# 1 Sets Functions and Relations Mathematical Induction Equivalence of Sets and Countability The Real Numbers...

Size: px
Start display at page:

Download "1 Sets Functions and Relations Mathematical Induction Equivalence of Sets and Countability The Real Numbers..."

Transcription

1 Contents 1 Sets Functions nd Reltions Mthemticl Induction Equivlence of Sets nd Countbility The Rel Numbers Sequences nd Convergence The Arithmetic of Sequences Subsequences Convergence of Sequences nd Series Open sets, Closed sets Limits of Functions 29 4 Continuitiy Uniform Continuity Properties of Continuous Functions Differentition The Men Vlue Theorem L Hôpitl s Rule The Riemnn Integrl Integrbility i

2 ii CONTENTS 6.2 Arithmetic of Integrls The Fundmentl Theorem

3 0 CONTENTS

4 Chpter 1 Sets A set is colletion of mthemticl objects. Sets, themselves, re mthemticl objects. One wy to think of sets is s jrs or continers tht hve mthemticl objects, possibly including other sets, inside of them. If X is set nd x is some mthemticl object, we write x X to denote the fct tht x is in the collection X. We red this s * x is n element of X or s x is in X. If this is not the cse, we write, insted, x X nd red it s x is not in X. We sy two sets re equl if they hve exctly the sme elements. Tht is, X = Y if nd only if for every mthemticl object x, we hve x X if nd only if x Y. There re two stndrd wys of describing sets. In the first, we simply list the elements of the set between brces ({}). So, for exmple, the empty set = {}. This is the set tht hs no elements in it. We could lso spek of the set A = {1, 2, 3, 4} where 1 A, but 5 A. Some cre is needed since, for exmple, { } is set with exctly one element in it (nmely, the empty set), so { }. In this wy, we hve the set N = {1, 2, 3, } of ll nturl numbers nd the set Z = {, 2, 1, 0, 1, 2, } of ll integers. The second wy to describe set is by considering ll x tht hve some property P (x). For this we write the set A = {x P (x)}. Then x A if nd only if P (x) is true. Now, in this P (x) should be sttement tht depends on x nd 1

5 2 CHAPTER 1. SETS is either true or flse for ech mthemticl object x. So, nother description of the emptyset is = {x x x}. In this, x x is our property P (x) nd this is flse for ll objects x. In this wy, we cn define Q = {x : there re m, n Z with x = m n nd n 0}, the set of ll rtionl numbers. Then 1 3 Q, for exmple. Definition 1.1 Let A nd B be sets. We sy A is subset of B nd write A B if for ll objects x, x A implies tht x B. In other words, everything in A is in B. Directly from the definitions of equlity of sets nd of subsets, we hve tht A = B if nd only if both A B nd B A. This is very common wy to how tht two sets re equl. Some cution with lnguge is good thing t this point. If A nd B re sets, the phrse A is contined in B could men either A B or A B. These re very different concepts nd it should lwys be mde cler which of these two possibilities is ment. For exmple {1} {1, 2}, but {1} {1, 2}. If X is set, we define the power set of X by P(X) = {A A X}. Hence, P(X) is the collection of ll subsets of X. Notice tht P(X) nd X P(X). As n exmple, if X = {1, 2}, then P(X) = {, {1}, {2}, {1, 2}}. Notice tht, here, we hve {1} P(X), but 1 P(X). Next if A nd B re sets, we define A B = {x x A nd x B} nd A B = {x x A or x B}. We cll these sets the intersection nd the union of A nd B, respectively. We hve the following: Proposition 1.2 Let A, B, d C be sets. then 1. A B = B A 2. A B = B A 3. (A B) C = A (B C) 4. (A B) C = A (B C) 5. A (B C) = (A B) (A C)

6 1.1. FUNCTIONS AND RELATIONS 3 6. A (B C) = (A B) (A C) Next, if A nd B re sets, we define the complement of B in A by A \ B = {x x A nd x B}. We then hve the following: Proposition 1.3 Let A, B, nd C be sets. Then, 1. A \ (A \ B) = A B 2. A \ (B C) = (A \ B) (A \ C) 3. A \ (B C) = (A \ B) (A \ C) 4. (A B) \ C = (A \ C) (B \ C) 5. (A B) \ C = (A \ C) (B \ C) 1.1 Functions nd Reltions If x nd y re mthemticl objects, we define the ordered pir (x, y) = {{x}, {x, y}}. The only property of (x, y) tht is required is the following: Proposition 1.4 We hve tht (x, y) = (, b) if nd only if x = nd y = b. A funtion is defined to be set f of ordered pirs such tht whenever (x, y) f nd (x, z) f, we my conclude tht y = z. In other words, there is t most one second component for ech first component for elements of f. In this cse, if we hve (x, y) f, we write y = f(x). If f is function, we define the domin of f by dom f = {x there is y such tht (x, y) f} nd the imge of f by im f = {y there is n x such tht (x, y) f}. If, now, A = dom f nd im f B, we sy tht f is function from the set A to the set B nd write f : A B. Notice tht the domin of f must be the

7 4 CHAPTER 1. SETS set A, but the imge does not hve to be the set B. The imge need only be subset of B. Now, let A nd B be sets, nd define the cross product A B = {(x, y) : x A, y B} = {z : prticulr, if f : A B, then f A B. there re x A nd y B such tht z = (x, y)}. In Now suppose tht Λ is set nd f is function with Λ = dom f such tht f(λ) is set for ech λ Λ. Let s write A λ for f(λ). Then, we sy tht {A λ } λ Λ is n indexed fmily of sets with index set Λ. For exmple, we might tke Λ = N nd define A n = { m n : m Z}. Then A n consists of those frctions whose demonintors re n. It is very common for the index set of fmily of sets to be either N or some subset of N. As nother exmple, we might hve Λ = Z nd define A n = {x Z : x > n}. Then A 0 = N, for exmple. Now, suppose tht {A λ } λ Λ is n indexed fmily of sets. union nd intersection of this fmily by nd A λ = {x there exists λ Λ with x A λ } λ Λ A λ = {x x A λ for ll λ Λ}. λ Λ We define the As n exmple, suppose tht Λ = N nd A n = { m n m Z}. We clim tht n N A n = Q nd n N A n = Z. As nothr exmple, if Λ = Z nd A n = {x Z x > n}, then n Z A n = Z nd n Z A n =. In generl, we write n=1 A n = n N A n whenever {A n } n N is fmily of sets indexed by N with similr nottion for intersections. We now hve: Proposition 1.5 Suppose tht X is set nd tht {A λ } λ Λ is n indexed fmily of sets. Then ) X \ ( λ Λ A λ) = λ Λ (X \ A λ) b) X \ ( λ Λ A λ) = λ Λ (X \ A λ)

8 1.2. MATHEMATICAL INDUCTION 5 Definition 1.6 Suppose tht f : X Y is function. We sy tht f is one-toone if the only wy for f(x 1 ) = f(x 2 ) is for x 1 = x 2. Alterntively, function is ne-to-one when x 1 x 2 implies tht f(x 1 ) f(x 2 ). Next, we sy tht f is onto if whenever y Y, there is n x X with y = f(x). In other words, f : X Y is onto when im f = Y. Finlly, we sy tht f is one-to-one correspondence (or n equivlence, or bijection) if f is both one-to-one nd onto. Definition 1.7 Suppose tht f : X Y nd g : Y Z re functions. We define the composition, g f : X Z by setting g f(x) = g(f(x)) for ll x X. Definition 1.8 A reltion is ny set of ordered pirs. If R is reltion, sy R X Y, then we cn define the opposite reltion ˆR = {(x, y) (y, x) R}. Notice tht ˆR = R. Proposition 1.9 A function f : X Y is one-to-one exctly when ˆf is function. In this cse, dom ˆf = im f nd im ˆf = dom f. In prticulr, if f : X Y is bijection, then ˆf : Y X is lso bijection. In this cse, we write f 1 for ˆf. Proposition 1.10 Suppose tht f : X Y nd g : Y Z. then we hve ) If both f nd g re one-to-one, so is g f. b) If both f nd g re onto, so is g f. c) If both f nd g re bijections, so is g f. d) If g f is onto, so is g. e) If g f is one-to-one, so is f. 1.2 Mthemticl Induction We ssume property of the nturl numbers clled well-ordering. It sys tht if A N is non-empty, then there is smllest element of A. In other words, there is n A such tht x for every x A.

9 6 CHAPTER 1. SETS Theorem 1.11 Suppose tht P (n) is property tht nturl number cn either hve or not hve. Suppose 1. P (0) is true. 2. For every n N, if P (n) is true, then P (n + 1) follows. Then P (n) is true for ll n N. Proof: Let A = {n N : P (n) is true}. Then 0 A, so A is not empty. We hve to show tht A = N. Suppose this is not the cse, so N \ A. Let be the smllest element of N \ A. Then 0, so 1 N. But then 1 A, so P ( 1). By the second hypothesis, we cn conclude tht P (( 1) + 1) is true. This show tht A, contrry to the definition of. 1.3 Equivlence of Sets nd Countbility Definition 1.12 We sy two sets X nd Y re equivlent if there exists bijection f : X Y. In this cse, we write X Y. Theorem 1.13 Let X, Y, nd Z be sets. Then ) X X. b) If X Y, then Y X. c) If X Y nd Y Z, then X Z. Definition 1.14 We sy set A is finite if it is either empty or there is n n N such tht A {1, 2,, n}. We sy tht A is countbly infinite if A N nd tht A is countble if it is either finite or countbly infinite. Proposition 1.15 A set A is countble if nd only if there is subset B N with A B. Proof:

10 1.3. EQUIVALENCE OF SETS AND COUNTABILITY 7 The forwrd direction is cler from the defintion of countbility. So now suppose tht B N nd tht A B. To prove the result, it is enough to show tht eitherb =, B {1,, n} for some n N or tht B N. If B is not empty, then there is smllest element of B. We cll this element b 1. Now, if B = {b 1 }, then B {1}, so we re done. Otherwise, there is smllest element b 2 of B \ {b 1 }. If B = {b 1, b 2 }, then B {1, 2}, so we re done. We proceed by induction. Either this process stops t some point, in which cse B {1,, n} for some n N, or B = {b 1, b 2, }, so B N. Proposition 1.16 Subsets of countble sets re countble. Proposition 1.17 N N is countble. Proof: Define f : N N N by f(m, n) = 2 m 3 n. By unique fctoriztion, f is one-to-one. Thus N N im f N, so N N is countble. Theorem 1.18 A set A is countble if nd only if there is one-to-one mp f : A N. f : N A. Similrly, A is countble if nd only if there is n onto mp Proof: If f : A N is one-to-one, then A im f N, so A is countble. On the other hnd, if f : N A is onto, then for ech A, there is n n N with f(n ) =. Define g : A N by g() = n. Then g is one-to-one, so A is countble. Conversely, if A is countble, the definition provides f : A N which is one-to-one. If A is countbly infinite, then the definition lso gives n f tht is onto. Now, if A is finite, choose f : {1,, n} A which is one-to-one nd onto. Then define F : N A by F (k) = f(k) if k n nd F (k) = f(1) if k > n. Then F is onto. Theorem 1.19 Supose tht {A n } n=1 is fmily of countble sets indexed by N. Then A = n=1 A n is countble.

11 8 CHAPTER 1. SETS Proof: Choose, for ech n, n onto function f n : N A n. Now define f : N N A by f(n, m) = f n (m). Then f is onto, so A is countble. It is good question whether every set is countble. Tht this is not the cse follows from the following. Proposition 1.20 Let X be set. Then X P(X). Proof: Suppose tht f : X P(X) is ny mp. We will show tht f cnnot be onto. Hence there is no bijection from X to P(X). Let A = {x X : x f(x)}. Then A X, so A P(X). Suppose, though tht f(y) = A. There re two choices for y. Either y A, in which cse the definition of A shows tht y f(y) = A, which is contrdiction, or y A = f(y), n which cse y A. Since either cse leds to contrdiction, no such y cn exist, so f is not onto. We will lter show tht the set of rel numbers, R is uncountble. 1.4 The Rel Numbers We ssume the existence of set R nd two opertions + : R R R nd : R R R. We usully write x+y nd x y for +(x, y) nd (x, y), resp. We lso ssume there is reltion < R R where we write x < y for (x, y) <. We further ssume the following properties: 1. For ll x, y, z R, we hve (x + y) + z = x + (y + z). 2. For ll x, y R, we hve x + y = y + x. 3. There is n element 0 R such tht x + 0 = x for ll x R. 4. For ech x R, there is y R such tht x + y = For ll x, y, z R, we hve (x y) z = x (y z). 6. For ll x, y R, x y = y x. 7. There is n element 1 R with 0 1 nd x 1 = x for ll x R.

12 1.4. THE REAL NUMBERS 9 8. For ech x 0 in R, there is y R with x y = For ll x, y, z R, we hve x (y + z) = x y + x z. 10. If x < y nd z R, then x + z < y + z. 11. If 0 < x nd y < z in R, then x y < x z. 12. If x, y, z R nd x < y nd y < z, then x < z. 13. If x, y R, exctly one of the following is true: x < y,y < x, or x = y. We write x y if x < y or x = y. Similrly, we write x > y is y < x nd x y if x > y or x = y. Notice tht ll the xioms so fr re stisied by Q with the usul opertions. This mens tht the finl xiom is our only wy to distinguish Q from R. Proposition 1.21 ) The element 0 bove is unique. b) For ech x, the element y such tht x + y = 0 is unique. We cll it x. c) The element 1 boce is unique. d) For ech x 0, the element y with x y = 1 is unique. We cll it 1 x. e) If x < y, then y < x. f) If 0 < x < y, then 0 < 1 y < 1 x. g) 0 < 1. h) If x R, then 0 x 2. Definition 1.22 For < b in R, define [, b] = {x R : x b}. Similrly, define (, b), [, b), nd (, b]. Also, define [, ) = {x R : x}. Similrly define (, ), (, ), nd (, ]. To understnd our finl xiom, we need some terminology. Suppose tht A R. We sy tht A is bounded bove if there is n x R such tht x for ll A. We lso sy tht x is n upper bound for A in this cse.

13 10 CHAPTER 1. SETS We sy tht x 0 R is lest upper bound for A if it is n upper bound for A nd if x is ny upper bound for A, we hve x 0 x. In other words, x 0 is the smllest possible upper bound for A. Our xiom is then Lest Upper Bound Property: Every non-empty subset of R with n upper bound in R hs lest upper bound in R. Proposition 1.23 For ech x R, there is n n N with x n. Proof: Assume not. Then there is n x R with n < x for ll n N. In other words, x is n upper bound for N. By our xiom, there is lest upper bound for N. Cll it x 0. Then n x 0 for ll n N. But, since x 0 1 < x 0, x 0 is NOT n upper bound for N. This mens tht there is n n 0 N with x 0 1 < n 0. But then x 0 < n 0 + 1, which contrdicts tht x 0 is n upper bound for N. Proposition 1.24 If x R, there is n n Z such tht n x < n + 1. Proof: First ssume tht x 0. By the previous result, there is n n N with x < n. By the fct tht N is well-ordered, there is smllest n 0 N such tht x < n 0. But then n 0 1 x, so n = n 0 1 works in the result. Now suppose tht x < 0. Then x > 0, so there is n n 0 Z with n 0 x < n But then n 0 1 < x n 0. If x n 0, let n = n 0 1. Otherwise, let n = n 0. Proposition 1.25 Let < b in R. Then, there is n r Q with < r < b. Proof: First, b > 0, so 1/(b ) > 0, so there is n n N with 1/(b ) < n. now find m Z such tht m < n b m + 1. Then we hve 1 < n (b ) = nb n, so n < nb 1 m < nb, so < m/n < b. let r = m/n. Proposition 1.26 There is no x Q with x 2 = 2 (here, 2 = by definition). There is n x R with x 2 = 2. In fct, for ech 0, there is n x 0 with x 2 =. Proof: First ssume tht 1. Consider the set A = {x R : x > 0, x 2 }. Notice tht 1 A, so A is non-empty. Also, suppose tht x R. Then either

14 1.4. THE REAL NUMBERS 11 x <, x =, or x >. In the lst cse, though x 2 = x x > x > 1 =, so x A. In other words, x A implies tht x, so is n upper bound for A. Using our xiom, let x 0 be the lest upper bound for A. The clim is tht x 2 0 =. To show this, we note tht the only other possibilities re tht x 2 0 < or x 2 0 >. We eliminte these two possibilities. Cse I: x 2 0 <. Let ε = min{1, ( x 2 0)/(2x 0 + 1)}. Then ε > 0,so x 0 + ε > x 0. But (x 0 + ε) 2 = x x 0 ε + ε 2 x x 0 ε + ε x (2x 0 + 1)ε x ( x 2 0) =. This mens tht x 0 + ε A, which contrdicts the fct tht x 0 is n upper bound for A. Cse II: x 2 0 >. In this cse, let ε = (x 2 0 )/(2x 0. Then (x 0 ε) 2 = x 2 0 2x 0 ε + ε 2 > x 2 0 2x 0 ε = x 2 0 (x 2 0 ) =. Now, if x > x 0 ε, we hve tht x 2 > (x 0 ε) 2 >, so x A. in other words, if x A, x x 0 ε. This shows tht x 0 ε is n upper bound for A. But, x 0 ε < x 0, which contrdicts tht x 0 is the lest upper bound for A. The only other possibility is tht x 2 0 =. This is our result for 1. Now, if = 0, x = 0 works for x 2 =. If 0 < < 1, then 1/ > 1, so by the previous cse there is n x 0 with x 2 0 = 1/. But now (1/x 0 ) 2 =, so we hve the result in this cse lso. Proposition 1.27 Let x < y in R. x < z < y. Then there is z R \ Q such tht Proof: Since x < y, x 2 < y 2. By previous result, there is r Q with x 2 < r < y 2. Then x < r + 2 < y. Then, with z = r + 2, we hve the result. Finly, we mke definition. Definition 1.28 For x R, we define x to be x if x 0 nd to be x if x < 0. Notice tht x 0 in every cse nd tht x x x. Proposition 1.29 We hve the folowing:

15 12 CHAPTER 1. SETS ) xy = x y b) x + y x + y c) x < is equivlent to < x <. d) If b R, then x b < is equivlent to b < x < b +.

16 Chpter 2 Sequences nd Convergence Definition 2.1 A sequence is function with domin N. We will be concerned with rel-vlued sequences, which re functions : N R. Insted of (n), it is trditionl to write n nd to write { n } n=1 for the sequence. Typiclly, formul is given for n in terms of n N. For exmple, if we let = { 1 n } n=1, we hve n = 1 n, so for exmple 3 = 1 3. As nother exmple, let b = {1 + ( 1) n } n=1. Then b 1 = 0, b 2 = 2, b 3 = 0,. Definition 2.2 Let { n } n=1 be sequence nd A R. We sy tht { n } n=1 converges to A if for every ε > 0, there is n N N such tht whenever n N, we hve n A < ε. This will be written s n A. As n exmple, we show tht { 1 n } n=1 converges to 0. Suppose tht ε > 0. Choose N N such tht 1 ε < N. Then, whenever n N, we hve 1 n 0 = 1 n 1 N < ε. On the other hnd, we show tht { 1 n } n=1 does NOT converge to 1. To do this, we need to negte the definition of convergence. Hence, we need to show tht there is n ε > 0 such tht for ll N N, there is n n N with 1 n 1 ε. I tke ε = 1 2. For ny N N, pick n mx{2, N}. Then 1 n 1 = 1 1 n = ε. In this definition, ε represents how close we wnt the terms ot he sequence to be to A nd N is how fr out in the sequence we hve to go to gurntee 13

17 14 CHAPTER 2. SEQUENCES AND CONVERGENCE tht the sequence is within this degree of closeness to A. It is cler tht if N works for given ε, ny lrger nturl number will lso. The first thing we wnt to do is show tht sequence cnnot converge to two different rel numbers simultneously. Proposition 2.3 Suppose tht the sequence { n } n=1 converges to both A nd B. Then A = B. Proof: Suppose not, in other words, suppose tht A B. Then ε = A B /3 > 0. Since { n } n=1 converges to A, there is n N 1 N such tht whenever n geqn 1 we hve n A < ε. SImilrly, there is n N 2 N such tht whenever n N 2, we hve n B < ε. Now, to obtin our contrdiction, let n N be chosen so tht n N 1, N 2. We then hve tht A B A n + n B ε + ε = 2ε = 2 A B. 3 This is contrdiction since A B > 0. Definition 2.4 We sy sequence { n } n=1 converges if there exists n A R such tht { n } n=1 converges to A. By the previous result, if such n A exists, it is unique. As nother exmple, we show tht the sequence { n = 1 + ( 1) n } n=1 does not converge. In fct, suppose tht n A nd let ε = 1. Then, there is n N N such tht whenever n N we hve n A < ε. Let n be ny odd integer with n N. Then n + 1 N lso, so we hve n A, n+1 A < ε. But n = 0 nd n+1 = 2, so A < ε nd 2 A < ε. But this shows tht 2 = 2 = 2 A + A < ε + ε = = 2, contrdiction. We will find mny more cses of sequences tht fil to converge in this chpter. 2.1 The Arithmetic of Sequences Proposition 2.5 Suppose tht n A nd b n B. Then n + b n A + B.

18 2.1. THE ARITHMETIC OF SEQUENCES 15 Proof: Let ε > 0. Choose N 1 N such tht whenever n N 1, we hve n A < ε/2. Similrly, choose N 2 N such tht whenever n N 2, we hve b n B < ε/2. Let N mx{n 1, N 2 }. Then, if n geqn, we hve ( n + b n ) (A + B) = n A + b n B n A + b n B < ε/2 + ε/2 = ε. It should be pointed out tht essentilly the sme proof shows tht n b n A B. Now let s consider wht it will tke to show tht n b n AB. Clerly, we hve to consider the expression n b n AB. The quetion is how to mke this smll (i.e., less thn ε) given the knowledge tht n A nd b n B re both smll. After some plying round, nd remembering trick used in the proof of the product rule for derivtives, we dd nd subtrct n B nd rgue tht n b n AB = n b n n B + n B AB n b n B + B n A. Now, in both terms we hve some expressions tht re smll. In the second, the expression is multiplied by B, so we cn del with it by using ε/ B insted of ε in the defintion of convergence of n. But the first term hs n, which depends on n nd so cnnot simply be divided off the ε. But, if we cn show tht n never gets too lrge, we cn still mke things work. We give definition first. Definition 2.6 We sy sequence { n } n=1 is bounded if there exists M R such tht n M for ll n N. As n exmple, the sequence {( 1) n } n=1 is bounded (we cn tke M = 1), but it not convergent. However, we do hve the following positive result: Proposition 2.7 Suppose tht { n } n=1 converges. Then it is bounded. Proof: Suppose tht n A. In the definition of convergence, tke ε = 1 nd find N N such tht whenever n N, we hve n A < ε = 1. Then, whenever n N, we lso hve n = n A + A n A + A < 1 + A. Now, if we let M = mx{ 1, 2,, N 1, 1 + A }, we hve tht n M for ll n N. Now we cn prove Proposition 2.8 Suppose tht n A nd b n B. Then n b n AB.

19 16 CHAPTER 2. SEQUENCES AND CONVERGENCE Proof: Let ε > 0. Since { n } n=1 converges (to A), it is bounded. Let M R with n M for ll n N. We my ssume M 0 (if it ws, M = 1 works just s well). Now, find N 1 N such tht whenever n N 1 we hve n A < ε/2( B + 1). We use B + 1 insted of B just in cse B = 0. Also, pick N 2 N such tht whenever n N 2, we hve b n B < ε/2m. Let N = mx{n 1, N 2 }. Then, for ny n N, we hve n b n AB = n b n n B + n B AB n b n B + B n A < M ε/2m + B ε/2( B + 1) < ε/2 + ε/2 = ε nd we re done. If we try to pply this result to the sequence { ( 1)n n } n=1, by writing n = ( 1) n nd b n = 1 n, we find tht it doesn t work becuse {( 1)n } n=1 doesn t converge. Nonetheless, we cn show tht { ( 1)n n } n=1 converges to 0 s follows: Proposition 2.9 Suppose tht { n } n=1 is bounded nd b n 0. Then n b n 0. Proof: Let ε > 0 nd M R such tht n M for ll n N. We my ssume M 0. Choose N N such tht whenever n N, we hve b n 0 < eps/m. Then, for these sme n, we hve n b n 0 = n b n M b n 0 < M ε/m = ε. If { n } n=1 is not bounded, this result fils. For exmple, tke n = n nd b n = 1/n. Then b n 0, yet n b n 1. Of course, our next step is to show tht division works for limits of sequences. We hve to be sure both tht the sequence we divide by is never 0 nd tht the limit is not 0 just to be sure everything we wnt is defined. Our first step is to show tht reciprocls work well. So, suppose tht n 0 for ll n nd tht n A with A 0. How do we show tht 1 n 1 A? We hve to consider 1 n 1 A = A n n A. Now, the numertor converges to 0 by ssumption, so if we 1 cn show tht n A is bounded, we will be done by the following result. Proposition 2.10 Let { n } n=1 be sequence. Then n A if nd only if n A 0.

20 2.1. THE ARITHMETIC OF SEQUENCES 17 This is n ssigned homework problem. Another useful, if esy result: Proposition 2.11 Suppose tht n b n for ll n N nd suppose tht b n 0. Then n 0. Proof: Suppose ε > 0 nd choose N N such tht n N gives b n 0 < ε. Then n 0 = n b n < ε. Lemm 2.12 Suppose tht n 0 for ll n N nd n A with A 0. Then { 1 n } n=1 is bounded. Proof: In the definition of convergence, choose ε = A /2 > 0 nd find N N so tht n N implies tht n A < ε = A /2. Then, for these n, we hve A = A n + n n A + n < A /2+ n, so n > A A /2 = A /2. But then 1 n < 2/ A. Now, let M = mx{ 1 1,, 1 N 1, 2/ A } to se tht 1 n M for ll n N. Proposition 2.13 Suppose tht n 0 for ll n N nd tht n A with A 0. Then 1 n 1 A. Proof: We need to show tht 1 n 1 1 A 0. So, n 1 A = A n = 1 n A A 1 n n A. Since { 1 n } n=1 is bounded nd 1 A is constnt, nd n A 0, we obtin our result. Proposition 2.14 Suppose tht n 0 for ll n N nd n A 0. Suppose lso tht b n B. Then bn n B A. Proof: b n n = b n 1 n B 1 A = B A. It is often the cse tht none of our results so fr pplies to given sequence, but the sequence cn be mnipulted so tht these results pply. Exmple: Let n = 2n+1 3n+2. Since neither {2n + 1} n=1 nor {3n + 2} n=1 converge, the result on division of convergent sequences does not pply. However, if we multiply nd divide by 1 2n+1 n, we see tht 3n+2 = 2+ 1 n 3+ 2 n. SInce 1 n 0, we see

21 18 CHAPTER 2. SEQUENCES AND CONVERGENCE tht n = 2 nd n = 3, so we find tht 2n+1 3n Much more complicted results cn be obtined in this wy. Finlly, we give result showing how convergence of sequences reltes to the order properties of R. Proposition 2.15 Suppose tht n A, b n B nd n b n for ll n N. Then A B. Proof: Suppose not, tht is B < A. Let ε = (A B)/2 > 0. Find N N such tht whenever n N, we hve both n A < ε nd b n B < ε. Then, for every n N, we hve b n < B + ε = B + (A B)/2 = (A + B)/2 = A (A B)/2 = A ε < n, in contrdiction to our ssumption. It should be pointed out tht the previous result is flse if is replced by < throughout. In fct, let n = 1 1 n nd b n = n. Then n < b n for every n. But, n 1 nd b n 1, so we hve equlity of the limits. 2.2 Subsequences Definition 2.16 Suppose tht { n } n=1 is sequence nd n 1 < n 2 < is sequence of nturl numbers. sequence { n } n=1. We sy tht { nk } k=1 is subsequence of the For exmple, suppose tht n = ( 1) n. If we choose n k = 2k, we get tht nk = ( 1) 2k = 1 for ll k N. Alterntively, if we chose m k = 2k 1, then mk = 1 for ll k N. So, even though the originl sequence fils to converge, these prticulr subsequences converge. Since the originl sequence is subsequence of itself, it is cler tht some subsequences fil to converge. A positive result in this direction is the following. Proposition 2.17 Suppose tht n A. Then for ny subsequence, nk A. Proof: First notice tht since n 1 N, we hve 1 n 1. Also, if k n k for some k N, we hve tht k + 1 n k + 1 n k+1. Thus, indiction shows tht k n k for every k N.

22 2.2. SUBSEQUENCES 19 Now, let ε > 0 nd find N N such tht whenever n N, we hve n A < ε. Then, if k N, we hve n k k N, so nk A < ε. Thus nk A. This is ctully very hndy wy to show tht some sequences fil to converge. If we cn find two subsequences tht converge to different limits, the whole sequence cnnot converge t ll. For exmple, suppose tht n = ( 1) n (1 + 1 n ). Then, it is esy to see tht 2k = k 2k+1 = ( k+1 ) 1. Hence { n} n=1 cnnot converge. 1 while Definition 2.18 Let { n } n=1 be sequence nd A R. We sy tht A is cluster point of { n } n=1 if for ll ε > 0, the set {n N : n A < ε} is infinite. For exmple, 1 is cluster point of the sequence {( 1) n } n=1. On the other hnd, the sequence {n} n=1 hs no cluster points t ll. More interestingly, we know tht Q is countble set, so there is n onto function : N Q. This function is, by definition sequence. Furthermore, if A R is ny rel number nd ε > 0, we know tht there re infinitely mny rtionl numbers in the open intervl (A ε, A + ε), so the set {n N : n A < ε} is infinite. Hence every rel number is cluster point of this sequence! Theorem 2.19 A point A is cluster point of the sequence { n } n=1 if nd only if there is subsequence { nk } k=1 such tht n k A. Proof: First ssume tht there is subsequence with nk A. Let ε > 0 nd choose N so tht k N implies tht nk A < ε. Then every n k with k N is in the set {n N : n A < ε}, so this set is infinite. This A is cluster point of the sequence { n } n=1. Conversely, suppose tht A is cluster point of { n } n=1. We need to find subsequence which converges to A. First, tke ε = 1 in the definition of cluster point to see tht there is n n 1 such tht n1 A < 1 (in fct, there re infinitely mny!). Now, suppose tht n 1 < n 2 < n k hve been chosen. Choose ε = 1 k+1 in the definition of cluster point to see tht the set {n N : n A < 1 k+1 } is

23 20 CHAPTER 2. SEQUENCES AND CONVERGENCE infinite. In prticulr, there is n n k+1 in this set nd with n k < n k+1. In this wy, we inductively find n 1 < n 2 < such tht nk A < 1 k for ll k N. Since 1 k 0, we see tht n k A 0, so nk A. Theorem 2.20 (Bolzno-Weierstrss Theorem) Every bounded sequence hs cluster point. Proof: Suppose tht { n } n=1 is bounded sequence with n M for ll n N. Let x 0 = M nd y 0 = M. Then x 0 n y 0 for ll n N. In prticulr, the set {n N : x 0 n y 0 } is infinite. Let z 0 = (x 0 + y 0 )/2 nd notice tht {n N : x 0 n y 0 } = {n N : x 0 n z 0 } {n N : z 0 n y 0 }. Hence t lest one of the two sets on the right is infinite. If the first one is, let x 1 = x 0 nd y 1 = z 0. If not, let x 1 = z 0 nd y 1 = y 0. Then, we hve tht {n N : x 1 n y 1 } is infinite nd tht y 1 x 1 = (y 0 x 0 )/2. If we ssume tht we hve chosen x 0 x k < y k y 0 such tht {n N : x k n y k } is infintie nd such tht y k x k = (y 0 x 0 )/2 k, then let z k = (x k +y k )/2 so tht z k x k = y k z k = (y k x k )/2 = (y 0 x 0 )/2 k+1. Now {n N : x k n y k } = {n N : x k n z k } {n N : z k n y k }, so t lest one of the sets on the right is infinite. if it is the first, let x k+1 = x k nd y k+1 = z k, nd otherwise let x k+1 = z k nd y k+1 = y k. In this wy, the induction continues. In this wy, we obtin sequences {x n } n=1 nd {y n } n=1 such tht x 0 x 1 x k y k y 1 y 0 such tht the set {n N : x k n y k } is infinite nd such tht y k x k = (y 0 x 0 )/2 k for ech k N. There re now two things to notice: the first is tht the set {x k : k N} is bounded bove (by y 0, for exmple). The second is tht y k x k = (y 0 x 0 )/2 k 0. Let A = sup{x k : k N}. The clim is tht A is cluster point of the originl sequence { n } n=1. To see this, let ε > 0. Then A ε is not n upper

24 2.3. CONVERGENCE OF SEQUENCES AND SERIES 21 bound for {x k : k N}, so there is K 1 N with A ε < x K1 A. Notice tht for ll k K 1, we hve A ε < x K1 leqx k A. Next, there is K 2 such tht k K 2 implies tht y k x k < ε. Choose ny k K 1, K 2. Then, we hve A ε < x k < y k = x k + (y k x k ) < x k + ε A + ε. In other words, [x k, y k ] (A ε, A + ε). But then, {n N : x k n y k } {n N : n A < ε}. Since the first set is infinite, so is the second. This proves tht A is cluster point, proving the result. Corollry 2.21 (Sequentil comleteness of bounded sets) Every bounded sequence hs convergent subsequence. Theorem 2.22 A bounded sequence converges if nd only if every convergent subsequence converges to the sme vlue. Proof: One direction is esy since if { n } n=1 converges to A, every subsequence converges to A lso. Hence every convergent subsequence converges to A. Now, suppose tht { n } n=1 is bounded sequence nd tht every convergent subsequence converges to A. We need to show tht n A. Assume not. Then, there is n ε > 0 such tht for every N N, there is n N with n A ε. Pick n 1 such tht n1 A ε. Next, pick n 2 n with n2 A ε. Continue in the wy to find n 1 < n 2 < with nk A ε for ll k N. The problem is tht the sequence { nk } k=1 probbly does not converge. However, it is bounded sequence, so it hs subsequence { nkj } which does converge. But this is subsequence of { n } n=1 which does not converge to A, which violtes our ssumption. 2.3 Convergence of Sequences nd Series Definition 2.23 We sy sequence { n } n=1 is non-decresing if n n+1 for ll n N. Similrly, we sy the sequence is non-incresing if n+1 n for ll n N. Finlly, we sy { n } n=1 is monotone if it is either non-decresing or non-incresing.

25 22 CHAPTER 2. SEQUENCES AND CONVERGENCE Theorem 2.24 A non-decresing sequence is convergent if nd only if it is bounded. Proof: We know tht every convergent sequence is bounded, so one direction is lredy known. For the converse, ssume tht { n } n=1 is non-decresing nd bounded. Consider the set { n : n N}. It is non-empty nd bounded bove, so it hs lest upper bound. Let A = sup{ n : n N}. We clim tht n A. In fct, suppose tht ε > 0. Since A ε is not n upper bound, there is N N such tht A ε < N A. But then, for ech n N, we hve A ε < N n A < A + ε, so n A < ε. As n exmple, let n = n. Using induction, it is esy to see tht n = n. Hence, n 2. But now, consider b n = n!. Then {b n } n=1 is clerly non-decresing nd b n n < 2, so it is bounded lso. Hence {b n } n=1 converges. This is very useful wy to show sequences correspoonding to infinite sums re convergent. Definition 2.25 We sy tht sequence { n } n=1 is Cuchy if whenever ε > 0, there is N N such tht whenever n, m N, we hve n m < ε. The difference between convergence to point nd being Cuchy is tht for n A, we know the limit point (A), nd the terms of the sequence re getting close to A. For Cuchy sequences, we only know tht the terms re getting closer to ech other. Proposition 2.26 Every convergent sequence is Cuchy. Proof: Suppose tht n A nd tht ε > 0. Choose N N such tht for ll n N we hve n A < ε/2. Then, whenever m, n N, we hve n m = ( n A) ( m A) n A + m A < ε/2 + ε/2 = ε. Proposition 2.27 A Cuchy sequence with convergent subsequence is, itself, convergent.

26 2.3. CONVERGENCE OF SEQUENCES AND SERIES 23 Proof: Suppose tht { n } n=1 is Cuchy nd tht nk A. We show tht n A. Let ε > 0. Choose N 1 N such tht m, n N implies tht n m < ε/2. Also, choose N 2 N such tht k N 2 implies tht nk A < ε/2. Now, let n N be ny nturl number. Choose ny k N, N 2. Then n k N so n A n nk + nk A < ε/2 + ε/2 = ε nd we re done. Proposition 2.28 Every Cuchy sequence is bounded. Proof: Suppose tht { n } n=1 is Cuchy equence. Choose ε = 1 in the definition of Cuchy to find N N such tht whenever m, n N we hve n m < 1. Then, for ech n N we hve n N + N n < N +1. Now let M = mx{ 1,, N 1, N +1} to see tht n M for ll n N. Theorem 2.29 A sequence is convergent if nd only if it is Cuchy. Proof: We hve shown tht convergent sequences re Cuchy. Conversely, A Cuchy sequence is bounded, nd so hs convergent subsequence. But then, the originl sequence must converge. While the previus proof ws esy, it required the very difficult Bolzno- Weiestrss theorem for its proof. This mkes it mjor result. Definition 2.30 Let { n } n=1 be sequence. We cn construct nother sequence {s n } n=1 of prtil sums by setting s n = n. We sy the series n=1 n converges if nd only if {s n } n=1 converges. Furthermore, if s n A, we write n=1 n = A. Proposition 2.31 If n=1 n converges, then n 0. Proof: Let {s n } n=1 be the sequence of prtil sums nd ssume tht s n A. Then n = s n s n 1 A A = 0. Proposition 2.32 Suppose tht n 0 for ll n N. Then n=1 n converges if nd only if the sequence of prtil sums is bounded.

27 24 CHAPTER 2. SEQUENCES AND CONVERGENCE Proof: Notice tht the sequence of prtil sums {s n } n=1 is non-decresing since n 0. Hence, it converges if nd only if it is bounded. Proposition 2.33 Supose tht n b n nd tht n=1 b n converges. Then n=1 n converges. Proof: Let {s n } n=1 be the sequence of prtil sums for n=1 n nd {t n } n=1 the sequence of prtil sums for n=1 b n. Then, since {t n } n=1 converges, it is Cuchy. We show tht {s n } n=1 is lso Cuchy, so it, in turn, converges. Let ε > 0. Choose N so tht m, n N implies tht t n t m < ε. We my ssume m < n. But then s n s m = m+1 + n m+1 + n b m+1 + b n = t n t m < ε nd we re done. Theorem 2.34 (Alternting Series Test) Suppose tht { n } n=1 is non-incresing with n 0 for ll n. Then n=1 ( 1)n+1 n converges if nd only if n 0. Proof: Let {s n } n=1 be the sequence of prtil sums nd let x n = s 2n 1 nd y n = s 2n. Then, x n+1 = s 2n+1 = S 2n 1 2n + 2n+1 x n nd y n+1 = s 2n+2 = s 2n + 2n+1 2n+2 s 2n = y n since n is non-incresing. Thus, {y n } n=1 is non-decresing nd {x n } n=1 is non-incresing. Next, y n = s 2n = n = 1 ( 2 3 ) ( 4 5 ) 2n 1, so {y n } n=1 is bounded bove. Similrly, {x n } n=1 is bounded below. Hence, both {x n } n=1 nd {y n } n=1 converge. Let x n A nd y n B. Then x n y n = 2n 0, so A B = 0, so A = B. Thus, s 2n nd s 2n 1 both convege to A, so s n A. In other words, our series converges. As n exmple, ( 1) n+1 n=1 n converrges even though n=1 1 n does not. 2.4 Open sets, Closed sets Definition 2.35 We sy tht subset U R is open if, for every x U, there is n ε > 0 such tht (x ε, x + ε) U. We sy subset C R is closed if R \ C is open.

28 2.4. OPEN SETS, CLOSED SETS 25 For exmple, let R nd set U = (, + ). We show tht U is open. In fct, if x U, then < x. Let ε = x > 0. Then (x ε, x+ε) = (, 2x ) (, + ) = U. Similrly, (, ) is lso open. Proposition 2.36 Suppose tht U nd V re open sets. Then U V is open. Proof: Let x U V. Then, there is n ε 1 > 0 such tht (x ε 1, x + ε 2 ) U nd n ε 2 > 0 such tht (x ε 2, x + ε 2 ) V. Let ε = min{ε 1, ε 2 }. Then (x ε, x + ε) U V. In prticulr, if < b, then (, b) = (, + ) (, b) is open. Hence open intervls re open. Proposition 2.37 Suppose tht {U α } α I Then α I U α is open. is n indexed fmily of open sets. Proof: Suppose tht x U = U α. Then there is n α I with x U α. Since U α is open, ther eis n ε > 0 such tht (x ε, x + ε) U α U. Thus, every union of open intervls is n open set. Next, if R, then [, + ) is closed becuse R\[, + ) = (, ) is open. Also, (, ] is closed. We hve the following using DeMorgn s theorem. Proposition 2.38 Suppose tht C nd D re closed sets. Then C D is closed. Proposition 2.39 Supose tht {C α } is n indexed fmily of closed sets. Then Cα is closed. The investigtion of open nd closed sets is prt of the subject mtter of topology. It turns out to be very useful to use convergence of sequences to determine when set is open or closed. The following provides route to do this. Theorem 2.40 Let C R. The following re equivlent: ) C is closed. b) Whenever { n } n=1 is sequence with n C for ll n N nd n x, then we cn conclude tht x C.

29 26 CHAPTER 2. SEQUENCES AND CONVERGENCE Proof: Suppose tht C is closed nd tht { n } n=1 is sequence with n C nd n x. Suppose lso tht x C. Then x R \ C, which is n open set. Hence, there is n ε > 0 with (x ε, x + ε) R \ C. Since n x, there is n N N such tht whenever n N, we hve n x < ε, which implies tht x ε < n < x + ε. But this implies tht n R \ C for ll n N, in contrdiction to our ssumption. For the converse, suppose tht C is not closed. Then R \ C is not open, which mens there is x R\C such tht whenever ε > 0, the set (x ε, x+ε) is not contined in R \ C. But this mens tht for ech ε > 0, there is n C with x < ε. Now, for ech n N, choose n C with n x < 1 n. Then, n x nd n C for ll n N but x C. In other words, if C is not closed, the second condition fils lso. Definition 2.41 Let A R nd let x R. We sy tht x is n ccumultion point of A if for ech ε > 0, the set A (x ε, x + ε) is infinite. As n exmple 1 is n ccumultion point of the set (0, 1) since (0, 1) (1 ε, 1 + ε) = (1 ε, 1) is infinite for every ε > 0. Similrly, every rel number is n ccumultion point of the set Q nd of the set R \ Q. Theorem 2.42 Let A R nd x R. Then the following re equivlent: ) x is n ccumultion point of A. b) For ech ε > 0, there is n element of A (x ε, x + ε) different from x. c) There is sequence { n } n=1 of distinct points in A with n x. Proof: ) implies b):suppose tht x is n ccumultion point of A. Then for ech ε > 0, the set A (x ε, x + ε) is infinite, so hs t lest one point different from x. c) implies ): If { n } n=1 is sequence of distinct points in A with n x, let ε > 0. Then, there is n N N such tht n x < ε for ll n N. In other

30 2.4. OPEN SETS, CLOSED SETS 27 words, the infinite set { n : n N} A (x ε, x + ε). Thus A (x ε, x + ε) is infinite for every ε > 0, so x is n ccumultion point of A. b) implies c): Let ε 1 = 1. There is n 1 A with 1 x nd 1 x < 1. Now, let ε 2 = min{ 1 2, 1 x }. Then ε 2 > 0 nd so there is n 2 x with 2 x < ε 2. Notice tht 1 2 by construction. Inductively, suppose we hve 1, k A ll distinct nd different from x with j x < 1 j for 1 j k. Set ε k+1 = min{ 1 k+1, 1 x,, k x }. Then ε k+1 > 0 so there is n k+1 A with k+1 x nd k+1 x < ε k+1. Then k+1 is different thn ll the previous j nd k+1 x < 1 k+1, so the induction continues. This gives us sequence { n } n=1 of distinct terms with n A for ll n nd n x. An importnt consequence of this is the following: Proposition 2.43 Assume A R is bounded bove. Then sup A is n element of A or n ccumultion point of A. Proof: Suppose tht x = sup A A. Then, for ech ε > 0, there is n A with x ε < < x < x + ε. Hence, x is n ccumultion point of A. Theorem 2.44 (Bolzno-Weierstrss Theorem) Every bounded infinie set hs n ccumultion point. Proof: Let A [ M, M] be infinite. Inductuvely, pick sequence { n } n=1 of distinct terms with n A for ll n. This sequence is then bounded, so it hs convergent subsequence. This subsequence lso hs distinct terms, so its limit is n ccumultion point of A. Theorem 2.45 Let A R nd let A be the set of ccumultion points of A. Then A A is closed. Proof: So, let A R nd set B = A A. We show tht R \ B is open. So, suppose tht x R \ B. Then x A, so there is n ε > 0 so tht A (x ε, x + ε) contins no point other thn x. But, since x A, we must

31 28 CHAPTER 2. SEQUENCES AND CONVERGENCE hve A (x ε, x + ε) =. But then, no point of (x ε, x + ε) cn be n ccumultion point of A, so A (x ε, x + ε) = lso. In other words, (x ε, x + ε) B =, so (x ε, x + ε) R \ B. This shows tht R \ B is open, so B is closed. Corollry 2.46 Let A R. Then A is closed if nd only if A A. Proof: If A is closed nd x A, then there is sequence { n } n=1 in A of distinct terms with n x. By our chrcteriztion of closed sets vi sequences, x A. Hence A A. Conversely, if A A, then A = A A is closed. Corollry 2.47 Suppose A is closed nd bounded bove. Then sup A A. Exercises: 1. Suppose tht 0 < r < 1. Show tht the sequence {r n } n=1 converges. Then find its limit. Wht hppens if 1 < r < 0? 2. Suppose tht 1 < r. Show tht {r n } n=1 is unbounded. Hint: if, not, show it converges. Wht would it hve to converge to? 3. Suppose tht r 1. Show tht 1 + r + r 2 + r r n 1 = 1 rn 1 r. 4. Suppose tht r < 1. Show tht n=1 rn 1 = 1 1 r. 5. Suppose tht 0 <, b. Show tht lim n ( n + b n ) 1/n = mx{, b}. ( 6. Let x 1 = 1 nd x n+1 = 1 2 x n + 2 x n ). Show tht x 2 n x2 n 2 2. From this, show tht x n 2.

32 Chpter 3 Limits of Functions In this chpter, we wnt to consder limits of functions. In prticulr, we wnt to justify clcultions like those done in elementry clsses such s x 2 4 lim x 2 x 2 = lim x 2 (x 2)(x + 2) x 2 = lim x 2 x + 2 = 4. Sevrl things should be pointed out in this clcultion. First, nd most significntly, the function involved, f(x) = (x 2 4)/(x 2) is not defined t the plce we re pproching in the limit (i.e. x = 2). However, the point x = 2 is n ccumultion point of the domin of f, i.e. {x R : x 2}. Next, the second inequlity depends on the fct tht the x-vlues we tke re not equl to 2, while the lst limit seems to plug in x = 2. These chrcteristics of limits mde the forml definition reltively lte occurence. We give it here. Definition 3.1 Let f : D R be function with D R nd suppose tht is n ccumultion point of D. We sy tht f hs limit L R s x pproches nd write lim x f(x) = L if for ll ε > 0, there is δ > 0 such tht whenever x D with 0 < x < δ, we hve f(x) L < ε. We sy tht the limit of f s x pproches exists if there is n L R such tht the limit of f s x pproches is L. 29

33 30 CHAPTER 3. LIMITS OF FUNCTIONS Another thing to point out is tht since we use 0 < x < ε, the vlue of f() is irrelevnt for the existence of the limit even when D. As n exmple, we show tht lim x 2 x 2 = 4. Let ε > 0 nd choose δ = min{1, ε/5}. Then δ > 0. Implicitly in this exmple, we hve D = R, so 2 is n ccumultion point of D. So, suppose tht x R with 0 < x 2 < δ. Then, first, x x < = 3, so x = 5. Then, we hve f(x) L = x 2 4 = (x 2)(x + 2) x 2 5 < ε s required. Proposition 3.2 Suppose tht the limit of f s x pproches is both L 1 nd L 2. Then L 1 = L 2. In other words, limits re unique. We hve implicitly used this when we write lim x f(x) = L, so some cre is required here. Proof: Suppose L 1 L 2 nd set ε = L 1 L 2 /2. Then pick δ 1 > 0 such tht whenever x D with 0 < x < δ 1, we hve f(x) L 1 < ε. Smilrly, pick δ 2 > 0 such tht whenever x D with 0 < x < δ 2, we hve f(x) L 2 < ε. Let δ = min{δ 1, δ 2 }. Since is n ccumultion point of D, there is n x D with x nd x < δ. For this x, we hve f(x) L 1 < ε nd f(x) L 2 < ε. But this implies tht L 1 L 2 = (f(x) L 2 ) (f(x) L 1 ) < ε + ε = 2ε = L 1 L 2, contrdction. Exmples: ) We lwys hve tht lim x x =. In fct, if ε > 0, simply choose δ = ε. Then, if x D nd 0 < x < δ, we hve x < ε. b) let D = R. We show tht lim x 3 x 2 + x = 12. In fct, let ε > 0. Choose δ = min{1, ε/8}. Then, if x D with 0 < x 3 < δ, we hve first of ll tht x + 4 = (x 3) + 7 x < = 8. Now, this shows tht (x 2 + x) 12 = (x 3)(x + 4) = x 3 x + 4 < 8 x 3 < ε. c) Let D = R \ {0} nd let f(x) = cos( 1 x ). Notice tht 0 is n ccumultion point of D. We clim tht f does not hve limit s x pproches 0. In fct, suppose to the contrry tht lim x 0 f(x) = L nd choose ε = 1. Then, there would be δ > 0 such tht whenever 0 < x 0 < δ, we

34 31 hve f(x) L < ε. But, we hve tht the sequence 1 πn 0, so there is n N N such tht whenever n N, we hve 1 πn 0 < δ. But then one of N nd N + 1 is odd nd the other is even. Thus one of cos(πn) nd cos(π(n + 1)) is 1 nd the other is 1. letting x = 1 πn 1 nd x = π(n+1) shows tht L 1 < ε nd L ( 1) < ε. But then, 2 = 1 ( 1) 1 L + L ( 1) < 2ε = 2. This is contrdiction. d) Let D = (0, 1) nd define f(x) = 0 if x is irrtionl nd f(x) = 1 q if x = p q is rtionl nd in lowest terms. We clim tht lim x f(x) = 0 for ll [0, 1]. In fct, let [0, 1] nd let ε > 0. There is n N N such tht 1 N < ε. Now consider the set of frctions of the form p q where q N. There re only finitely mny such frctions. Let δ = min{ p q : q N, p q } be the smllest distnce from to one of these frctions. Then δ > 0. Now suppose tht x D nd 0 < x < δ. Either x is irrtionl, in which cse f(x) 0 = 0 0 = 0 < ε, or x is rtionl, but x = p q where q > N. In the ltter cse, f(x) 0 = 1 q 0 = 1 q < 1 N < ε. At this point, we cn prove ll of the lgebric properties of limits for functions in wy tht is similr to how we showed them for sequences in the lst chpter. The following result is n exmple of this technique. Proposition 3.3 Let f, g : D R nd n ccumultion point of D. Suppose tht lim x f(x) = L 1 nd lim x g(x) = L 2. Then lim x f(x) + g(x) = L 1 + L 2. Proof: Let ε > 0. Choose δ 1 > 0 such tht whenever x D with 0 < x < δ 1 we hve f(x) L 1 < ε/2. Similrly, find δ 2 > 0 such tht whenever x D with 0 < x < δ 2 we hve g(x) L 2 < ε/2. let δ = min{δ 1, δ 2 }, so tht δ > 0. Now, if x D nd 0 < x < δ, we hve f(x) + g(x) (L 1 + L 2 ) f(x) L 1 + g(x) L 2 < ε/2 + ε/2 = ε, s required. Just s for sequences, we cn define boundedness of functions: Definition 3.4 Let f : D R. We sy tht f is bounded if there is n M R such tht f(x) M for ll x D.

35 32 CHAPTER 3. LIMITS OF FUNCTIONS Proposition 3.5 Suppose tht f : D R nd is n ccumultion point of D where lim x f(x) exists. Then, there is n δ > 0 such tht f is bounded on the set D ( δ, + δ). It is possible to prove ll the results for the lgebric properties of limits of functions vi techniques similr to those used for the properties of limits of sequences. However, it turns out to be more helpful to use the results from sequences to do this. For this we need the following, very importnt result. Theorem 3.6 Let f : D R nd nd ccumultion point of D. The following re equivlent. ) The limit lim x f(x) = L. b) Whenever { n } n=1 is sequence in D with n for ll n N nd n, we hve tht f( n ) L. Proof: ) implies b): Suppose tht lim x f(x) = L. Also, suppose tht n D, n, nd n. We need to show tht f( n ) L. So let ε > 0. The, there is δ > 0 such tht whenever x D with 0 < x < δ, we hve f(x) L < ε. But then, there is n N N such tht whenever n N, we hve n < δ. But now, for these n, we hve tht n, so 0 < n < δ, which implies tht f( n ) L < ε. In other words, f( n ) L. b) implies ): Suppose tht f does not hve limit of L s x D pproches. Then, there is n ε > 0 such tht for every δ > 0, there is n x D with 0 < x < δ yet f(x) L ε. Let δ = 1 n nd find x n D with 0 < x < 1 n but such tht f(x n) L ε. Then x n with x n for ll n, but {f(x n )} n=1 does not converge to L. This contrdicts our ssumption nd shows tht lim x f(x) = L. As n exmple of the use of this result, we show tht limits of products work the right wy. Proposition 3.7 Let f, g : D R nd n ccumultion point of D. Assume tht lim x f(x) = L nd lim x g(x) = M. Then lim x f(x)g(x) = LM. Similrly, if g(x) 0 for ll x D nd M 0, then lim x f(x) g(x) = L M.

36 33 Proof: Suppose tht { n } n=1 in D with n for ll n. Then, f( n ) L nd g( n ) M, so by the result on products of sequences, f( n )g( n ) LM. Since this hppens for ll sequences { n } n=1, we hve tht lim x f(x)g(x) exists nd equls LM. The result for quotients follows similrly. The next result is bit more subtle. Insted of ssuming convergence to specific point, it merely ssumes convergence. Theorem 3.8 Assume f : D R nd is n ccumultion point of D. The following re equivlent: ) The limit of f s x pproches exists. b) Whenever { n } n=1 is sequence in D with n nd n, then the sequence {f( n )} n=1 converges. Proof: ) implies b): This follows directly from Theorem 3.6. b) implies ): The difficulty here is tht we only ssume tht the sequences {f( n )} n=1 converge, not tht they ll converge to the sme thing. Since is n ccumultion point of D, there is t lest one sequence { n } n=1 in D with n for ll n nd n. By ssumption, {f( n )} n=1 converges, sy f( n ) L. Now, ssume tht {b n } n=1 is ny other sequence in D with b n for ll n nd b n. We need to show tht f(b n ) L lso (we only know this sequence converges, NOT tht it converges to L). To do this, let {c n } n=1 be defined by c 2n 1 = n nd c 2n = b n, so c n lterntes between n nd b n. Then c n D nd c n neqx for ll n. Also, c n. By ssumption {f(c n )} n=1 converges, sy f(c n ) L. But now, {f( n )} n=1 is subsequence of {f(c n )} n=1, so f( n ) L lso. Hence, L = L. But now, {f(b n )} n=1 is lso subsequence of {f(c n )} n=1, so f(b n ) L = L. Now we cn pply Theorem 3.6 gin to see tht lim x f(x) = L. Exercises: 1. Show tht lim x 3 x 2 9 x 3 = 6 using the ε, δ definition. 2. Show tht lim x 2 x 2 + 4x + 5 = 17 using the ε, δ definition.

37 34 CHAPTER 3. LIMITS OF FUNCTIONS 3. Suppose tht f, g : D R nd is n ccumultion point of D. Suppose tht g is bounded nd tht lim x f(x) = 0. Show tht lim x f(x)g(x) = 0 using the ε, δ definition. 4. Suppose tht f, g, h : D R nd is n ccumultion point of D. Suppose tht f(x) g(x) h(x) for ll x D nd tht lim x f(x) = lim x h(x) = L. Show tht lim x g(x) = L. 5. Suppose tht f : D R nd is n ccumultion point of D. Suppose lso tht for ech ε > 0 there is δ > 0 such tht whenever x, y D with x y nd x y < δ, then f(x) f(y) < ε. Show tht lim x f(x) exists. Hint: Suppose tht n D with n. Show tht {f( n )} is Cuchy.

### The Regulated and Riemann Integrals

Chpter 1 The Regulted nd Riemnn Integrls 1.1 Introduction We will consider severl different pproches to defining the definite integrl f(x) dx of function f(x). These definitions will ll ssign the sme vlue

### Lecture 1. Functional series. Pointwise and uniform convergence.

1 Introduction. Lecture 1. Functionl series. Pointwise nd uniform convergence. In this course we study mongst other things Fourier series. The Fourier series for periodic function f(x) with period 2π is

### W. We shall do so one by one, starting with I 1, and we shall do it greedily, trying

Vitli covers 1 Definition. A Vitli cover of set E R is set V of closed intervls with positive length so tht, for every δ > 0 nd every x E, there is some I V with λ(i ) < δ nd x I. 2 Lemm (Vitli covering)

p-adic Egyptin Frctions Contents 1 Introduction 1 2 Trditionl Egyptin Frctions nd Greedy Algorithm 2 3 Set-up 3 4 p-greedy Algorithm 5 5 p-egyptin Trditionl 10 6 Conclusion 1 Introduction An Egyptin frction

### Math 61CM - Solutions to homework 9

Mth 61CM - Solutions to homework 9 Cédric De Groote November 30 th, 2018 Problem 1: Recll tht the left limit of function f t point c is defined s follows: lim f(x) = l x c if for ny > 0 there exists δ

### UNIFORM CONVERGENCE. Contents 1. Uniform Convergence 1 2. Properties of uniform convergence 3

UNIFORM CONVERGENCE Contents 1. Uniform Convergence 1 2. Properties of uniform convergence 3 Suppose f n : Ω R or f n : Ω C is sequence of rel or complex functions, nd f n f s n in some sense. Furthermore,

### Advanced Calculus: MATH 410 Notes on Integrals and Integrability Professor David Levermore 17 October 2004

Advnced Clculus: MATH 410 Notes on Integrls nd Integrbility Professor Dvid Levermore 17 October 2004 1. Definite Integrls In this section we revisit the definite integrl tht you were introduced to when

### MAA 4212 Improper Integrals

Notes by Dvid Groisser, Copyright c 1995; revised 2002, 2009, 2014 MAA 4212 Improper Integrls The Riemnn integrl, while perfectly well-defined, is too restrictive for mny purposes; there re functions which

### Lecture 3 ( ) (translated and slightly adapted from lecture notes by Martin Klazar)

Lecture 3 (5.3.2018) (trnslted nd slightly dpted from lecture notes by Mrtin Klzr) Riemnn integrl Now we define precisely the concept of the re, in prticulr, the re of figure U(, b, f) under the grph of

### Lecture 3. Limits of Functions and Continuity

Lecture 3 Limits of Functions nd Continuity Audrey Terrs April 26, 21 1 Limits of Functions Notes I m skipping the lst section of Chpter 6 of Lng; the section bout open nd closed sets We cn probbly live

### The Henstock-Kurzweil integral

fculteit Wiskunde en Ntuurwetenschppen The Henstock-Kurzweil integrl Bchelorthesis Mthemtics June 2014 Student: E. vn Dijk First supervisor: Dr. A.E. Sterk Second supervisor: Prof. dr. A. vn der Schft

### Infinite Geometric Series

Infinite Geometric Series Finite Geometric Series ( finite SUM) Let 0 < r < 1, nd let n be positive integer. Consider the finite sum It turns out there is simple lgebric expression tht is equivlent to

### a n = 1 58 a n+1 1 = 57a n + 1 a n = 56(a n 1) 57 so 0 a n+1 1, and the required result is true, by induction.

MAS221(216-17) Exm Solutions 1. (i) A is () bounded bove if there exists K R so tht K for ll A ; (b) it is bounded below if there exists L R so tht L for ll A. e.g. the set { n; n N} is bounded bove (by

### Farey Fractions. Rickard Fernström. U.U.D.M. Project Report 2017:24. Department of Mathematics Uppsala University

U.U.D.M. Project Report 07:4 Frey Frctions Rickrd Fernström Exmensrete i mtemtik, 5 hp Hledre: Andres Strömergsson Exmintor: Jörgen Östensson Juni 07 Deprtment of Mthemtics Uppsl University Frey Frctions

### Riemann is the Mann! (But Lebesgue may besgue to differ.)

Riemnn is the Mnn! (But Lebesgue my besgue to differ.) Leo Livshits My 2, 2008 1 For finite intervls in R We hve seen in clss tht every continuous function f : [, b] R hs the property tht for every ɛ >

### 7.2 The Definite Integral

7.2 The Definite Integrl the definite integrl In the previous section, it ws found tht if function f is continuous nd nonnegtive, then the re under the grph of f on [, b] is given by F (b) F (), where

### 1 i n x i x i 1. Note that kqk kp k. In addition, if P and Q are partition of [a, b], P Q is finer than both P and Q.

Chpter 6 Integrtion In this chpter we define the integrl. Intuitively, it should be the re under curve. Not surprisingly, fter mny exmples, counter exmples, exceptions, generliztions, the concept of the

### IMPORTANT THEOREMS CHEAT SHEET

IMPORTANT THEOREMS CHEAT SHEET BY DOUGLAS DANE Howdy, I m Bronson s dog Dougls. Bronson is still complining bout the textbook so I thought if I kept list of the importnt results for you, he might stop.

### Handout: Natural deduction for first order logic

MATH 457 Introduction to Mthemticl Logic Spring 2016 Dr Json Rute Hndout: Nturl deduction for first order logic We will extend our nturl deduction rules for sententil logic to first order logic These notes

### Review of Riemann Integral

1 Review of Riemnn Integrl In this chpter we review the definition of Riemnn integrl of bounded function f : [, b] R, nd point out its limittions so s to be convinced of the necessity of more generl integrl.

### Lecture 1: Introduction to integration theory and bounded variation

Lecture 1: Introduction to integrtion theory nd bounded vrition Wht is this course bout? Integrtion theory. The first question you might hve is why there is nything you need to lern bout integrtion. You

### 7.2 Riemann Integrable Functions

7.2 Riemnn Integrble Functions Theorem 1. If f : [, b] R is step function, then f R[, b]. Theorem 2. If f : [, b] R is continuous on [, b], then f R[, b]. Theorem 3. If f : [, b] R is bounded nd continuous

### Math 360: A primitive integral and elementary functions

Mth 360: A primitive integrl nd elementry functions D. DeTurck University of Pennsylvni October 16, 2017 D. DeTurck Mth 360 001 2017C: Integrl/functions 1 / 32 Setup for the integrl prtitions Definition:

### Improper Integrals. Type I Improper Integrals How do we evaluate an integral such as

Improper Integrls Two different types of integrls cn qulify s improper. The first type of improper integrl (which we will refer to s Type I) involves evluting n integrl over n infinite region. In the grph

### Math 554 Integration

Mth 554 Integrtion Hndout #9 4/12/96 Defn. A collection of n + 1 distinct points of the intervl [, b] P := {x 0 = < x 1 < < x i 1 < x i < < b =: x n } is clled prtition of the intervl. In this cse, we

### Properties of the Riemann Integral

Properties of the Riemnn Integrl Jmes K. Peterson Deprtment of Biologicl Sciences nd Deprtment of Mthemticl Sciences Clemson University Februry 15, 2018 Outline 1 Some Infimum nd Supremum Properties 2

### and that at t = 0 the object is at position 5. Find the position of the object at t = 2.

7.2 The Fundmentl Theorem of Clculus 49 re mny, mny problems tht pper much different on the surfce but tht turn out to be the sme s these problems, in the sense tht when we try to pproimte solutions we

### Riemann Sums and Riemann Integrals

Riemnn Sums nd Riemnn Integrls Jmes K. Peterson Deprtment of Biologicl Sciences nd Deprtment of Mthemticl Sciences Clemson University August 26, 2013 Outline 1 Riemnn Sums 2 Riemnn Integrls 3 Properties

### Theoretical foundations of Gaussian quadrature

Theoreticl foundtions of Gussin qudrture 1 Inner product vector spce Definition 1. A vector spce (or liner spce) is set V = {u, v, w,...} in which the following two opertions re defined: (A) Addition of

### Rudin s Principles of Mathematical Analysis: Solutions to Selected Exercises. Sam Blinstein UCLA Department of Mathematics

Rudin s Principles of Mthemticl Anlysis: Solutions to Selected Exercises Sm Blinstein UCLA Deprtment of Mthemtics Mrch 29, 2008 Contents Chpter : The Rel nd Complex Number Systems 2 Chpter 2: Bsic Topology

### Exam 2, Mathematics 4701, Section ETY6 6:05 pm 7:40 pm, March 31, 2016, IH-1105 Instructor: Attila Máté 1

Exm, Mthemtics 471, Section ETY6 6:5 pm 7:4 pm, Mrch 1, 16, IH-115 Instructor: Attil Máté 1 17 copies 1. ) Stte the usul sufficient condition for the fixed-point itertion to converge when solving the eqution

### Calculus II: Integrations and Series

Clculus II: Integrtions nd Series August 7, 200 Integrls Suppose we hve generl function y = f(x) For simplicity, let f(x) > 0 nd f(x) continuous Denote F (x) = re under the grph of f in the intervl [,x]

### Presentation Problems 5

Presenttion Problems 5 21-355 A For these problems, ssume ll sets re subsets of R unless otherwise specified. 1. Let P nd Q be prtitions of [, b] such tht P Q. Then U(f, P ) U(f, Q) nd L(f, P ) L(f, Q).

### Riemann Sums and Riemann Integrals

Riemnn Sums nd Riemnn Integrls Jmes K. Peterson Deprtment of Biologicl Sciences nd Deprtment of Mthemticl Sciences Clemson University August 26, 203 Outline Riemnn Sums Riemnn Integrls Properties Abstrct

### The First Fundamental Theorem of Calculus. If f(x) is continuous on [a, b] and F (x) is any antiderivative. f(x) dx = F (b) F (a).

The Fundmentl Theorems of Clculus Mth 4, Section 0, Spring 009 We now know enough bout definite integrls to give precise formultions of the Fundmentl Theorems of Clculus. We will lso look t some bsic emples

### The final exam will take place on Friday May 11th from 8am 11am in Evans room 60.

Mth 104: finl informtion The finl exm will tke plce on Fridy My 11th from 8m 11m in Evns room 60. The exm will cover ll prts of the course with equl weighting. It will cover Chpters 1 5, 7 15, 17 21, 23

### Advanced Calculus: MATH 410 Uniform Convergence of Functions Professor David Levermore 11 December 2015

Advnced Clculus: MATH 410 Uniform Convergence of Functions Professor Dvid Levermore 11 December 2015 12. Sequences of Functions We now explore two notions of wht it mens for sequence of functions {f n

### 8 Laplace s Method and Local Limit Theorems

8 Lplce s Method nd Locl Limit Theorems 8. Fourier Anlysis in Higher DImensions Most of the theorems of Fourier nlysis tht we hve proved hve nturl generliztions to higher dimensions, nd these cn be proved

### Further Topics in Analysis

Further Topics in Anlysis Lecture Notes 2012/13 Lecturer: Prof. Jens Mrklof Notes by Dr. Vitly Moroz School of Mthemtics University of Bristol BS8 1TW Bristol, UK c University of Bristol 2013 Contents

### f(x) dx, If one of these two conditions is not met, we call the integral improper. Our usual definition for the value for the definite integral

Improper Integrls Every time tht we hve evluted definite integrl such s f(x) dx, we hve mde two implicit ssumptions bout the integrl:. The intervl [, b] is finite, nd. f(x) is continuous on [, b]. If one

### f(x)dx . Show that there 1, 0 < x 1 does not exist a differentiable function g : [ 1, 1] R such that g (x) = f(x) for all

3 Definite Integrl 3.1 Introduction In school one comes cross the definition of the integrl of rel vlued function defined on closed nd bounded intervl [, b] between the limits nd b, i.e., f(x)dx s the

### UNIFORM CONVERGENCE MA 403: REAL ANALYSIS, INSTRUCTOR: B. V. LIMAYE

UNIFORM CONVERGENCE MA 403: REAL ANALYSIS, INSTRUCTOR: B. V. LIMAYE 1. Pointwise Convergence of Sequence Let E be set nd Y be metric spce. Consider functions f n : E Y for n = 1, 2,.... We sy tht the sequence

### Chapter 6. Infinite series

Chpter 6 Infinite series We briefly review this chpter in order to study series of functions in chpter 7. We cover from the beginning to Theorem 6.7 in the text excluding Theorem 6.6 nd Rbbe s test (Theorem

### a n+2 a n+1 M n a 2 a 1. (2)

Rel Anlysis Fll 004 Tke Home Finl Key 1. Suppose tht f is uniformly continuous on set S R nd {x n } is Cuchy sequence in S. Prove tht {f(x n )} is Cuchy sequence. (f is not ssumed to be continuous outside

### MAT 215: Analysis in a single variable Course notes, Fall Michael Damron

MAT 215: Anlysis in single vrible Course notes, Fll 2012 Michel Dmron Compiled from lectures nd exercises designed with Mrk McConnell following Principles of Mthemticl Anlysis, Rudin Princeton University

### STUDY GUIDE FOR BASIC EXAM

STUDY GUIDE FOR BASIC EXAM BRYON ARAGAM This is prtil list of theorems tht frequently show up on the bsic exm. In mny cses, you my be sked to directly prove one of these theorems or these vrints. There

### How do we solve these things, especially when they get complicated? How do we know when a system has a solution, and when is it unique?

XII. LINEAR ALGEBRA: SOLVING SYSTEMS OF EQUATIONS Tody we re going to tlk bout solving systems of liner equtions. These re problems tht give couple of equtions with couple of unknowns, like: 6 2 3 7 4

### Convex Sets and Functions

B Convex Sets nd Functions Definition B1 Let L, +, ) be rel liner spce nd let C be subset of L The set C is convex if, for ll x,y C nd ll [, 1], we hve 1 )x+y C In other words, every point on the line

### Unit #9 : Definite Integral Properties; Fundamental Theorem of Calculus

Unit #9 : Definite Integrl Properties; Fundmentl Theorem of Clculus Gols: Identify properties of definite integrls Define odd nd even functions, nd reltionship to integrl vlues Introduce the Fundmentl

Qudrtic Forms Recll the Simon & Blume excerpt from n erlier lecture which sid tht the min tsk of clculus is to pproximte nonliner functions with liner functions. It s ctully more ccurte to sy tht we pproximte

### Review of basic calculus

Review of bsic clculus This brief review reclls some of the most importnt concepts, definitions, nd theorems from bsic clculus. It is not intended to tech bsic clculus from scrtch. If ny of the items below

### PARTIAL FRACTION DECOMPOSITION

PARTIAL FRACTION DECOMPOSITION LARRY SUSANKA 1. Fcts bout Polynomils nd Nottion We must ssemble some tools nd nottion to prove the existence of the stndrd prtil frction decomposition, used s n integrtion

### Improper Integrals. The First Fundamental Theorem of Calculus, as we ve discussed in class, goes as follows:

Improper Integrls The First Fundmentl Theorem of Clculus, s we ve discussed in clss, goes s follows: If f is continuous on the intervl [, ] nd F is function for which F t = ft, then ftdt = F F. An integrl

### Homework 11. Andrew Ma November 30, sin x (1+x) (1+x)

Homewor Andrew M November 3, 4 Problem 9 Clim: Pf: + + d = d = sin b +b + sin (+) d sin (+) d using integrtion by prts. By pplying + d = lim b sin b +b + sin (+) d. Since limits to both sides, lim b sin

### Overview of Calculus I

Overview of Clculus I Prof. Jim Swift Northern Arizon University There re three key concepts in clculus: The limit, the derivtive, nd the integrl. You need to understnd the definitions of these three things,

### Improper Integrals, and Differential Equations

Improper Integrls, nd Differentil Equtions October 22, 204 5.3 Improper Integrls Previously, we discussed how integrls correspond to res. More specificlly, we sid tht for function f(x), the region creted

### Bases for Vector Spaces

Bses for Vector Spces 2-26-25 A set is independent if, roughly speking, there is no redundncy in the set: You cn t uild ny vector in the set s liner comintion of the others A set spns if you cn uild everything

### Main topics for the First Midterm

Min topics for the First Midterm The Midterm will cover Section 1.8, Chpters 2-3, Sections 4.1-4.8, nd Sections 5.1-5.3 (essentilly ll of the mteril covered in clss). Be sure to know the results of the

### Notes on length and conformal metrics

Notes on length nd conforml metrics We recll how to mesure the Eucliden distnce of n rc in the plne. Let α : [, b] R 2 be smooth (C ) rc. Tht is α(t) (x(t), y(t)) where x(t) nd y(t) re smooth rel vlued

### 38 Riemann sums and existence of the definite integral.

38 Riemnn sums nd existence of the definite integrl. In the clcultion of the re of the region X bounded by the grph of g(x) = x 2, the x-xis nd 0 x b, two sums ppered: ( n (k 1) 2) b 3 n 3 re(x) ( n These

### MATH 101A: ALGEBRA I PART B: RINGS AND MODULES 35

MATH 101A: ALGEBRA I PART B: RINGS AND MODULES 35 9. Modules over PID This week we re proving the fundmentl theorem for finitely generted modules over PID, nmely tht they re ll direct sums of cyclic modules.

### Chapter 0. What is the Lebesgue integral about?

Chpter 0. Wht is the Lebesgue integrl bout? The pln is to hve tutoril sheet ech week, most often on Fridy, (to be done during the clss) where you will try to get used to the ides introduced in the previous

### Prof. Girardi, Math 703, Fall 2012 Homework Solutions: 1 8. Homework 1. in R, prove that. c k. sup. k n. sup. c k R = inf

Knpp, Chpter, Section, # 4, p. 78 Homework For ny two sequences { n } nd {b n} in R, prove tht lim sup ( n + b n ) lim sup n + lim sup b n, () provided the two terms on the right side re not + nd in some

### ODE: Existence and Uniqueness of a Solution

Mth 22 Fll 213 Jerry Kzdn ODE: Existence nd Uniqueness of Solution The Fundmentl Theorem of Clculus tells us how to solve the ordinry differentil eqution (ODE) du = f(t) dt with initil condition u() =

### Coalgebra, Lecture 15: Equations for Deterministic Automata

Colger, Lecture 15: Equtions for Deterministic Automt Julin Slmnc (nd Jurrin Rot) Decemer 19, 2016 In this lecture, we will study the concept of equtions for deterministic utomt. The notes re self contined

### SUMMER KNOWHOW STUDY AND LEARNING CENTRE

SUMMER KNOWHOW STUDY AND LEARNING CENTRE Indices & Logrithms 2 Contents Indices.2 Frctionl Indices.4 Logrithms 6 Exponentil equtions. Simplifying Surds 13 Opertions on Surds..16 Scientific Nottion..18

### 7 Improper Integrals, Exp, Log, Arcsin, and the Integral Test for Series

7 Improper Integrls, Exp, Log, Arcsin, nd the Integrl Test for Series We hve now ttined good level of understnding of integrtion of nice functions f over closed intervls [, b]. In prctice one often wnts

### Math 426: Probability Final Exam Practice

Mth 46: Probbility Finl Exm Prctice. Computtionl problems 4. Let T k (n) denote the number of prtitions of the set {,..., n} into k nonempty subsets, where k n. Argue tht T k (n) kt k (n ) + T k (n ) by

### Advanced Calculus I (Math 4209) Martin Bohner

Advnced Clculus I (Mth 4209) Spring 2018 Lecture Notes Mrtin Bohner Version from My 4, 2018 Author ddress: Deprtment of Mthemtics nd Sttistics, Missouri University of Science nd Technology, Roll, Missouri

### (e) if x = y + z and a divides any two of the integers x, y, or z, then a divides the remaining integer

Divisibility In this note we introduce the notion of divisibility for two integers nd b then we discuss the division lgorithm. First we give forml definition nd note some properties of the division opertion.

### UniversitaireWiskundeCompetitie. Problem 2005/4-A We have k=1. Show that for every q Q satisfying 0 < q < 1, there exists a finite subset K N so that

Problemen/UWC NAW 5/7 nr juni 006 47 Problemen/UWC UniversitireWiskundeCompetitie Edition 005/4 For Session 005/4 we received submissions from Peter Vndendriessche, Vldislv Frnk, Arne Smeets, Jn vn de

### Recitation 3: More Applications of the Derivative

Mth 1c TA: Pdric Brtlett Recittion 3: More Applictions of the Derivtive Week 3 Cltech 2012 1 Rndom Question Question 1 A grph consists of the following: A set V of vertices. A set E of edges where ech

### ARITHMETIC OPERATIONS. The real numbers have the following properties: a b c ab ac

REVIEW OF ALGEBRA Here we review the bsic rules nd procedures of lgebr tht you need to know in order to be successful in clculus. ARITHMETIC OPERATIONS The rel numbers hve the following properties: b b

### Review of Calculus, cont d

Jim Lmbers MAT 460 Fll Semester 2009-10 Lecture 3 Notes These notes correspond to Section 1.1 in the text. Review of Clculus, cont d Riemnn Sums nd the Definite Integrl There re mny cses in which some

### II. Integration and Cauchy s Theorem

MTH6111 Complex Anlysis 2009-10 Lecture Notes c Shun Bullett QMUL 2009 II. Integrtion nd Cuchy s Theorem 1. Pths nd integrtion Wrning Different uthors hve different definitions for terms like pth nd curve.

### set is not closed under matrix [ multiplication, ] and does not form a group.

Prolem 2.3: Which of the following collections of 2 2 mtrices with rel entries form groups under [ mtrix ] multipliction? i) Those of the form for which c d 2 Answer: The set of such mtrices is not closed

### Theory of the Integral

Spring 2012 Theory of the Integrl Author: Todd Gugler Professor: Dr. Drgomir Sric My 10, 2012 2 Contents 1 Introduction 5 1.0.1 Office Hours nd Contct Informtion..................... 5 1.1 Set Theory:

### THE EXISTENCE-UNIQUENESS THEOREM FOR FIRST-ORDER DIFFERENTIAL EQUATIONS.

THE EXISTENCE-UNIQUENESS THEOREM FOR FIRST-ORDER DIFFERENTIAL EQUATIONS RADON ROSBOROUGH https://intuitiveexplntionscom/picrd-lindelof-theorem/ This document is proof of the existence-uniqueness theorem

### Math 113 Exam 2 Practice

Mth Em Prctice Februry, 8 Em will cover sections 6.5, 7.-7.5 nd 7.8. This sheet hs three sections. The first section will remind you bout techniques nd formuls tht you should know. The second gives number

### MATH 174A: PROBLEM SET 5. Suggested Solution

MATH 174A: PROBLEM SET 5 Suggested Solution Problem 1. Suppose tht I [, b] is n intervl. Let f 1 b f() d for f C(I; R) (i.e. f is continuous rel-vlued function on I), nd let L 1 (I) denote the completion

### Chapter 1: Fundamentals

Chpter 1: Fundmentls 1.1 Rel Numbers Types of Rel Numbers: Nturl Numbers: {1, 2, 3,...}; These re the counting numbers. Integers: {... 3, 2, 1, 0, 1, 2, 3,...}; These re ll the nturl numbers, their negtives,

### Duality # Second iteration for HW problem. Recall our LP example problem we have been working on, in equality form, is given below.

Dulity #. Second itertion for HW problem Recll our LP emple problem we hve been working on, in equlity form, is given below.,,,, 8 m F which, when written in slightly different form, is 8 F Recll tht we

### SOLUTIONS FOR ANALYSIS QUALIFYING EXAM, FALL (1 + µ(f n )) f(x) =. But we don t need the exact bound.) Set

SOLUTIONS FOR ANALYSIS QUALIFYING EXAM, FALL 28 Nottion: N {, 2, 3,...}. (Tht is, N.. Let (X, M be mesurble spce with σ-finite positive mesure µ. Prove tht there is finite positive mesure ν on (X, M such

### 11 An introduction to Riemann Integration

11 An introduction to Riemnn Integrtion The PROOFS of the stndrd lemms nd theorems concerning the Riemnn Integrl re NEB, nd you will not be sked to reproduce proofs of these in full in the exmintion in

### Math 1B, lecture 4: Error bounds for numerical methods

Mth B, lecture 4: Error bounds for numericl methods Nthn Pflueger 4 September 0 Introduction The five numericl methods descried in the previous lecture ll operte by the sme principle: they pproximte the

### Properties of Integrals, Indefinite Integrals. Goals: Definition of the Definite Integral Integral Calculations using Antiderivatives

Block #6: Properties of Integrls, Indefinite Integrls Gols: Definition of the Definite Integrl Integrl Clcultions using Antiderivtives Properties of Integrls The Indefinite Integrl 1 Riemnn Sums - 1 Riemnn

### Math 4200: Homework Problems

Mth 4200: Homework Problems Gregor Kovčič 1. Prove the following properties of the binomil coefficients ( n ) ( n ) (i) 1 + + + + 1 2 ( n ) (ii) 1 ( n ) ( n ) + 2 + 3 + + n 2 3 ( ) n ( n + = 2 n 1 n) n,

### Finite Automata Theory and Formal Languages TMV027/DIT321 LP4 2018

Finite Automt Theory nd Forml Lnguges TMV027/DIT321 LP4 2018 Lecture 10 An Bove April 23rd 2018 Recp: Regulr Lnguges We cn convert between FA nd RE; Hence both FA nd RE ccept/generte regulr lnguges; More

### Integral points on the rational curve

Integrl points on the rtionl curve y x bx c x ;, b, c integers. Konstntine Zeltor Mthemtics University of Wisconsin - Mrinette 750 W. Byshore Street Mrinette, WI 5443-453 Also: Konstntine Zeltor P.O. Box

### AQA Further Pure 1. Complex Numbers. Section 1: Introduction to Complex Numbers. The number system

Complex Numbers Section 1: Introduction to Complex Numbers Notes nd Exmples These notes contin subsections on The number system Adding nd subtrcting complex numbers Multiplying complex numbers Complex

### Homework 4. (1) If f R[a, b], show that f 3 R[a, b]. If f + (x) = max{f(x), 0}, is f + R[a, b]? Justify your answer.

Homework 4 (1) If f R[, b], show tht f 3 R[, b]. If f + (x) = mx{f(x), 0}, is f + R[, b]? Justify your nswer. (2) Let f be continuous function on [, b] tht is strictly positive except finitely mny points

### Calculus I-II Review Sheet

Clculus I-II Review Sheet 1 Definitions 1.1 Functions A function is f is incresing on n intervl if x y implies f(x) f(y), nd decresing if x y implies f(x) f(y). It is clled monotonic if it is either incresing

### n f(x i ) x. i=1 In section 4.2, we defined the definite integral of f from x = a to x = b as n f(x i ) x; f(x) dx = lim i=1

The Fundmentl Theorem of Clculus As we continue to study the re problem, let s think bck to wht we know bout computing res of regions enclosed by curves. If we wnt to find the re of the region below the

### A BRIEF INTRODUCTION TO UNIFORM CONVERGENCE. In the study of Fourier series, several questions arise naturally, such as: c n e int

A BRIEF INTRODUCTION TO UNIFORM CONVERGENCE HANS RINGSTRÖM. Questions nd exmples In the study of Fourier series, severl questions rise nturlly, such s: () (2) re there conditions on c n, n Z, which ensure

### 1. Gauss-Jacobi quadrature and Legendre polynomials. p(t)w(t)dt, p {p(x 0 ),...p(x n )} p(t)w(t)dt = w k p(x k ),

1. Guss-Jcobi qudrture nd Legendre polynomils Simpson s rule for evluting n integrl f(t)dt gives the correct nswer with error of bout O(n 4 ) (with constnt tht depends on f, in prticulr, it depends on

### 5.7 Improper Integrals

458 pplictions of definite integrls 5.7 Improper Integrls In Section 5.4, we computed the work required to lift pylod of mss m from the surfce of moon of mss nd rdius R to height H bove the surfce of the

### LECTURE. INTEGRATION AND ANTIDERIVATIVE.

ANALYSIS FOR HIGH SCHOOL TEACHERS LECTURE. INTEGRATION AND ANTIDERIVATIVE. ROTHSCHILD CAESARIA COURSE, 2015/6 1. Integrtion Historiclly, it ws the problem of computing res nd volumes, tht triggered development

### 20 MATHEMATICS POLYNOMIALS

0 MATHEMATICS POLYNOMIALS.1 Introduction In Clss IX, you hve studied polynomils in one vrible nd their degrees. Recll tht if p(x) is polynomil in x, the highest power of x in p(x) is clled the degree of