad (for a > ) are ifiities Moreover, the first case i the it (6) tells us a p p + a p p + + a + a 0 = if p > q ad a b q q + b q q p 0 + + b + b 0 O the other had, sice x = l is equivalet to (x l) = 0, we have {x } coverges to l {x l} is a ifiitesimal For example, the it (4) tells us that { } is a ifiitesimal Exercise 6 { } How to characterize a positive ifiity {x } i terms of the ifiitesimal? x Exercise 7 Explai the ifiities! = for ay a 0 a! a = if a + b 0 + b 3 = + 4 = Some properties of fiite its ca be exteded to ifiities ad ifiitesimals For example, if x = + ad y = +, the (x + y ) = + The property ca be deoted as the arithmetic rule (+ ) + (+ ) = + Moreover, if x =, y = 0, ad y < 0 for big, the x y = Thus we have aother arithmetic rule = Commo sese suggests more arithmetic rules such as 0 c c+ =, c = (for c 0), =, 0 = (for c 0), c = 0, x a p
0 CHAPTER LIMIT AND CONTINUITY where c is a fiite umber ad represets a sequece coverget to c We must be careful i applyig arithmetic rules ivolvig ifiities ad ifiitesimals For example, we have = 0, =, = 0, = +, = 0, This shows that 0 has o defiite value 0 Example 4 By the (exteded) arithmetic rule, we have = 0 ( + 3)( ) = ( + 3) ( ) = = ( + ) = + = (+ ) + 0 = + = ( ) = Exercise 8 Prove the properties of ifiities 0 + = + (bouded)+ = : If {x } is bouded ad y =, the (x + y ) = mi{+, + } = + : If x = y = +, the mi{x, y } = + 3 Sadwich rule: If x y ad y = +, the x = + 4 (> c > 0) (+ ) = + : If x > c for some costat c > 0 ad y = +, the x y = + Exercise 9 Show that it is ot ecessarily true that + = by costructig examples of sequeces {x } ad {y } that diverge to but oe of the followig holds (x + y ) = (x + y ) = + 3 {x + y } is bouded ad diverget Exercise 30 Show that oe caot make a defiite coclusio o 0 by costructig examples of sequeces {x } ad {y }, such that x = 0 ad y = but oe of the followig holds x y = x y = 0 3 x y = 4 {x y } is bouded ad diverget Exercise 3 Provide couterexamples to the wrog extesios of the arithmetic rules + =, (+ ) (+ ) = 0, 0 = 0, 0 =, 0 = +