Advanced Analysis. Min Yan Department of Mathematics Hong Kong University of Science and Technology

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1 Advaced Aalysis Mi Ya Departmet of Mathematics Hog Kog Uiversity of Sciece ad Techology September 3, 009

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3 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 Property Limit of Trigometric Fuctio Limit of Expoetial Fuctio 5 36 More Property Order of Ifiity ad Ifiitesimal Additioal Exercise 6 4 Cotiuous Fuctio 63 4 Defiitio 63 4 Uiformly Cotiuous Fuctio Maximum ad Miimum Itermediate Value Theorem 7 45 Ivertible Cotiuous Fuctio Iverse Trigometric ad Logarithmic Fuctios Additioal Exercise 79 Differetiatio 8 Approximatio ad Differetiatio 8 Approximatio 8 Differetiatio 84 3 Derivative 85 3

4 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 Additioal Exercise 35 3 Itegratio 39 3 Riema Itegratio 40 3 Riema Sum 40 3 Itegrability Criterio Itegrability of Cotiuous ad Mootoe Fuctios Properties of Itegratio Additioal Exercise 54 3 Atiderivative 58 3 Fudametal Theorem of Calculus 59 3 Atiderivative Itegratio by Parts Chage of Variable 7 35 Additioal Exercise Topics o Itegratio Itegratio of Ratioal Fuctios Improper Itegratio Riema-Stieltjes Itegratio Bouded Variatio Fuctio Additioal Exercise 0 4 Series 03 4 Series of Numbers 04 4 Sum of Series 04 4 Compariso Test Coditioal Covergece 44 Rearragemet of Series 5 45 Additioal Exercise 9

5 CONTENTS 5 4 Series of Fuctios 4 Uiform Covergece 4 Properties of Uiform Covergece 8 43 Power Series Fourier Series Additioal Exercise 44 5 Multivariable Fuctio 49 5 Limit ad Cotiuity 50 5 Limit i Euclidea Space 50 5 Topology i Euclidea Space Multivariable Fuctio Cotiuous Fuctio 6 55 Multivariable Map Additioal Exercise 67 5 Multivariable Algebra 69 5 Liear Trasform 69 5 Biliear ad Quadratic Form 7 53 Multiliear Map ad Polyomial Additioal Exercises 87 6 Multivariable Differetiatio 89 6 Differetiatio 90 6 Differetiability ad Derivative 90 6 Partial Derivative 9 63 Rules of Differetiatio Directioal Derivative 98 6 Iverse ad Implicit Fuctio 30 6 Iverse Differetiatio 30 6 Implicit Differetiatio Hypersurface Additioal Exercise 3 63 High Order Differetiatio Quadratic Approximatio High Order Partial Derivative Taylor Expasio Maximum ad Miimum Costraied Extreme Additioal Exercise Multivariable Itegratio Riema Itegratio Volume i Euclidea Space Riema Sum Properties of Itegratio Fubii Theorem Volume i Vector Space 356

6 6 CONTENTS 76 Chage of Variable Improper Itegratio Additioal Exercise 37 7 Itegratio o Hypersurface Rectifiable Curve Itegratio of Fuctio o Curve Itegratio of -Form o Curve Surface Area Itegratio of Fuctio o Surface Itegratio of -Form o Surface Volume ad Itegratio o Hypersurface Stokes Theorem Gree Theorem Idepedece of Itegral o Path Stokes Theorem Gauss Theorem 45

7 Chapter Limit ad Cotiuity 7

8 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

9 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

10 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 = 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

11 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

12 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 = Exercise 4 Prove! > The use this to prove ( )! = 0

13 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

14 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 = = = ( ( + 0 ) ( + ) ( + = = 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

15 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

16 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 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 =,

17 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 / + )( + ) ( + ) ( ) ( + )( 5) 4 ( ) ( ) 5! Exercise Compute the its a a, where a + 6 ( + + 3)(! + 5 ) ( + )! cos + si ( + ) ( ) 4 5

18 8 CHAPTER LIMIT AND CONTINUITY ( + a) +b+c 3 a + b + c, where a, b, c > 0 ( ) 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 ( ) = ( ) + 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)

19 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 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

20 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 = +