MAS111 Convergence and Continuity

Transcription

1 MAS Covergece ad Cotiuity Key Objectives At the ed of the course, studets should kow the followig topics ad be able to apply the basic priciples ad theorems therei to solvig various problems cocerig covergece of sequeces ad cotiuity of fuctios. Real umbers: defiitio of algebraic ad trascedetal umbers, provig basic iequalities, fidig supremum ad ifimum of a set of real umbers, statig the completeess axiom. Sequeces: defiitio of limit, provig results cocerig limits of sequeces, fidig the limit of a bouded mootoe sequece, proof ad applicatio of the sadwich theorem, proof ad applicatio of the Bolzao-Weierstrass Theorem, calculatio of limits. Series: defiitio of covergece, applicatio of the compariso test, root test ad ratio test for covergece, geometric ad harmoic series, alteratig series ad absolute covergece, power series, fidig radius ad domai of covergece, statig the power series expasio of si x, cos x ad exp x, calculatio of sums of simple series. Real fuctios: defiitio of the limit of a fuctio, defiitio of oe-sided limits, use of the sadwich theorem, calculatio of limits of fuctios. Cotiuous fuctios: defiitio of cotiuity, derivatio of basic properties of cotiuous fuctios o closed itervals, statemet ad applicatio of the Itermediate Value Theorem, provig results cocerig the roots of polyomials.

2 Real Numbers R is a complete ordered field. Axioms for additio: A. x, y R, x + y = y + x. A2. x, y, z R, (x + y + z = x + (y + z. A3. 0 R, called zero, such that x + 0 = x for all x R. A4. x, ( x R such that x + ( x = 0. Axioms for multiplicatio: M. x, y R, xy = yx. M2. x, y, z R, (xyz = x(yz. M3. R, called oe, such that x = x. M4. it 0 ad x R\{0}, x R such that xx =. Distributive law: x, y, z R, x(y + z = xy + xz. Axioms for order: O. x, y R, exactly oe of the followig holds: x < y, x = y, y < x. 02. x, y, z R, 03. x, y, z R, 04. x, y R, x < y ad y < z = x < z. x < y = x + z < y + z. 0 < x ad 0 < y = 0 < xy. Completeess axiom: If A is a o-empty subset of R ad has a upper boud, the it has a least upper boud. 2

3 Theorem.. (Archimedes Priciple Let x, y R ad let x > 0. The there exists N such that x > y. Corollary.2. Let x, y R be such that x < y. The there exists a ratioal umber r R such that x < r < y. Corollary.3. There exists a uique umber l R such that l > 0 ad l 2 = 2. Propositio.4. Let A be a o-empty subset of R with a upper boud. Let l R. The followig coditios are equivalet: (i l = sup A; (ii l is a upper boud of A such that ε > 0, a A with l ε < a. Propositio.5. Let A be a o-empty subset of R with a lower boud. Let l R. The followig coditios are equivalet: (i l = if A; (ii l is a lower boud of A such that ε > 0, a A with a < l + ε. Propositio.6. Let p/q be a ratioal root of a x + a x + + a x + a o = 0 where ad a 0,, a Z with a 0 a 0. If p ad q are coprime, the p a 0 ad q a. Questio.7. What is a algebraic umber? a trascedetal umber? Questio.8. What are the sup ad if of { } x + x : x R? 2 Questio.9. Show that (i x a < b a b < x < a + b. (ii ( + x + x for all N ad x. Theorem.0. Let A ad B be o-empty bouded subsets of R. The (i sup(a + B = sup A + sup B ; (ii if(a + B = if A + if B. Theorem.. (Arithmetic-Geometric mea iequality For ay o-egative real umbers a, a 2,, a, we have (a a 2 a a + a a. 3

4 2 Sequeces A (real sequece is a fuctio x : N R. We study the behaviour of x( for large. We ofte write x for x( ad deote a sequece x : N R by listig its image: x, x 2, x 3,, x, or by or simply, (x. (x, (x N Defiitio 2.. Let (x be a sequece. We say that (x coverges to the limit l if ε > 0, N N such that N = x l < ε i which case, we say that the sequece (x is coverget or, coverges. We say that (x is diverget or diverges if it is ot coverget. Theorem 2.2. A sequece ca coverge to at most oe limit. Notatio We deote the limit of (x, if it exists, by We also write lim x. x l as or simply, x l to mea that (x coverges to the limit l. Theorem 2.3. lim = 0. Proof. By Archimedes Priciple. A sequece (x is said to be bouded above if there exists some costat K R such that x K for all. A sequece (x is said to be bouded below if there exists some costat K R such that x K for all. A sequece (x is said to be bouded if it is both bouded above ad below which is equivalet to sayig that there exists some costat K R such that x K for all. Theorem 2.4. Every coverget sequece is bouded. 4

5 Example 2.5. (i Theorem 2.4 does ot say that a bouded sequece coverges, ideed, the followig sequece is bouded ad diverget:, 0,, 0,. (ii The sequece, 2, 3,,, diverges because it is ubouded. Theorem 2.6. Let (a a ad (b b. The we have (i a + b a + b; (ii a b a b; (iii a b ab; (iv a b a b if b, b 0. Theorem 2.7. (Sadwich Theorem Give that a x b ad that both sequeces (a ad (b coverge to l, the the sequece (x also coverges to l Example 2.8. Let 0 < a <. The a 0. Defiitio 2.9. A sequece (a is called icreasig if m = a a m ; it is called decreasig if m = a a m. Further, (a is called mootoe if it is either icreasig or decreasig. Example 2.0. Give a > 0, the sequece the limit lim a? a is a mootoe sequece. What is Theorem 2.. Let the sequece (a be icreasig ad bouded above. coverges, moreover, we have lim a = sup{a : N}. The it Theorem 2.2. Let the sequece (a be decreasig ad bouded below. coverges, moreover, we have The it lim a = if{a : N}. Example 2.3. The sequece 3, 3 + 3, ,... coverges. What is the limit? 5

6 Example 2.4. The sequece (( + is icreasig ad bouded by 3. Therefore it coverges ad the limit is deoted by e. Theorem 2.5. Give a a as, we have a + + a a. Defiitio 2.6. Let (x be a sequece. A subsequece of (x is ay sequece of the form x, x 2, x 3,, x k, where k N ad Example 2.7. < 2 < 3 < < k <. (i (x is a subsequece of itself. (ii (x 2 ad (x 2+ are subsequeces of (x. (iii x 2, x 5, x 6, x 23, x 3, x 3, x 3, x 3, x 3, is ot a subsequece of (x. (iv, 3, 5, 5, 5, 6, 7, 8, is a subsequece of, 2, 3, 4, 5, 5, 5, 5, 5, 6, 7, 8,. Propositio 2.8. Let (x be a sequece. Give that both subsequeces (x 2 ad (x 2+ coverge to the same limit l, we also have x l as. Theorem 2.9. Every sequece has a mootoe subsequece. Theorem (Bolzao-Weierstrass Theorem Every bouded sequece has a coverget subsequece. Defiitio 2.2. A sequece (x is called a Cauchy sequece if Notatio: ε > 0, N N such that, m N x x m < ε. lim x x m = 0.,m Theorem (Cauchy Criterio for Covergece A real sequece (x is coverget if, ad oly if, it is a Cauchy sequece. 6

7 3 series The symbols a = a + a a +, called a (real series, deote the (real sequece (s of partial sums, where s = a + a a is called the -th partial sum of the series. We say that a series a coverges or, is coverget, if the sequece (s of partial sums coverges, i which case, the limit of (s is called the sum of the series ad we write a = lim s = lim (a + a a. A series a is said to be diverget or, diverge, if it is ot coverget. CAUTION. Do ot cofuse the sequece with the series the latter is the followig sequece : a, a 2, a, a + a a +, a, a + a 2, a + a 2 + a 3,, a + a a,!!! Example 3.. The series 2 = coverges ad the sum is. We compute the -th partial sum : s = = ( 2 2 = 2 2 which coverges to as, that is, 2 =. 7

8 Example 3.2. The series coverges for r <. What is the sum? Example 3.3. ( + ( + 2 = 2. Example 3.4. The harmoic series is diverget. Ideed, its sequece (s of partial sums is ubouded ad hece diverges. For ay K > 0, pick m > 2K, the for ay > 2 m, we have ( s = + ( ( ( m m > ( ( ( 2 + (2 + (4 + + (2 m m = m 2 > K which shows that (s is ubouded. Theorem 3.5. If a series a coverges, the a 0 as. CAUTION. If a 0, the series r a NEED NOT coverge! See Example 3.4. Theorem 3.6. Let a 0 for all N. The partial sums s. Example 3.7. The series a coverges if it has bouded 2 is coverget because s < 2 for all. Theorem 3.8. (Geeral Priciple of covergece A series oly if, for ay ε > 0, there exists N N such that m > > N = a + + a a m < ε. a coverges if, ad 8

9 The followig are three very useful tests for covergece of series. Theorem 3.9. (Compariso Test Let 0 a b for all (or, from some owards. If the series coverges, the a coverges. Equivaletly, if a diverges, the b b diverges. Theorem 3.0. (Root Test If a > 0 ad a r < for some fixed r, ad for all (or, from some owards, the a coverges. If a from some owards, the a diverges. Theorem 3.. (Ratio Test If a > 0 ad a + a r < for some fixed r, ad for all ( or, from some owards, the a coverges. If from some owards, the a + a a diverges. Example 3.2. What ca you say about the covergece or divergece of the followig series; A series of the form x! ; log ; ( ; 2 + si. ( + a with a 0 is called a alteratig series. Theorem 3.3. (Leibiz A alteratig series ( + a is coverget if the sequece (a decreases to 0, that is, (a is decreasig ad lim a = 0. Example 3.4. The series ( + coverges. 9

10 Defiitio 3.5. A series absolutely, if the series the we say that a a coverges. If is said to be absolutely coverget or, coverge a coverges, but a diverges, a coverges coditioally. Propositio 3.6. If a coverges, the a coverges; i other words, every absolutely coverget series is coverget. Cauchy Product Theorem 3.7. Let The the series Proof. Write A = The we have a ad c = p+q= b be absolutely coverget series. Defie a p b q = a 0 b + a b + + a b 0. c coverges absolutely ad ( ( c = a b. a ad B = b. Let s = a 0 + a + + a, t = b 0 + b + + b. w := s t AB as. Sice ( ( 0 c c ( a a ( b b a b, the series c coverges because it has bouded partial sums. 0

11 Next, we show ( ( c = a b. First, assume that a, b 0 for all. The the iequalities w [ 2 ] c c w ad the fact that both sequeces w ad w [ 2 ] coverge to AB yield c = lim (c c = AB. Fially, make o assumptio o a ad b. For ay x R, we ca write x = x + x with x +, x 0. Ideed, we ca let x + = max(x, 0 ad x = mi(x, 0. Now we have c = a p b q = (a + p a p (b + q b q p+q= = p+q= p+q= a + p b + q p+q= := x y u + v. a + p b q p+q= a p b + q + p+q= a p b q By the above argumets, we have ( ( x = u = ( a + ( b + a b +,, ( y = ( v = a + ( ( b a b,. Hece we have c = = x y u + ( a + a ( b + v b = AB.

12 4 Power series A series of the form a (x a is called a power series (i xcetred at a with coefficiets a. We ofte cosider the power series a x = a 0 + a x + a 2 x 2 + cetred at 0. Theorem 4.. Give ay power series holds: (i a x coverges oly at x = 0; (ii a x, oe of the followig three coditios a x coverges absolutely for all x R; (iii there exists R > 0 such that diverges for x > R. a x coverges absolutely for x < R ad Defiitio 4.2. Whe coditio (iii above occurs, the umber R > 0 is called the radius of covergece of the power series a x. For coditios (i ad (ii above, we defie the radius of covergece R to be 0 ad, respectively. Thus, every power series has a radius of covergece R. Defiitio 4.3. The domai of covergece of a power series a x is the set If a power series { x R : } a x coverges. a x has radius of covergece R, the its domai of covergece is a iterval with edpoits R ad R, for example, ( R, R or [ R, R, or some other form. Note that the domai of covergece always cotais the ope iterval ( R, R. 2

13 Give a power series a x, we ca compute its radius of covergece R by the followig formulae: R = = lim lim a + a a provided the above limits exist. Example 4.4. The power series x has radius of covergece R = lim /( + / = lim + =. The series coverges for x (,. At x =, the series diverges. At x =, ( the series coverges. So the domai of covergece is [,. Example 4.5. Fid the radius of covergece of each of the followig series: x!, ( x 2 (2!, x, (3 + ( x. 3

14 5 Limits of fuctios Defiitio 5.. Let a S R. Let f : S\{a} R be a fuctio. We say that f(x teds to l as x teds to a (i S if ε > 0, δ > 0 such that x S ad 0 < x a < δ = f(x l < ε. Usually, S is a ope iterval, we ofte omit the words i S if it is uderstood ad also say that f has a limit at a i the above situatio. Lemma 5.2. (Uiqueess of limit l = l. If f(x teds to l ad l as x teds to a, the Defiitio 5.3. If f(x teds to l as x teds to a, the we call l the limit of f(x as x teds to a ad write lim f(x = l. x a We ote that f eed ot be defied at a. Example 5.4. The fuctio f(x = x si x is defied o R\{0} ad Ideed, let ε > 0. Choose δ = ε. The lim x si x 0 x = 0. 0 < x 0 < δ = x < δ = x si x 0 = x si x x < δ = ε. The followig two propositios are very useful for computig limits of fuctios. Propositio 5.5. Give lim x a f(x = l ad lim x a g(x = ρ, we have lim(f + g(x = l + ρ; x a lim(fg(x = lρ; x a lim x a ( (x = f l if l 0. Further, if f(x g(x for all x, the l ρ. Propositio 5.6. If h(x f(x g(x ad Example 5.7. lim f(x = l. x a x lim x x = 2 ; lim si x x 0 x =. lim h(x = lim g(x = l, the x a x a 4

15 Oe-sided limits Defiitio 5.8. Let a S R ad let f : S\{a} R be a fuctio. We say that f(x teds to the limit l as x teds to a from the left ( or icreases to a (i S if ε > 0, δ > 0 such that x S ad a δ < x < a = f(x l < ε. Usually, S is a iterval with a ed poit a, we ofte omit the words i S if it is uderstood ad deote the limit by lim f(x = l, or lim f(x = l x a x a which is called the left-had limit of f at a. We defie the right-had limit of a fuctio likewise. Defiitio 5.9. Let a S R ad let f : S\{a} R be a fuctio. We say that f(x teds to the limit l as x teds to a from the right ( or decreases to a (i S if ε > 0, δ > 0 such that x S ad a < x < a + δ = f(x l < ε. Usually, S is a iterval with a ed poit a, we ofte omit the words i S if it is uderstood ad deote the limit by lim f(x = l, or lim f(x = l x a + x a which is called the right-had limit of f at a. Example 5.0. lim ( x [x] = 0, lim exp x 0 x = 0, lim x 0 + exp lim x ( +[x] = ; does ot exist. x Theorem 5.. A fuctio has a limit at a if, ad oly if, it has equal oe-sided limits at a, i other words, the followig two coditios are equal: (i (ii lim f(x = l ; x a lim f(x = lim f(x = l. x a x a + Applicatio of limits : Derivatives Let f : (a, b R be a fuctio ad let c (a, b. We say that f is differetiable at c if the followig limit exists f(x f(c lim x c x c i which case, the limit is called the derivative of f at c, ad is deoted by f (c. 5

16 Example 5.2. Let f(x = x. The f (c = c for all c R. More geerally, we have: Theorem 5.3. Let f(x = a x be a power series with radius of covergece R. The f is differetiable at every poit x ( R, R with derivative f (x = a x. Example 5.4. exp (x = ( + x + x2 2! + = + x + x2 2! si (x = (x x3 3! + = x2 + = cos (x. 2! + = exp (x ; Propositio 5.5. The expoetial fuctio exp : R (0, is a bijectio. Proof. See Appedix. Defiitio 5.6. The iverse exp : (0, R of the expoetial fuctio exp is called the atural logarithmic fuctio ad is deoted by Therefore we have for x i appropriate domais. log : (0, R. log exp(x = exp log(x = x 6

17 6 Cotiuous fuctios Let (a, b be a ope iterval i R. We iclude the case of (a, b = (, = R. Defiitio 6.. Let f : (a, b R. We say that f is cotiuous at a poit c (a, b if lim f(x = f(c, x c i other words, ε > 0, δ > 0 such that x c < δ = f(x f(c < ε. Example 6.2. Let f : (a, b R be a differetiable fuctio. The f is cotiuous. CAUTION. The coverse is false! The expoetial fuctio exp is differetiable, hece it is cotiuous. Propositio 6.3. Let f, g : (a, b R be cotiuous fuctios. The f + g, f g ad fg are cotiuous fuctios o (a, b. Further, if g(x 0 for every x (a, b, the the quotiet f g is cotiuous o (a, b. Proof. This follows from the arithmetics of limits i Propositio 5.5. Propositio 6.4. Let f, g : R R be cotiuous fuctios. The their composite f g : R R is also a cotiuous fuctio. Example 6.5. The Gaussia fuctio exp( x 2 is cotiuous. Propositio 6.6. Let f : R R be cotiuous at a poit c ad f(c > 0. The there exists δ > 0 such that f(x > 0 for all x (c δ, c + δ. Remark. We have similar result to the above for f(c < 0. A useful criterio of cotiuity is that a fuctio is cotiuous if, ad oly if, it preserves covergece of sequeces. Theorem 6.7. Let f : R R be a fuctio ad let c R. The followig coditios are equivalet: (i f is cotiuous at c ; (ii if lim x = c, the lim f(x = f(c. 7

18 Example 6.8. We have lim = because 0 < 2. The fuctio log is cotiuous at the poit, therefore lim log( = log = 0, or log lim = 0. Fially, cotiuous fuctios have the followig importat properties. Theorem 6.9. (Itermediate Value Theorem Let f : [a, b] R be a cotiuous fuctio such that f(af(b < 0. The there is a poit t (a, b satisfyig f(t = 0. Example 6.0. The followig equatio has o ratioal root, but has a egative root: 3x 5 + x 4 + x 3 + x 2 + x + = 0. Example 6.. Let f : [0, ] [0, ] be a cotiuous fuctio. The there is a poit t [0, ] such that f(t = t. Defiitio 6.2. A fuctio f : S R is said to be bouded if there is a costat M such that f(x M for all x S. Theorem 6.3. Let f : [a, b] R be a cotiuous fuctio. The f is bouded. Example 6.4. I the above theorem, it is importat that f is defied o a closed iterval. The theorem is false for cotiuous fuctios defied o ope itervals, for istace, the fuctio f(x = is cotiuous o the ope iterval o (0,, but is x ubouded! The followig theorem says that a cotiuous fuctio o a closed iterval achieves its supremum ad ifimum. Theorem 6.5. Let f : [a, b] R be a cotiuous fuctio. s, t [a, b] such that f(s = sup{f(x : x [a, b]}, f(t = if{f(x : x [a, b]}. The there exist 8

19 Appedix Theorem. The expoetial fuctio exp : R (0, is a cotiuous bijectio. Proof. We first show that exp(x + y = exp(x exp(y (x, y R. This follows from the product formula i Theorem 3.7: ( ( x y x k y k exp(x exp(y = =!! k!( k! k=0 ( ( ( = x k y k (x + y =! k! I particular, we have = exp(x + y. k=0 exp(x exp( x = exp(0 = ( ad exp(x 0 for all x R. Sice exp(x > 0 for x > 0, we have exp(x > 0 for all x R. By the defiitio of exp, we have exp(x > exp(y > exp(0 for x > y > 0. It follows from ( that exp( y > exp( x. Hece exp is strictly icreasig ad ijective. To show that exp is surjective, we apply the Itermediate Value Theorem sice exp is cotiuous. Let r (0,. If r =, the r = exp(0. If r >, the exp(r > r > exp(0 implies that r = exp(x for some x (0, r by Itermediate Value Theorem. If r <, the r > implies that = exp(y for some y R ad r therefore r = = exp( y. This proves surjectivity of exp. exp(y 9