202_LectNt - Chapter 1. Logic To reason correctly, we have...

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Unformatted text preview: Chapter 1. Logic To reason correctly, we have to follow some rules. These rules of reasoning are what we called logic. We will only need a few of these rules, mainly to deal with taking opposite of statements and to handle conditional statements. We will use the symbol (or :) to denote the word “not”. Also, we will use the symbol 8 to denote “for all”, “for any”, “for every”. Similarly, the symbol 9 will denote “there is (at least one)”, “there exists”, “there are (some)” and usually followed by “such that”. The symbols 8 and 9 are called quantifiers. Negation. Below we will look at rules of negation (i.e. taking opposite). They are needed when we do indirect proofs (or proofs by contradiction). For any expression p; we have . p/ D p: Examples. (1) expression : opposite expression : rule : (2) expression : opposite expression : rule : statement : quantified statement : opposite statement : quantified opposite statement : (4) statement : quantified statement : opposite statement : quantified opposite statement : z }| { p x > 0 and x < 1 x 0 or x 1 z }| { q . p and q / D . p/ or . q/ x < 0 or x > 1 x 0 and x 1 . p or q / D . p/ and . q/ (3) For every x 0; x has a square root. (True) 8x 0 .x has a square root/: There exists x 0 such that x does not have a square root. (False) 9x 0 .x has a square root/: 0; there is y 0 such that y 2 D x : (True) For every x 8x 9x 0; 9y 0 . y 2 D x /: There exists x 0 such that for every y 0; y 2 6D x : (False) 0; 8y 0 . y2 D x /: From examples (3) and (4), we see that the rule for negating statements with quantifiers is first switch every 8 to 9 and every 9 to 8, then negate the remaining part of the statement. If-then Statements. If-then statements occur frequently in mathematics. We will need to know some equivalent ways of expressing an if-then statement to do proofs. The statement “if p; then q ” may also be stated as “ p implies q ”, “ p only if q ”, “ p is sufficient for q ”, “q is necessary for p” and is commonly denoted by “ p H) q ”. For example, the statement “if x D 3 and y D 4; then x 2 C y 2 D 25” may also be stated as “x D 3 and y D 4 are sufficient for x 2 C y 2 D 25” or “x 2 C y 2 D 25 is necessary for x D 3 and y D 4”. Example. (5) statement : opposite statement : rule : If x 0; then jx j D x : (True) x 0 and jx j 6D x : (False) . p H) q / D p and . q / Remark. Note p H) q D . . p H) q // . p and . q // D . p / or . q/ D . p / or q : D 3 For the statement “if p; then q ” . p H) q /; there are two related statements: the converse of the statement is “if q , then p” (q H) p) and the contrapositive of the statement is “if . q /; then . p/” ( q H) p). Examples. (6) statement : converse : contrapositive : (7) If x D 3; then x 2 D 9: (True) If x 2 D 9; then x D 3: (False, as x may be 3.) If x 2 6D 9; then x 6D 3: (True) 6 (True) 3 (True) 3 (True) statement : converse : contrapositive : statement : converse : contrapositive : x D 3 H) 2x D 2x D 6 H) x D 2x 6D 6 H) x 6D (8) If jx j D 3; then x D 3: (False, as x may be 3.) If x D 3; then jx j D 3: (True) If x 6D 3; then jx j 6D 3: (False, as x may be 3.) Remarks. Examples (6) and (7) showed that the converse of an if-then statement is not the same as the statement nor the opposite of the statement in general. Examples (6), (7) and (8) showed that an if-then statement and its contrapositive are either both true or both false. In fact, this is always the case because by the remark on the last page, . q/ H) . p/ D D D D . q / or . q or . p/ . p/ or q p H) q : p/ So an if-then statement and its contrapositive statement are equivalent. Finally, we introduce the terminology “ p if and only if q ” to mean “if p; then q ” and “if q ; then p”. The statement “ p if and only if q ” is the same as “ p is necessary and sufficient for q ”. We abbreviate “ p if and only if q ” by “ p () q ”. So p () q means p H) q and q H) p”. The phrase “if and only if” is often abbreviated as “iff”. Caution! Note 8 8 D 8 8 and 9 9 D 9 9 ; but 8 9 6D 9 8 : For example, “every student is assigned a number” is the same as “8 student, 9 number such that the student is assigned the number.” This statement implies different students may be assigned possibly different numbers. However, if we switch the order of the quantifiers, the statement becomes “9 number such that 8 student, the student is assigned the number.” This statement implies there is a number and every student is assigned that same number! 4 Chapter 2. Sets To read and write mathematical expressions accurately and concisely, we will introduce the language of sets. A set is a collection of “objects” (usually numbers, ordered pairs, functions, etc.) If object x is in a set S , then we say x is an element (or a member) of S and write x 2 S : If x is not an element of S , then we write x 62 S : A set having finitely many elements is called a finite set, otherwise it is called an infinite set. The empty set is the set having no objects and is denoted by ;: A set may be shown by listing its elements enclosed in braces (eg. f1; 2; 3g is a set containing the objects 1; 2; 3; the positive integer N D f1; 2; 3; : : :g, the integer ZD f: : : ; 2; 1; 0; 1; 2; : : :g, the empty set ; D fg) or by description enclosed in braces (eg. the rational numbers Q D f m : m 2 Z n 2 Ng; the real numbers R D fx : x is a real numberg ; n and the complex numbers C D fx C iy : x ; y 2 Rg:) In describing sets, the usual convention is to put the form of the objects on the left side of the colon and to state the conditions on the objects on the right side of the colon. The set will consist of all elements satisifying all the conditions. It is also common to use a vertical bar in place of colon in set descriptions. Examples. (i) The closed interval with endpoints a ; b is [a ; b] D fx : x 2 R and a (ii) The set of square numbers is f1; 4; 9; 16; 25; : : :g D fn 2 : n 2 Ng: x b g: (iii) The set of all positive real numbers is RC D fx : x 2 R and x > 0g: (If we want to emphasize this is a subset of R; we may stress x is real in the form of the objects and write RC D fx 2 R : x > 0g: If numbers are always taken to mean real numbers, then we may write simply RC D fx : x > 0g:) (iv) The set of points (or ordered pairs) on the line `m with equation y D mx is f.x ; mx / : x 2 Rg: For sets A; B ; we say A is a subset of B (or B contains A) iff every element of A is also an element of B : In that case, we write A B : (For the case of the empty set, we have ; S for every set S :) Two sets A and B are equal if and only if they have the same elements (i.e. A D B means A B and B A:) So A D B if and only if (x 2 A () x 2 B ). If A B and A 6D B ; then we say A is a proper subset of B and write A B : (For example, if A D f1; 2g; B D f1; 2; 3g; C D f1; 1; 2; 3g; then A B is true, but B C is false. In fact, B D C: Repeated elements are counted only one time so that C has 3 elements, not 4 elements.) For a set S ; we can collect all its subsets. This is called the power set of S and is denoted by P . S / or 2S : For examples, P .;/ D f;g; P .f0g/ D f;; f0gg and P .f0; 1g/ D f;; f0g; f1g; f0; 1gg: For a set with n elements, its power set will have 2n element. This is the reason for the alternative notation 2S for the power set of S : Power set is one operation of a set. There are a few other common operations of sets. Definitions. For sets A1 ; A2 ; : : : ; An ; (i) their union is A1 A2 An D fx : x 2 A1 or x 2 A2 or or x 2 An g; and x 2 An g; and x n 2 An g; (ii) their intersection is A1 \ A2 \ (iii) their Cartesian product is A1 A2 An D f.x 1 ; x 2 ; : : : ; x n / : x 1 2 A1 and x 2 2 A2 and \ An D fx : x 2 A1 and x 2 A2 and (iv) the complement of A2 in A1 is A1 n A2 D fx : x 2 A1 and x 62 A2 g: Examples. (i) f1; 2; 3g f3; 4g D f1; 2; 3; 4g; f1; 2; 3g \ f2; 3; 4g D f2; 3g; f1; 2; 3g n f2; 3; 4g D f1g: (ii) [ 2; 4] \ N D f1; 2; 3; 4g; [0; 2] (iv) R [1; 5] [4; 6] D [0; 6]: (iii) .[0; 7] \ Z n fn 2 : n 2 Ng D f0; 1; 2; 3; 4; 5; 6; 7g n f1; 4; 9; 16; 25; : : :g D f0; 2; 3; 5; 6; 7g: / R R D f.x ; y ; z / : x ; y ; z 2 Rg; Q A; .R n Q/ D f.a ; b/ : a is rational and b is irrationalg: A 5 Remarks. (i) For the case of the empty set, we have A ;D AD; A \ ; D ; D ; \ A; ;D;D; A; A n ; D A and ; n A D ;: (ii) The notions of union, intersection and Cartesian product may be extended to infinitely many sets similarly. The union is the set of objects in at least one of the sets. The intersection is the set of objects in every one of the sets. The Cartesian product is the set of ordered tuples such that the i -th coordinate must belong to the i -th set. n (iii) The set A1 notation A1 A2 A2 A3 An may be written as k D1 Ak : If for every positive integer k ; there is a set Ak ; then the 1 k D1 may be abbreviated as Ak or x 2S the union of all the sets Ax ’s for all x 2 S is denoted by Cartesian product. Examples. (i) .[1; 2] (ii) \h n2N Ax : Similar abbreviations exist for intersection and k 2N Ak : If for every x 2 S ; there is a set Ax ; then [2; 3] [3; 4] h [4; 5] h / \ ZD [1; C1/ \ ZD N: D [0; 1]: 0; 1 C h 1 1 1 1 D [0; 2/ \ 0; 1 \ 0; 1 \ 0; 1 \ n 2 3 4 (iii) For every k 2 N; let Ak D f0; 1g; then A1 A2 A3 D f.x 1 ; x 2 ; x 3 ; : : :/ : each x k is 0 or 1 for k D 1; 2; 3; : : :g: (iv) For each m 2 R; let `m be the line with equation y D mx on the plane, then and \ m 2R `m D f.0; 0/g: m 2R `m D R2 n f.0; y / : y 2 R; y 6D 0g (v) Show that if A B and C D ; then A \ C B \ D: Reason. For every x 2 A \ C; we have x 2 A and x 2 C: Since A B and C D ; we have x 2 B and x 2 D ; which imply x 2 B \ D : Thus, we see that every element in A \ C is also in B \ D : Therefore, A \ C B \ D : (vi) Show that . A B/ n C D . A n C/ . B n C /: Reason. For every x 2 . A B / n C; we have x 2 A B and x 62 C: So either x 2 A or x 2 B : In the former case, x 2 A n C or in the latter case, x 2 B n C: So x 2 . A n C / . B n C /: Hence, . A B / n C . A n C / . B n C /: Conversely, for every x 2 . A n C / . B n C /; either x 2 A n C or x 2 B n C: In the former case, x 2 A and x 62 C or in the latter case, x 2 B and x 62 C: In both cases, x 2 A B and x 62 C: So x 2 . A B / n C: Hence, . A n C / . B n C / . A B / n C: Combining with the conclusion of the last paragraph, we have . A B / n C D . A n C / . B n C /: We shall say that sets are disjoint iff their intersection is the empty set. Also, we say they are mutually disjoint iff the intersection of every pair of them is the empty set. A relation on a set E is any subset of E E : The following is an important concept that is needed in almost all branches of mathematics. It is a tool to divide (or partition) the set of objects we like to study into mutually disjoint subsets. Definition. An equivalence relation R on a set E is a subset R of E (a) (reflexive property) for every x 2 E ; .x ; x / 2 R ; (b) (symmetric property) if .x ; y / 2 R ; then . y ; x / 2 R ; (c) (transitive property) if .x ; y /; . y ; z / 2 R ; then .x ; z / 2 R : We write x y if .x ; y / 2 R : For each x 2 E ; let [x ] D f y : x containing x : Note that every x 2 [x ] by (a) so that [x ] D E : If x x2E E such that y g: This is called the equivalence class y ; then [x ] D [ y ] because by (b) and (c), z 2 [x ] () z x () z y () z 2 [ y ]: If x 6 y ; then [x ] \ [ y ] D ; because assuming z 2 [x ] \ [ y ] will lead to x z and z y ; which imply x y ; a contradiction. So every pair of equivalence classes are either the same or disjoint. Therefore, R partitions the set E into mutually disjoint equivalence classes. 6 Examples. (1) (Geometry) For triangles T1 and T2 ; define T1 T2 if and only if T1 is similar to T2 : This is an equivalence relation on the set of all triangles as the three properties above are satisfied. For a triangle T ; [T ] is the set of all triangles similar to T : (2) (Arithmetic) For integers m and n ; define m n if and only if m n is even. Again, properties (a), (b), (c) can easily be verified. So this is also an equivalence relation on Z There are exactly two equivalence classes, namely : [0] D f: : : ; 4; 2; 0; 2; 4; : : :g (even integers) and [1] D f: : : ; 5; 3; 1; 1; 3; 5; : : :g (odd integers). Two integers in the same equivalence class is said to be of the same parity. (3) Some people think that properties (b) and (c) imply property (a) by using (b), then letting z D x in (c) to conclude .x ; x / 2 R : This is false as shown by the counterexample that E D f0; 1g and R D f.1; 1/g; which satisfies properties (b) and (c), but not property (a). R fails property (a) because 0 2 E ; but .0; 0/ 62 R as 0 is not in any ordered pair in R : A function (or map or mapping) f from a set A to a set B (denoted by f : A ! B ) is a method of assigning to every a 2 A exactly one b 2 B : This b is denoted by f .a / and is called the value of f at a : Thus, a function must be well-defined in the sense that if a D a 0 ; then f .a / D f .a 0 /: The set A is called the domain of f (denoted by dom f ) and the set B is called the codomain of f (denoted by codom f ). We say f is a B -valued function (eg. if B D R; then we say f is a real-valued function.) When the codomain B is not emphasized, then we may simply say f is a function on A: The image or range of f (denoted by f . A/ or im f or ran f ) is the set f f .x / : x 2 Ag: (To emphasize this is a subset of B ; we also write it as f f .x / 2 B : x 2 Ag:) The set G D f x ; f .x / : x 2 Ag is called the graph of f : Two functions are equal if and only if they have the same graphs. In particular, the domains of equal functions are the same set. ; Examples. The function f : Z ! R given by f .x / D x 2 has dom f D Z codom f D R: Also, ran f D f0; 1; 4; 9; 16; : : :g: This is different from the function g : R ! R given by g .x / D x 2 because dom g D R 6D dom f : Also, a function may have more than one parts in its definition, eg. the absolute value function h : R ! R defined by x if x 0 : Be careful in defining functions. The following is bad: let x n D . 1/n and i .x n / D n : The h .x / D x if x < 0 rule is not well-defined because x 1 D 1 D x 3 ; but i .x 1 / D 1 6D 3 D i .x 3 /: Definitions. (i) The identity function on a set S is IS : S ! S given by IS .x / D x for all x 2 S : n (ii) Let f : A ! B ; g : B 0 ! C be functions and f . A/ B 0 : The composition of g by f is the function g f : A ! C defined by .g f /.x / D g . f .x // for all x 2 A: A: The function f jC : C ! B defined by f jC .x / D f .x / for every x 2 C (iii) Let f : A ! B be a function and C is called the restriction of f to C: (iv) A function f : A ! B is surjective (or onto) iff f . A/ D B : (v) A function f : A ! B is injective (or one-to-one) iff f .x / D f . y / implies x D y : (vi) A function f : A ! B is a bijection (or a one-to-one correspondence) iff it is injective and surjective. (vii) For an injective function f : A ! B ; the inverse function of f is the function f f 1 . y / D x () f .x / D y : 1 : f . A/ ! A defined by Remarks. A function f : A ! B is surjective means f . A/ D B ; which is the same as saying every b 2 B is an f .a / for at least one a 2 A: In this sense, the values of f do not omit anything in B : We will loosely say f does not omit any element of B for convenience. However, there may possibly be more than one a 2 A that are assigned the same b 2 B : Hence, the range of f may repeat some elements of B : If A and B are finite sets, then f surjective implies the number of elements in A is greater than or equal to the number of elements in B : Next, a function f : A ! B is injective means, in the contrapositive sense, that x 6D y implies f .x / 6D f . y /; which we may loosely say f does not repeat any element of B : However, f may omit elements of B as there may possibly be elements in B that are not in the range of f : So if A and B are finite sets, then f injective implies the number of elements in A is less than or equal to the number of elements in B : Therefore, a bijection from A to B is a function whose values do not omit nor repeat any element of B : If A and B are finite sets, then f bijective implies the number of elements in A and B are the same. 7 Remarks (Exercises). (a) Let f : A ! B be a function. We have f is a bijection if and only if there is a function g : B ! A such that g f D I A and f g D I B : (In fact, for f bijective, we have g D f 1 is bijective.) (b) If f : A ! B and h : B ! C are bijections, then h f : A ! C is a bijection. (c) Let A; B be subsets of R and f : A ! B be a function. If for every b 2 B ; the horizontal line y D b intersects the graph of f exactly once, then f is a bijection. Example. Show that f : [0; 1] ! [3; 4] defined by f .x / D x 3 C 3 is a bijection. Method 1 If f .x / D f . y /; then x 3 C 3 D y 3 C 3; which implies x 3 D y 3 : Taking cube roots of both sides, we get p x D y : Hence f is injective. Next, for every y 2 [3; 4]; solving the equation x 3 C 3 D y for x ; we get x D 3 y 3: Since y 2 [3; 4] implies y 3 2 [0; 1]; we see x 2 [0; 1]: Then f .x / D y : Hence f is surjective. Therefore, f is bijective. Method 2. p 3 Define g : [3; 4] ! [1; 2] by g . y / D 3 y 3: For every x 2 [0; 1] and y 2 [3; 4]; .g f /.x / D g .x 3 C 3/ D p p 3 C 3/ .x 3 D x and . f g /. y / D f . 3 y 3/ D . 3 y 3/3 C 3 D y : By remark (a) above, f is a bijection. p To deal with the number of elements in a set, we introduce the following concept. For sets S1 and S2 ; we will define S1 S2 and say they have the same cardinality (or the same cardinal number) if and only if there exists a bijection from S1 to S2 : This is easily checked to be an equivalence relation on the collection of all sets. For a set S ; the equivalence class [ S ] is often called the cardinal number of S and is denoted by card S or j S j: This is a way to assign a symbol for the number of elements in a set. It is common to denote card ; D 0; , for a positive integer n ; card f1; 2; : : : ; n g D n ; card N D @0 (read aleph-naught) and card R D c (often called the cardinality of the continuum). 8 Chapter 3. Countability Often we compare two sets to see if they are different. In case both are infinite sets, then the concept of countable sets may help to distinguish these infinite sets. Definitions. A set S is countably infinite iff there exists a bijection f : N ! S (i.e. N and S have the same cardinal number @0 :) A set is countable iff it is a finite or countably infinite set. A set is uncountable iff it is not countable. Remarks. Suppose f : N ! S is a bijection. Then f is injective means f .1/; f .2/; f .3/; : : : are all distinct and f is surjective means f f .1/; f .2/; f .3/; : : :g D S : So n 2 N $ f .n / 2 S is a one-to-one correspondence between N and S : Therefore, the elements of S can be listed in an “orderly” way (as f .1/; f .2/; f .3/; : : : ) without repetition or omission. Conversely, if the elements of S can be listed as s1 ; s2 ; : : : without repetition or omission, then f : N ! S defined by f .n / D sn will be a bijection as no repetition implies injectivity and no omission implies surjectivity. Bijection Theorem. Let g : S ! T be a bijection. S is countable if and only if T is countable. (Reasons. The finite set case is clear. For infinite sets, it is true because S countable implies there is a bijective function f : N ! S ; which implies h D g f : N ! T is bijective, i.e. T is countable. For the converse, h is bijective implies f D g 1 h is bijective.) Remarks. Similarly, taking contrapositive, S is uncountable if and only if T is uncountable. Basic Examples. (1) N is countably infinite (because the identity function IN.n / D n is a bijection). (2) Zis countably infinite because the following function is a bijection (one-to-one correspondence): ND f f f ZD # 1; 2; 0; 1; # # 3; # 4; # 5; # 6; # 7; # 8; 4; # 9; # 1; 2; n 2 2; 3; 3; ::: ::: 4; : : : g g: The function f : N ! Zis given by f .n / D by g .m / D if n is even 2m if m > 0 : Just check g f D IN and f g D IZ : 1 2m if m 0 (3) N N D f.m ; n /: m ; n 2 Ng is countably infinite. .1 ; 1 / (Diagonal Counting Scheme) Using the diagram on the right, define f : N ! N N by f .1/ D .1; 1/, f .2/ D .2; 1/, f .3/ D .1; 2/, .2; 1/ f .4/ D .3; 1/, f .5/ D .2; 2/, f .6/ D .1; 3/, : : : , then f is injective because no ordered pair is repeated. Also, f is surjective because .3 ; 1 / ! mX 2 Cn .m C n 2/.m C n 1/ .m ; n / D f kCn D f Cn : 2 k D0 .4 ; 1 / . n 2 1 / if n is odd and its inverse function g : Z! N is given .1; 2/ .2; 2/ .3; 2/ : : : : : : .1 ; 3 / .2 ; 3 / : : : : : : : : : : : : .1; 4/ : : : : : : : : : : : : : : : : : : : : : (4) The open interval .0; 1/ D fx : x 2 R and 0 < x < 1g is uncountable. Also, R is uncountable. f .1/ f .2/ f .3/ f .4/ D D D D : : : 0:a11 a12 a13 a14 : : : 0:a21 a22 a23 a24 : : : 0:a31 a32 a33 a34 : : : 0:a41 a42 a43 a44 : : : : : : Suppose .0; 1/ is countably infinite and f : N ! .0; 1/ is a bijection as shown on the left. Consider the number x whose decimal representation is 2 if ann D 1 0:b1 b2 b3 b4 : : :, where bn D . Then 0 < x < 1 and x 6D f .n / 1 if ann 6D 1 for all n because bn 6D ann . So f cannot be surjective, a contradiction. Next R is uncountable because tan .x 1 / provides a bijection from .0; 1/ onto R: 2 To determine the countability of more complicated sets, we will need the theorems below. 9 Countable Subset Theorem. Let A uncountable, then B is uncountable.) B : If B is countable, then A is countable. (Taking contrapositive, if A is An is countable. In general, if S is countable s 2S Countable Union Theorem. If An is countable for every n 2 N; then n2N (say f : N ! S is a bijection) and As is countable for every s 2 S ; then As D countable union of countable sets is countable.) n2N A f .n/ is countable. (Briefly, Product Theorem. If A, B are countable, then A B D f.a ; b/: a 2 A; b 2 B g is countable. In fact, if A1 ; A2 ; : : : ; An are countable, then A1 A2 An is countable (by mathematical induction). (Sketch of Reasons. For the countable subset theorem, if B is countable, then we can list the elements of B and to count the elements of A, we can skip over those elements of B that are not in A: For the countable union theorem, if we list the elements of A1 in the first row, the elements of A2 in the second row, : : : ; then we can count all the elements by using the diagonal counting scheme. As for the product theorem, we can imitate the example of N N and also use the diagonal counting scheme.) Examples. (5) Q D o m m : m 2 Z : For every n 2 N; the function f n : Z! Sn given by f n .m / D n n is a bijection (with fn 1 m D m ), so Sn is countable by the bijection theorem. Therefore, Q is countable by the n countable union theorem. (Then subsets of Q like Zn f0g; N f0g; Q \ .0; 1/ are also countable.) 1 Sn ; where Sn nD1 D n (6) R n Q is uncountable. (In fact, if A is uncountable and B is countable, then A n B is uncountable as A n B countable implies . A \ B / . A n B / D A countable by the countable union theorem, which is a contradiction). (7) C D fx C iy : x ; y 2 Rg contains R and R is uncountable, so by the countable subset theorem, C is uncountable. (8) Show that the set A D fr m : m 2 N; r 2 .0; 1/g is uncountable, but the set B D fr m : m 2 N; r 2 Q \ .0; 1/g is countable. Solution. Taking m D 1; we see that .0; 1/ p A: Since .0; 1/ is uncountable, A p uncountable. Next we will is observe that B D Bm ; where Bm D fr m : r 2 Q \ .0; 1/g D fr m g for each m 2 N: Since m 2N r 2Q\.0;1/ p Q \ .0; 1/ is countable and fr m g has 1 element for every r 2 Q \ .0; 1/; Bm is countable by the countable union theorem. Finally, since N is countable and Bm is countable for every m 2 N; B is countable by the countable union theorem. (9) Show that the set L of all lines with equation y D mx C b; where m ; b 2 Q; is countable. Solution. Note that for each pair m ; b of rational numbers, there is a unique line y D mx C b in the set L : So the function f : Q Q ! L defined by letting f .m ; b/ be the line y D mx C b (with f 1 sending the line back to .m ; b/) is a bijection. Since Q Q is countable by the product theorem, so the set L is countable by the bijection theorem. (10) Show that if An D f0; 1g for every n 2 N; then A1 A2 A3 is uncountable. (In particular, this shows that the product theorem is not true for infintely many countable sets.) Solution. Assume A1 A2 A3 D f.a1 ; a2 ; a3 ; : : :/ : each ai D 0 or 1g is countable and f : N ! A1 A2 A3 is a bijection. Following example (4), we can change the n -th coordinate of f .n / (from 0 to 1 or from 1 to 0) to produce an element of A1 A2 A3 not equal to any f .n /; which is a contradiction. So it must be uncountable. (11) Show that the power set P .N/ of all subsets of N is uncountable. Solution. As in example (10), let An D f0; 1g for every n 2 N: Define g : P .N/ ! A1 A2 A3 by 1 if m 2 S : (For example, g .f1; 3; 5; : : :g/ D .1; 0; 1; 0; 1; : : :/:) Note g . S / D .a1 ; a2 ; a3 ; : : :/; where am D 0 if m 62 S g has the inverse function g 1 .a1 ; a2 ; a3 ; : : :/ D fm : am D 1g: Hence g is a bijection. Since A1 A2 A3 is uncountable, so P .N/ is uncountable by the bijection theorem. 10 p p (12) Show that the set S of all nonconstant polynomials with integer coefficients is countable. Solution. For n 2 N; the set of Sn of all polynomials of degree n with integer coefficients is countable because the function f : Sn ! .Zn f0g/ Z Zdefined by f .an x n C an 1 x n 1 C C a 0 / D .a n ; a n 1 ; : : : ; a 0 / is a bijection and .Zn f0g/ Z Zis countable by the product theorem. So, S D Sn is countable by n2N the countable union theorem. (13) Show that there exists a real number, which is not a root of any nonconstant polynomial with integer coefficients. Solution. For every nonconstant polynomial f with integer coefficients, let R f denotes the set of roots of f : Then R f has at most (deg f ) elements, hence R f is countable. Let S be the set of all nonconstant polynomials with R f is the set of all roots of nonconstant integer coefficients, which is countable by the last example. Then polynomials with integer coefficients. It is countable by the countable union theorem. Since R is uncountable, Rn R f is uncountable by the fact in example (6). So there exist uncountably many real numbers, which are not roots of any nonconstant polynomial with integer coefficients. Remarks. Any number which is a root of a nonconstant polynomial with integer coefficients is called an algebraic number. A number which is not a root of any nonconstant polynomial with integer coefficients is called a transcendental number. transcendental numbers? If so, are there finitely many or countably many such numbers? Since every rational number a is the root of the polynomial bx a ; every rational number is algebraic. There are irrational numbers like b p 2; which are algebraic because they are the roots of x 2 2: Using the identity cos 3 D 4 cos3 3 cos ; the irrational number cos 20 is easily seen to be algebraic as it is a root of 8x 3 6x 1: Example (13) showed that there are only countably many algebraic numbers and there are uncountably many transcendental real numbers. In a number theory course, it will be shown that and e are transcendental. Theorem. (1) (Injection Theorem) Let f : A ! B be injective. If B is countable, then A is countable. (Taking contrapositive, if A is uncountable, then B is uncountable.) (2) (Surjection Theorem) Let g : A ! B be surjective. If A is countable, then B is countable. (Taking contrapositive, if B is uncountable, then A is uncountable.) (Reasons. For the first statement, observe that the function h : A ! f . A/ defined by h .x / D f .x / is injective (because f is injective) and surjective (because h . A/ D f . A/). So h is a bijection. If B is countable, then f . A/ is countable by the countable subset theorem, which implies A is countable by the bijection theorem. For the second statement, observe that B D g . A/ D x2 A f 2S f 2S fg .x /g: If A is countable, then it is a countable union of countable sets. By the countable union theorem, B is countable.) Examples. (14) Show Q is countable by using the injection theorem. Solution. Define f : Q ! Z N by f .x / D .m ; n /; where m = n is the reduced fraction form of x : Then f is injective because f .x / D f .x 0 / D .m ; n / implies x D m = n D x 0 : Since Z N is countable by product theorem, so Q is countable by the injection theorem. (15) Let A1 be uncountable and A2 ; : : : ; An be nonempty sets. Show that A1 A2 An is uncountable. An ! A1 by g .x 1 ; x 2 ; : : : ; x n / D x 1 : Since A2 ; : : : ; An are nonempty, let Solution. Define g : A1 A2 a2 2 A2 ; : : : ; an 2 An : Then for every a1 2 A1 ; we have g .a1 ; a2 ; : : : ; an / D a1 so that g is surjective. Since A1 is uncountable, by the surjection theorem, A1 A2 An is uncountable. The following is a famous statement in mathematics. Continuum Hypothesis. If S is uncountable, then there exists at least one injective function f : R ! S ; i.e. every uncountable set has at least as many elements as the real numbers. In 1940, Kurt G¨ del showed that the opposite statement would not lead to any contradiction. In 1966, Paul Cohen o won the Fields’ Medal for showing the statement also would not lead to any contradiction. So proof by contradiction may not be applied to every statement. 11 Chapter 4. Series Definitions. A series is the summation of a countable set of numbers in a specific order. If there are finitely many numbers, then the series is a finite series, otherwise it is an infinite series. The numbers are called terms. The sum of the first n terms is called the n-th partial sum of the series. An infinite series is of the form a1 C a2 C a3 C : : : or we may write it as |{z} |{z} |{z} 1st term 2nd term 3rd term 1 X k D1 ak : The first partial sum is S1 D a1 . The second partial sum is S2 D a1 C a2 . The n th partial sum is Sn D a1 C a2 C : : : C an . Series are used frequently in science and engineering to solve problems or approximate solutions. (E.g. trigonometric or logarithm tables were computed using series in the old days.) 1 1 1 1 1 Examples.(1) 1C 2 C 1 C 1 C 16 C: : : D? ( Sn D 1C 1 C 1 C: : : C 21n D 2 21n ; 1C 2 C 1 C 1 C 16 C: : : D lim 2 n D 2.) 48 24 48 n!1 2 We say the series converges to 2, which is called the sum of the series. (2) 1 C 1 C 1 C 1 C 1 C 1 C : : : D 1 ( Sn D 1 C 1 C : : : C 1 D n , lim Sn D 1.) We say the series diverges (to 1). | {z } (3) 1 1C1 1C1 1C1 1 C : : : . ( Sn D n n n!1 1 0 if n is odd , lim Sn doesn’t exist.) We say the series diverges. if n is even n!1 n!1 n!1 Definitions. A series 1 X k D1 ak D a1 C a2 C a3 C : : : converges to a number S iff lim .a1 C a2 C : : : C an / D lim Sn D S : In that case, we may write 1 X k D1 ak D S and say S is the sum of the series. A series diverges to 1 iff the partial sum Sn tends to infinity as n tends to infinity. A series diverges iff it does not converge to any number. Remarks. (1) For every series 1 X k D1 ak , there is a sequence (of partial sums) f Sn g. Conversely, if the partial sum sequence S1 , : : : , ak D Sk Sk Sn 1 / D Sn . So f Sn g is the partial sum sequence of f Sn g is given, we can find the terms an as follows: a1 D S1 , a2 D S2 a1 C : : : C an D S1 C . S2 S1 / C : : : C . Sn 1 X k D1 1 for k > 1. Then ak . Conceptually, series and sequences are equivalent. So to study series, we can use facts about sequences. (2) Let N be a positive integer. because B D lim .a N C n!1 1 X k D1 ak converges to A if and only if 1 X kD N ak converges to B D A .a1 C C aN 1/ C an / D lim .a1 C a2 C n!1 C an / .a 1 C C aN 1/ D A .a1 C C a N 1 /: So to see if a series converges, we may ignore finitely many terms. Theorem. If 1 X k D1 ak converges to A and 1 X k D1 bk converges to B ; then 1 X k D1 .ak C bk / D A C B D 1 X k D1 ak C 1 X k D1 bk ; 1 X k D1 .a k bk / D A BD 1 X k D1 ak 1 X k D1 bk ; 1 X k D1 cak D c A D c 1 X k D1 ak for any constant c: For simple series such as geometric or telescoping series, we can find their sums. 12 Theorem (Geometric Series Test). We have 1 X r n C1 r D lim .1 C r C r C : : : C r / D lim D n!1 n!1 1 r k D0 k 2 n ( 1 1 if jr j < 1 : 1r doesn’t exist otherwise Example. 0:999 tions! D 9 9 9 C C C 10 100 1000 D 9 1 . / D 1 D 1:000 1 10 1 10 : So 1 has two decimal representa- Theorem (Telescoping Series Test). We have D lim .b1 n!1 1 X k D1 .bk bkC1 / D lim .b1 n!1 n!1 b2 / C .b2 b3 / C C .bn bnC1 / bnC1 / D b1 n!1 lim bnC1 converges if and only if lim bn is a number. Examples. (1) 1 X 1 k .k C 1/ k D1 51=.kC1// D .5 D 1 X p 1 k k D1 . 1 /D 1 k C1 p p 3 1 1 C 2 2 D5 1 1 C 3 3 1 C 4 D1 n!1 lim 1 D 1: nC1 (2) 1 X k D1 .51= k 5/ C . 5 5/ C k !1 lim 51=.kC1/ D 5 50 D 4: If a series is not geometric or telescoping, we can only determine if it converges or diverges. This can be done most of the time by applying some standard tests. If the series converges, it may be extremely difficult to find the sum! Theorem (Term Test). If k !1 1 X k D1 ak converges, then lim ak D 0. (If lim ak 6D 0, then the series k !1 k !1 1 X k D1 ak diverges.) If lim ak D 0, the series 1 X k D1 ak may or may not converge. Sk 1 / D S S D 0.) (Reason. Suppose 1 X k D1 ak converges to S : Then lim Sn D S and lim ak D lim . Sk n!1 k !1 k !1 Term test is only good for series that are suspected to be divergent! Examples. (1) 1 C 1 C 1 C 1 C : : :. Here ak D 1 for all k , so lim ak D 1. Series diverges. (2) (3) k !1 1 X 1 X k D1 1 1 1 1 cos. / D cos 1 C cos C cos C : : : diverges because lim cos. / D cos 0 D 1 6D 0. k !1 k 2 3 k k D1 cos k D cos 1 C cos 2 C cos 3 C : : : diverges because lim cos k 6D 0: (Otherwise, lim cos k D 0: Then p k !1 k !1 k !1 k !1 lim j sin k j D lim 1 2 1 cos2 k D 1 and 0 D lim j cos.k C 1/j D lim j cos k cos 1 k !1 k !1 sin k sin 1j D sin 1 6D 0; a contradiction.) C1 4 1 8 (4) 1 C : : :. Here ak D . 1k1 2 / for all k , so lim ak D 0: (Term test doesn’t apply!) Series converges k !1 by the geometric series test. 1111111 1 (5) 1 C C C C C C C C : : : C C : : :. We have lim ak D 0: (Term test doesn’t apply.) Series k !1 2 4 4 {z 4 4 | 8 | {z 2 }| }8 {z } diverges to 1 because S1 2 times 4 times 8 times S2 S3 and S2n 1 D n has limit 1: 13 For a nonnegative series 1 X k D1 ak (i.e. ak 0 for every k ), we have S1 S2 S3 number or equal to C1. So either or 1 X k D1 ak converges to a number or 1 X k D1 ak diverges to C1. (In short, either : : : and lim Sn must exist as a n!1 1 X k D1 ak D S 1 X k D1 ak D C1.) For nonnegative series, we have the following tests. Theorem (Integral Test). Let f : [1; C1/ ! R decrease to 0 as x ! C1. Then Z 1 1 X k D1 f .k / converges if and only if 1 f .x / dx < 1. (Note in general, 1 X k D1 f .k / 6D Z 1 1 f . x / d x :) Z 1 (Reason. This follows from f .2/ C f .3/ C : : : C f .n / C as shown in the figures below.) 1 f .x /d x f .1/ C f .2/ C : : : C f .n 1/ C f (1) f (2) f (3) ... 1 2 3 4 ... n 1 2 3 f (2) ... 4 ... n Examples. (1) Consider the convergence or divergence of 1 : 1 C k2 k D1 Z1 1 1 1 As x % 1; 1 C x 2 % 1; so & 0: Now d x D arctan x D 1 C x2 1 C x2 2 1 1 converges. 1 X 4 < 1: So 1 X 1 1 C k2 k D1 (2) Consider the convergence or divergence of 1 X 1 X 1 1 and . k ln k k .ln k /2 k D2 k D2 As x % 1; x ln x and x .ln x /2 % 1; so their reciprocals decrease to 0: Now Z 2 1 dx x ln x D ln.ln x / 1 2 D 1: So 1 X 1 diverges. Next k ln k k D2 Z 2 1 dx D x .ln x /2 1 X 11 1 1 D < 1: So converges. ln x 2 ln 2 k .ln k /2 k D2 1 X 1 1 1 1 D 1C C C C : : : converges if and only if p > 1. p p p k 2 3 4p k D1 (Reason. For p 0; the terms are at least 1, so the series diverges by term test. For p > 0; f .x / D x1p decreases Z1 Z1 1 1 x p C1 1 1 to 0 as x ! C1. Since dx D D d x D .ln x /j1 D 1 if p D 1 and if p > 1; 1 xp pC1 1 p1 xp 1 Z1 11 1 x p C1 dx D D 1 if p < 1, the integral test gives the conclusion.) p x pC1 1 1 Theorem (p-test). For a real number p; . p/ D Remarks. For even positive integer p; the value of . p/ was computed by Euler back in 1736. He got .2/ D 2 6 ; .4/ D 4 90 ; :::; .2n / D . 1/nC1 14 .2 /2n B2n ; :::; 2.2n /! kC1 Bm for k m m D0 1980’s, R. Apery was able to show .3/ was irrational. where B0 D 1 and .k C 1/ Bk D Theorem (Comparison Test). Given vk diverges, then (Reason. vk uk k1 X 1: The values of .3/; .5/; : : : are unknown. Only in the 0 for every k. If 1 X k D1 vk converges, then 1 X k D1 u k converges. If 1 X k D1 uk 1 X k D1 vk diverges. 0) uk 1 X k D1 vk 1 X k D1 uk 0. If 1 X k D1 vk is a number, then 1 X k D1 u k is a number. If 1 X k D1 u k D C1, then 1 X k D1 vk D C1.) Theorem (Limit Comparison Test). Given u k , vk > 0 for every k : If lim 1 X 1 1 X X v vk converge) or (both diverge to C1). If lim k D 0; then u k converges ) vk converges. If k !1 u k k D1 k D1 k D1 k D1 1 1 X X vk lim D 1; then u k diverges ) vk diverges. k !1 u k k D1 k D1 P P P vk (Sketch of Reason. For k large, L . For L > 0; vk Lu k D L u k . If one series converges, then the uk other also converges. If one diverges (to C1), so does the other. For L D 0; vk < u k eventually. For L D 1; vk > u k u k and eventually. So the last two statements follow from the comparison test.) Examples. Consider the convergence or divergence of the following series: 1 1 (1) cos k2 k k D1 1 X k !1 vk uk is a positive number L ; then either (both 1 X kC1 1 (4) sin . k2 1 k 2 C 5k k k D2 k D1 k D1 1 1 X1 X1 1 1 1 1 Solutions. (1) Since 0 < 2 cos < 2 and converges by p-test, cos converges. k k k k2 k2 k k D1 k D1 (2) 3k (3) (2) Since 0 < 3 2 k 1 X 1 X 1 X p 3k k2 p 1 for k p 2 and k 1 1 X 3 2 k D2 k diverges by the geometric series test, k C1 k 2 C5k p k k2 1 X k D2 3k k2 1 diverges. kC1 (3) When k is large, 2 k C 5k converges by p-test, (4) When k is large, k D1 p r 1p X kC1 k 2 C 5k D 3=2 . We compute lim k !1 k2 k D lim k !1 1 X1 k C 1 k2 D 1: Since k k 2 C 5k k 3 =2 k D1 converges by the limit comparison test. 1 1 1 sin is close to 0, so sin is close to because lim D 1 (i.e. sin as ! 0). !0 k k k 1 1 1 X1 X sin. k / sin 1 We compute lim D lim D 1. Since diverges by p-test, sin diverges by the limit 1 k !1 !0 k k k D1 k D1 k comparison test. For series with alternate positive and negative terms, we have the following test. Theorem (Alternating Series Test). If ck decreases to 0 as k ! 1 (i.e. c1 then 1 X k D1 c2 c3 ::: 0 and lim ck D 0), k !1 . 1/kC1 ck D c1 c2 C c3 c4 C c5 : : : converges. 15 (Reason. Since c1 c2 c3 : : : 0, we have 0 S2 S4 S6 : : : S5 S3 S1 : Since lim j Sn n!1 lim cn D 0, the distances between the partial sums decrease to 0 and so lim Sn must exist.) n!1 n!1 Sn 1 j D Examples. Both e k ln k k D2 k D1 and ek % 1; so 1=.k ln k / & 0 and e 1 X . 1/k and 1 X k k cos k converge by the alternating series test because as k % 1; k ln k % 1 & 0 and cos k D . 1/k : For series with arbitrary positive or negative term, we have the following tests. Theorem (Absolute Convergence Test). If jak j; we get 0 1 X k D1 jak j converges, then 1 X k D1 ak converges. (Reason. From jak j ak ak C jak j 2jak j: Since 1 X k D1 2jak j converges, so by the comparison test, 1 X k D1 .ak C jak j/ converges. Then 1 X k D1 1 X k D1 ak D 1 X k D1 .ak C jak j/ 1 X k D1 1 X k D1 jak j converges.) Definition. We say converges, but ak converges absolutely iff jak j converges. We say 1 X k D1 ak converges conditionally iff 1 X k D1 ak 1 X k D1 jak j diverges. Examples. Determine if the following series converge absolutely or conditionally (a) 1 X cos k k3 k D1 (b) cos k k3 1 X Solutions. (a) 1 X k D1 cos k . 1Ck k D1 1 X comparison test. So (b) 1 X 1 1 X1 X cos k 1 . Since converges by p-test, it follows that converges by the k3 k3 k3 k D1 k D1 k D1 Z 1 X 1 1 1 1 dx X1 X1 cos k D because cos k D . 1/k . D ln.1 C x / D1) diverges. 1Ck 1Ck 1Cx 1Ck 1 1 k D1 k D1 k D1 1 1 X cos k X 1 1 However, decreases to 0 as k ! C1. So by the alternating series test, D . 1/k 1Ck 1Ck 1Ck k D1 k D1 1 X cos k converges conditionally. 1Ck k D1 k !1 cos k converges absolutely by the absolute convergence test. k3 k D1 converges. Therefore Theorem (Ratio Test). If ak 6D 0 for every k and lim jakC1 =ak j exists, then 8 > >< > > > > > > < > > > > > >> : 1 ) 1 X 1 X 1 X k D1 k D1 ak converges absolutely akC1 lim D1 k !1 ak > > 1 1 ) ak may converge e:g: k2 k D1 k D1 ) 1 X ! 1 or diverge e:g: k k D1 1 X ! : ak diverges 16 akC1 akC1 akC2 akCn , then for k large, , , : : :, r , so jakCn j jak jr n and k !1 ak ak akC1 akCn 1 jak j C jakC1 j C jakC2 j C : : : jak j.1 C r C r 2 C r 3 C : : :/ which converges if r < 1 by the geometric series test and P P n akCn ak r diverges if r > 1 by the term test.) (Sketch of reason. Let r D lim Theorem (Root Test). If lim p k k !1 jak j exists, then k !1 lim p k 8 > > >< > > > > > < 1 ) 1 X 1 X 1 X k D1 k D1 ak converges absolutely ja k j > > > > > > > > :> D1 1 ) ak may converge e:g: k2 k D1 k D1 ) p k 1 X ! 1 or diverge e:g: k k D1 1 X ! : 1 ak diverges r: So jak j rk , P (Sketch of reason. Let r D lim k !1 p jak j, then for k large, k jak j jak j Pk r .) Examples. Consider the convergence or divergence of the following series: 1 1 X X k! 1 (1) (2) . 3k 2k kk k D1 k D1 Solutions. (1) Since lim k !1 1 3kC1 2kC1 1 3k 2k D lim k k !1 r 3kC1 1 3k 2k 3k 1 2k D lim 3 2kC1 k!1 1 . 2 /k 1 33 . 2 /kC1 3 D 1 X 1 1 < 1; by the ratio test, k 3 3 2k k D1 q converges. Alternatively, since lim k !1 D lim p k k !1 1 3k 2k D lim 1 1 k !1 test, 1 X k D1 3 k ./ 2k 3 D 1 < 1, by the root 3 1 3k 2k converges. D lim (2) Since lim .k C 1/! k k k !1 .k C 1/k C1 k ! 1 k !1 .1 C 1 /k k D 1 X k! 1 < 1; by the ratio test, converges. e kk k D1 Remarks. You may have observed that in example (1), the limit you got for applying the root test was the same as the limit you got for applying the ratio test. This was not an accident! akC1 p D r 2 R; then lim k ak D r: (This implies that the root test can be k !1 ak k !1 applied to more series than the ratio test.) Theorem. If ak > 0 for all k and lim Examples. (1) Let ak D k ; then lim k !1 p akC1 k C1 k D lim D 1: So, lim k D 1: k !1 k !1 ak k p k k! akC1 1 p k! 1 kk (2) Let ak D k ; then lim D as above. So lim k ak D lim D ; i.e. when k is large, k ! ; k !1 ak k !1 k !1 k k e e e which is a simple version of what is called Stirling’s formula. It is useful for estimating n ! when n is large. For 100 100 example, since log10 1:566; so 101:566; then we get 100! 10156:6; which has about 157 digits. e e Theorem (Summation by Parts). Let S j D j X k D1 ak D a1 C a2 C : : : C a j and 1bk D n1 X k D1 bkC1 bk D bkC1 .k C 1 / k bk ; then n X k D1 ak bk D Sn bn 17 Sk 1bk : (Reason. Note a1 D S1 and ak D Sk n X k D1 Sk 1 for k > 1. So, S 1 /b 2 C : : : C . S n b1 / Sn 1 /bn bn 1 /:/ ak bk D S1 b1 C . S2 D Sn bn S1 .b2 ::: Sn 1 .bn Example. Show that sin k converges. k k D1 1 1 1 Let ak D sin k and bk D : Using the identity sin m sin D cos.m k 2 2 Sk D sin 1 C sin 2 C C sin k D 1 X 1 / 2 1 cos.m C / ; we have 2 cos 1 2 cos.k C 1 / 2 2 sin 1 2 : This implies j Sk j 1 Sn for every k : Applying summation by parts and noting that lim D 0; we get n!1 n sin.1=2/ Sn n n1 X k D1 1 X 1 X k D1 n X sin k sin k D lim D lim n!1 n!1 k k k D1 k D1 ! Sk 1 kC1 1 k D 1 X k D1 Sk 1 k 1 kC1 : Now Sk 1 k 1 kC1 absolute convergence test, 1 X 1 X sin k D Sk k k D1 k D1 1 X1 1 sin.1=2/ kD1 k 1 k 1 kC1 1 k 1 D 1 by the telescoping series test. So by the sin.1=2/ converges. Inserting Parentheses and Rearrangements of Series. Definition. We say 1 X k D1 bk is obtained from 1 X k D1 ak by inserting parentheses iff there is a strictly increasing function C a p .2/ ; b3 D a p .2/C1 C C a p .3/; : : : : p : N f0g ! N f0g such that p.0/ D 0; b1 D a1 C C ap .1/ ; b2 D ap .1/C1 C (Note bn is the sum of kn D p.n / p.n 1/ terms.) Grouping Theorem. Let 1 X k D1 bk be obtained from 1 X k D1 ak by inserting parentheses. If 1 X k D1 ak converges to s ; then 1 X k D1 bk will converge to s : Next, if lim an D 0; kn is bounded and n!1 1 X k D1 bk converges to s ; then bk D lim tn D lim n!1 1 X k D1 ak will converge to s : (Reason. Let sn D let p.n / cr; j D p.n ap .i /Cr 0 n X k D1 ak and tn D n X k D1 bk : For the first part, 1 X k D1 p .n/ X k D1 1/ be bounded by M : For a positive integer j ; let p.i / j < p.i C 1/: For r D 1; 2; : : : ; M ; define 1 X if p.i / C r j : Then ak D lim s j D lim ti C lim .c1; j C C cM; j / D s C 0 C C 0 D s :) if p.i / C r > j j !1 i !1 j !1 k D1 n!1 ak D s : For the second part, Examples. (1) Since 1 X 1 111 1 DCCC C 2k 2 4 8 16 k D1 11 C 48 C converges to 1, so by the theorem, 1 1 1 1 C C C 128 256 512 1024 18 1 C 2 1 1 1 C C 16 32 64 C C D 1: (2) .1 1/ C converges to 0, but 1 1 C 1 1 C diverges by term test. So lim an D 0 is important. n!1 1111 111111 C C CC C converges to 0. However, the series Also, .1 1/ C 2222 333333 without parentheses diverges (as Sn2 D 1 and Sn2 Cn D 0) even though the terms have limit 0. So kn bounded is important. 1 2 C 1/ C .1 (3) Since 1 son test with 1 X 1 3 1 4 C D 1 X j D1 1 2j 1 11 C 23 1 2j 1 C 4 D 1 X 1 ), so by the theorem, 1 j2 j D1 1 converges (by the limit compari2 j .2 j 1 / j D1 1 X . 1/k C1 D converges to the same sum. k k D1 : N ! N such that bk D a .k/: Definition. 1 X k D1 bk is a rearrangement of 1 X k D1 ak iff there is a bijection 11111 Example. Given ln 2 D 1 C C C : : : (which converges conditionally). Consider the rearrangement 23456 111111 1 1 C : : :. Observe that 1C CC CC 3 |{z} | {z 7 |{z} | {z11 |{z} 25 49 6 | {z } } } 2C 1 2C 1 2C 1 .1 C 1 2 1 2 / C .1 3 C .1 3 1 4 1 4 1 2 / C .1 5 /C 1 6 C1 6 1 5 / C .1 7 C .1 7 1 8 1 8 1 4 / C : : : D ln 2 : : : D 1 ln 2 2 / C : : : D 3 ln 2: 2 1, 1 Riemann’s Rearrangement Theorem. Let ak 2 R and there is a rearrangement 1 X k D1 ak converge conditionally. For any x 2 R or x D 1 X k D1 a .k/ of ak 0 1 X k D1 ak such that 1 X k D1 a .k / D x : 0 if ak 0 : Then ak D pk jak j if ak < 0 qk and jak j D pk C qk : (Sketch of reason. Let pk D Now both 1 X k D1 pk ; 1 X k D1 if ak 0 and qk D if ak < 0 qk must diverge to C1: (If both converges, then their sum 1 X k D1 jak j will be finite, a contradiction. 1; a contradiction If one converges and the other diverges to C1; then 1 X k D1 ak D 1 X k D1 pk 1 X k D1 qk will diverges to also.) Let u n ; vn be sequences of real numbers having limits x and u n < vn ; u n < vnC1 ; v1 > 0: Now let P1 ; P2 ; : : : be the nonnegative terms of 1 X k D1 ak in the order they occur and Q 1 ; Q 2 ; : : : be the absolute value of the negative terms Pk ; in the order they occur. Since 1 X k D1 1 X k D1 Q k differ from 1 X k D1 pk ; 1 X Let m 1 ; k1 be the smallest integers such that P1 C C Pm 1 > v and P1 C C Pm 1 Q1 Q k1 < u 1 : Let m 2 ; k2 be the smallest integers such that P1 C C Pm 1 Q1 Q k1 C Pm 1 C1 C C Pm 2 > v2 and P1 C C Pm 1 Q 1 Q k1 C Pm 1 C1 C C Pm 2 Q k1 C1 Q k2 < u 2 and continue this way. This is possible since the sums of Pk and Q k are C1: Now if sn ; tn are the partial sums of this series P1 C C Pm 1 Q 1 Q k1 C whose last terms are Pm n ; Q kn ; respectively, then jsn vn j Pm n and jtn u n j Q kn by the choices of m n ; kn : Since Pn ; Q n have limit 0, so sn ; tn must have limit x : As all other partial sums are squeezed by sn and tn ; the series we constructed must have limit x :) Dirichlet’s Rearrangement Theorem. If ak 2 R and converges to the same sum as k D1 1 qk only by zero terms, they also diverges to C1: 1 X k D1 ak converges absolutely, then every rearrangement 1 X k D1 a .k/ 1 X k D1 ak . 19 (Reason. Define pk ; qk as in the last proof. Since pk ; qk Since a .k/ D p .k/ q .k/ ; we may view jak j; 1 X k D1 pk ; 1 X k D1 qk converge, say to p and q ; respectively. 1 X k D1 p .k/ as a rearrangement of the nonnegative terms of m X k D1 1 X k D1 ak and inserting 0; the p: zeros where a .k/ < 0: For any positive integer m ; the partial sum sm D partial sum sm is also increasing, hence As n ! 1; we get Example. p .k / 1 X k D1 pk D p: Since pk n X k D1 1 X k D1 p .k/ converges. Now, for every positive integer n ; pk 1 X k D1 p .k / 1 X k D1 p .k/ D p: Similarly, 1 1 C 2 22 1 X k D1 q .k/ D q : Then 1 X k D1 a .k / D p qD 1 X k D1 1 2 a k :) D 1 X k D1 . 1k /D 2 1 1 C 23 24 1 1 C6 7 2 {z 2 4 terms 1 C : : : converges (absolutely) to 25 1 1 1 C 16 5 2} | 2 1 . 3 1 1 C 14 15 2 2 1 1 C 12 13 2 {z 2 8 terms . 1 2 / 1 . 3 1 C::: 29 } 1 1 1 1 1 C 2C 4 C8 3 22 2 2 | {z 2 } | 2 terms 1 1 C 10 11 2 2 is a rearrangement of 1 X k D1 . 1k / , so it also converges to 2 Remarks. As a consequence of the rearrangement theorem, the sum of a nonnegative series is the same no matter how the terms are rearranged. Complex Series Complex numbers S1 ; S2; S3 ; : : : with Sn D u n C i vn are said to have limit lim Sn D u C i v iff lim u n D u n!1 n!1 and lim vn D v: A complex series is a series where the terms are complex numbers. The definitions of convergent, absolutely convergent and conditional convergent are the same. The remarks and the basic properties following the definitions of convergent and divergent series are also true for complex series. The geometric series test, telescoping series test, term test, absolute convergence test, ratio test and root test are also true for complex series. For z k D x k C iyk ; we have to x and n!1 1 X k D1 z k converges to z D x C iy if and only if 1 X k D1 x k converges 1 X k D1 yk converges to y : So complex series can be reduced to real series for study if necessary. Examples. (1) Note lim i n 6D 0 (otherwise 0 D lim ji n j D lim 1 is a contradiction). So n!1 n!1 n!1 1 X k D1 i k diverges by term test. 1 and converges by p-test implies converges absolutely. However, if k2 k2 k2 k D1 k D1 1 X zk z kC1 k 2 k2 jz j > 1; then lim D lim jz j D jz j > 1 implies diverges by the ratio test. k !1 .k C 1/2 z k k !1 .k C 1/2 k2 k D1 (2) If jz j 1; then z k2 k 1 X1 1 X zk 20 Chapter 5. Real Numbers Decimal representations and points on a line are possible ways of introducing real numbers, but they are not too convenient for proving many theorems. Instead we will introduce real numbers by its important properties. Axiomatic Formulation. There exists a set R (called real numbers) satisfying the following four axioms: (1) (Field Axiom) R is a field (i.e. R has two operations C and such that for any a , b, c 2 R, (ii) a C b D b C a , a b D b a , (iii) .a C b/ C c D a C .b C c/, .a b/ c D a .b c/, (i) a C b, a b 2 R, (iv) there are unique elements 0, 1 2 R with 1 6D 0 such that a C 0 D a , a 1 D a , (v) there is a unique element a 2 R such that a C . a / D 0; if a 6D 0; then there is a unique element a 1 such that a .a 1 / D 1: (vi) a .b C c/ D a b C a c.) (This axiom allows us to do algebra with equations. Define a b to mean a C . b/I ab to mean a bI a to b mean a .b 1 /: Also, define 2 D 1 C 1; 3 D 2 C 1; : : : :) (2) (Order Axiom) R has an (ordering) relation < such that for any a , b 2 R (i) exactly one of the following a < b, a D b, b < a is true, (ii) if a < b, b < c, then a < c, (iii) if a < b, then a C c < b C c, (iv) if a < b and 0 < c, then ac < bc. (This axiom allows us to work with inequalities. For example, using (ii) and (iii), we can see that if a < b and c < d ; then a C c < b C d because a C c < b C c < b C d : Also, we can get 0 < 1 (for otherwise 1 < 0 would imply by (iii) that 0 D 1 C . 1/ < 0 C . 1/ D 1; which implies by (iv) that 0 < . 1/. 1/ D 1; a contradiction). Now define a > b to mean b < a I a b to mean a < b or a D bI etc. Also, define closed interval [a ; b] D fx : a x bgI open interval .a ; b/ D fx : a < x < bgI etc. Part (i) of the order axiom implies any two real numbers can be compared. We define max.a1 ; : : : ; an / to be the maximum of a1 ; : : : ; an and similarly for minimum. Also, define jx j D max.x ; x /: Then x jx j and x j x j; i.e. jx j x jx j: Next jx j a if and only if x a and x a ; i.e. a x a : Finally, adding jx j x jx j and j y j y j y j; we get jx j j y j x C y jx j C j y j; which is the triangle inequality jx C y j jx j C j y j:) (3) (Well-ordering Axiom) N D f1; 2; 3; : : :g is well-ordered (i.e. for any nonempty subset S of N, there is m 2 S such that m x for all x 2 S . This m is the least element (or the minimum) of S ). (This axiom allows us to formulate the principle of mathematical induction later.) Definitions. For a nonempty subset S of R, S is bounded above iff there is some M 2 R such that x M for all x 2 S . Such an M is called an upper bound of S . The supremum or least upper bound of S (denoted by sup S or lub S ) Q Q is an upper bound M of S such that M M for all upper bounds M of S . (4) (Completeness Axiom) Every nonempty subset of R which is bounded above has a supremum in R. (This axiom allows us to prove results that have to do with the existence of certain numbers with specific properties, as in the intermediate value theorem.) 1 Examples. (1) For S D n : n 2 N D 1; 1 ; 1 ; : : : ; the upper bounds of S are all M 23 (2) For S D fx 2 R: x < 0g, the upper bounds of S are all M 0: So sup S D 0 62 S : 1: So sup S D 1 2 S : Definitions. N D f1; 2; 3; 4; : : :g is the natural numbers (or positive integers), ZD f: : : ; 3; 2; 1; 0; 1; 2; 3; : : :g is the integers, Q D f m : m 2 Zand n 2 Ng is the rational numbers and R n Q D fx 2 R: x 62 Qg is the irrational n numbers. Remarks (Exercises). The first three axioms are also true if R is replaced by Q: However, the completeness axiom is false for Q: For example, S D fx : x 2 Q; x > 0; x 2 < 2g is bounded above by 3 in Q; but it does not have a supremum in Q: 21 As above, we define S to be bounded below if there is some m 2 R such that m x for all x 2 S . Such an m is Q called a lower bound of S . The infimum or greatest lower bound (denoted by inf S or glb S ) of S is a lower bound m of S such that m m for all lower bounds m of S . Q inf S lower bounds are here S sup S upper bounds are here Remarks (Exercises). (1) Let B D f x : x 2 B g: (This is the reflection of B about 0.) If B is bounded below, then B is bounded above and inf B D sup. B /: Similarly, if B is bounded above, then B is bounded below and sup B D inf . B /: From these and the completeness axiom, we get the following statement. (Completeness Axiom for Infimum) Every nonempty subset of R which is bounded below has an infimum in R. -B B ( inf(-B) sup(-B) 0 infB A infA ) supA supB (2) For a set B , if it is bounded above and c 0; then let cB D fcx : x 2 B g: (This is the scaling of B by a factor of c:) We have sup cB D c sup B : If ; 6D A B , then inf B inf A whenever B is bounded below and sup A sup B whenever B is bounded above. c units infB B supB inf(c+B) c+B sup(c+B) (3) For c 2 R; let c C B D fc C x : x 2 B g: (This is a translation of B by c units.) It follows that B has a supremum if and only if c C B has a supremum, in which case sup.c C B / D c C sup B : The infimum statement is similar, i.e. inf.c C B / D c C inf B : More generally, if A and B are bounded, then letting A C B D fx C y : x 2 A; y 2 B g; we have sup. A C B / D sup A C sup B and inf. A C B / D inf A C inf B : If S is bounded above and below, then S is bounded. Note sup S , inf S may or may not be in S . Also, if S is bounded, then for all x 2 S ; jx j D max.x ; x / max.sup S ; inf S / (because x sup S and x inf S :) Conversely, if there is c 2 R such that for all x 2 S ; jx j c; then c x c so that S is bounded (above by c and below by c:) Simple Consequences of the Axioms. Theorem (Infinitesimal Principle). For x, y 2 R, x < y C " for all " > 0 if and only if x for all " > 0 if and only if y x .) Proof. If x y , then for all " > 0, x y : (Similarly, y "<x y D y C 0 < y C " by (iv) of the field axiom and (iii) of the order axiom. "0 x Dx Conversely, if x < y C " for all " > 0; then assuming x > y ; we get x y > 0 by (iii) of the order axiom. Let y ; then x D y C "0 : Since "0 > 0; we also have x < y C "0 : These contradict (i) of the order axiom. So y : The other statement follows from the first statement since y " < x is the same as y < x C ": Remarks. Taking y D 0, we see that jx j < " for all " > 0 if and only if x D 0. This is used when it is difficult to show two expressions a ; b are equal, but it may be easier to show ja bj < " for every " > 0: Theorem (Mathematical Induction). For every n 2 N, A.n / is a (true or false) statement such that A.1/ is true and for every k 2 N, A.k / is true implies A.k C 1/ is also true. Then A.n / is true for all n 2 N. 22 Proof. Suppose A.n / is false for some n 2 N. Then S D fn 2 N: A.n / is falseg is a nonempty subset of N. By the well-ordering axiom, there is a least element m in S : Then A.m / is false. Also, if A.n / is false, then m n . Taking contrapositive, this means that if n < m ; then A.n / is true. Now A.1/ is true, so m 6D 1 and m 2 N imply m 2. So m 1 1: Let k D m 1 2 N; then k D m 1 < m implies A.k / is true. By hypothesis, A.k C 1/ D A.m / is true, a contradiction. Theorem (Supremum Property). If a set S has a supremum in R and " > 0; then there is x 2 S such that sup S " < x sup S . Proof. Since sup S " < sup S , sup S " is not an upper bound of S . Then there is x 2 S such that sup S " < x . Since sup S is an upper bound of S , x sup S . Therefore sup S " < x sup S . Theorem (Infimum Property). If a set S has an infimum in R and " > 0; then there is x 2 S such that inf S C " > x inf S . Proof. Since inf S C " > inf S , inf S C " is not a lower bound of S . Then there is x 2 S such that inf S C " > x . Since inf S is a lower bound of S , x inf S . Therefore inf S C " > x inf S . Theorem (Archimedean Principle). For any x 2 R, there is n 2 N such that n > x . Proof. Assume there exists x 2 R such that for all n 2 N; we have n x : Then N D fn : n 2 Ng has an upper bound x . By the completeness axiom, N has a supremum in R. By the supremum property, there is n 2 N such that sup N 1 < n , which yields the contradiction sup N < n C 1 2 N. Question. How is Q contained in R? How is R n Q contained in R? Below we will show that Q is “dense” in R in the sense that between any two distinct real numbers x ; y , no matter how close, there is a rational number. Similarly, R n Q is “dense” in R: First we need a lemma. Lemma. For every x 2 R; there exists a least integer greater than or equal to x : (In computer science, this is called the ceiling of x and is denoted by dx e:) Similarly, there exists a greatest integer less than or equal to x : (This is denoted by [x ]: In computer science, this is also called the floor of x and is denoted by bx c:) Proof. By the Archimedean principle, there is n 2 N such that n > jx j: Then n < x < n : By (iii) of the order axiom, 0 < x C n < 2n : The set S D fk 2 N : k x C n g is a nonempty subset of N because 2n 2 S : By the well-ordering axiom, there is a least positive integer m x C n : Then m n is the least integer greater than or equal to x : So the ceiling of every real number always exist. Next, to find the floor of x ; let k be the least integer greater than or equal to less than or equal to x : x ; then k is the greatest integer Theorem (Density of Rational Numbers). If x < y, then there is m 2 Q such that x < m < y . n n Proof. By the Archimedean principle, there is n 2 N such that n > 1=. y x /: So ny n x > 1 and hence nx C 1 < ny : Let m D [nx ] C 1; then m 1 D [nx ] n x < [nx ] C 1 D m : So nx < m n x C 1 < ny ; i.e. x < m < y : n Theorem (Density of Irrational Numbers). If x < y, then there is w 2 R n Q such that x < w < y . y x Proof. Let w0 2 R n Q. By the density of rational numbers, there is m 2 Q such that jw0 j < m < jw0 j : (If m D 0; n n n y m m then pick another rational number between 0 and jw0 j : So we may take n 6D 0:) Let w D n jw0 j; then w 2 R n Q and x < w < y: Examples. (1) Let S D . 1; 3/ .4; 5]; then S is not bounded below and so S has no infimum. On the other hand, S is bounded above by 5 and every upper bound of S is greater than or equal to 5 2 S : So sup S D 5: (2) Let S D f 1 : n 2 Ng D f1; 1 ; 1 ; 1 ; : : :g: In the examples following the definition of supremum, we saw sup S D 1: n 234 1 Here we will show inf S D 0: (Note 0 62 S .) Since n > 0 for all n 2 N; 0 is a lower bound of S . So by the completeness axiom for infimum, inf S must exist. Assume S has a lower bound t > 0: By the Archimedean principle, there is n 2 N such that n > 1= t : Then t > 1= n 2 S ; a contradiction to t being a lower bound of S : So 0 is the greatest lower bound of S : 23 (3) Let S D [2; 6/ \ Q: Since 2 x < 6 for every x 2 S ; S has 2 as a lower bound and 6 as an upper bound. We will show inf S D 2 and sup S D 6: (Note 2 2 S and 6 62 S .) Since 2 2 S ; so every lower bound t satisfy t 2: Therefore inf S D 2: For supremum, assume there is an upper bound u < 6: Since 2 2 S ; so 2 u : By the density of rational numbers, there is a r 2 Q such that u < r < 6: Then r 2 [2; 6/ \ Q D S : As u < r contradicts u being an upper bound of S ; so every upper bound u 6: Therefore, sup S D 6: 24 Chapter 6. Limits Limit is the most important concept in analysis. We will first discuss limits of sequences, then limits of functions. Definitions. An (infinite) sequence in a set S (e.g. S D R or S D [0; 1]) is a list x 1 ; x 2 ; x 3 ; : : : of elements of S in a specific order. Briefly it is denoted by fx n g: (Mathematically it may be viewed as a function x : N ! S with x .n / D x n for n 2 N:) We say the sequence fx n g is bounded above iff the set fx 1 ; x 2 ; x 3 ; : : :g is bounded above. (Bounded below and bounded sequences are defined similarly.) We will also write supfx n g for the supremum of the set fx 1 ; x 2 ; x 3 ; : : :g and inf fx n g for the infimum of the set fx 1 ; x 2 ; x 3 ; : : :g: CAUTION: Since we seldom talk about a set with one element from now on, so notations like fx n g will denote sequences unless explicitly stated otherwise. For x ; y 2 R; the distance between x and y is commonly denoted by d .x ; y /; which equals jx y j: Below we will need a quantitative measure of what it means to be “close” for a discussion of the concept of limit. For " > 0; the open interval .c "; c C "/ is called the "-neighborhood of c: Note x 2 .c "; c C "/ if and only if d .x ; c/ D jx cj < "; i.e. every number in .c "; c C "/ has distance less than " from c: Limit of a sequence fx n g is often explained by saying it is the number the x n ’s are closer and closer to as n gets larger and larger. There are two bad points about this explanations. (1) Being close or large is a feeling! It is not a fact. It cannot be proved by a logical argument. (2) The effect of being close can accumulate to yield large separation! If two numbers having a distance less than or equal to 1 are considered close, then 0 is close to 1 and 1 is close to 2 and 2 is close to 3, : : : ; 99 is close to 100, but 0 is quite far from 100. So what is the meaning of close? How can limit be defined so it can be checked? Intuitively, a sequence fx n g gets close to a number x if and only if the distance d .x n ; x / goes to 0. This happens if and only if for every positive "; the distance d .x n ; x / eventually becomes less than ": The following example will try to make this more precise. 2n 2 1 gets close to 2. For " D 0:1; how soon (that is, for n2 C 1 what n ) will the distance d .x n ; 2/ be less than "? (What if " D 0:01? What if " D 0:001? What if " is an arbitrary positive number?) 2n 2 1 3 Solution. Consider d .x n ; 2/ D 2D 2 < ": Solving for n ; we get n 2 > .3="/ 1: If " D 0:1; then n2 C 1 n C1 p n > 29: So as soon as n 6; the distance between x n and 2 will be less than " D 0:1: p p (If " D 0:01; then n > 299: So n 18 will do. If " D 0:001; then n > 2999: So n 55 will do. p If 0 < " 3; then n [ .3="/ 1] C 1 will do. If " > 3; then since n23 1 < 3 < " for every n 2 N; so C n 1 will do. So for every " > 0; there is a K 2 N so that as soon as n K ; the distance d .x n ; 2/ will be less than ":) Note the value of K depends on the value of "I the smaller " is, the larger K will be. (Some people write K " to indicate K depends on ":) Example. As n gets large, intuitively we may think x n D Definition. A sequence fx n g converges to a number x (or has limit x ) iff for every " > 0, there is K 2 N such that for every n K ; it implies d .x n ; x / D jx n x j < " (which means x K ; x K C1 ; x K C2 ; : : : 2 .x "; x C "/:/ Remarks. (i) From the definition, we see that fx n g converges to x ; fx n x g converges to 0 and fjx n x jg converges to 0 are equivalent because in the definition, jx n x j is the same as j.x n x / 0j D jjx n x j 0j: (ii) To show fx n g converges to x means for every " > 0; we have to find a K as in the definition or show such a K exists. On the other hand, if we are given that fx n g converges to x , then for every " > 0; (which we can even choose for our convenience,) there is a K as in the definition for us to use. Let us now do a few more examples to illustrate how to show a sequence converges by checking the definition. Later, we will prove some theorems that will help in establishing convergence of sequences. 25 Examples. (1) Let vn D c: For every " > 0, let K D 1, then n (2) Let wn D c implies jwn K implies jvn cj D 0 < ". So fvn g converges to c: K 1 1 : For every " > 0, there exists an integer K > (by the Archimedean principle). Then n n " 1 1 cj D < ". So fwn g converges to c: n K (3) Let x n D n .cos n / n : Show that fx n g converges to 1 by checking the definition. K implies 1 Solution. For every " > 0; there exists an integer K > 1 C by the Archimedian principle. Then n " n cos n 1 1 . 1/ D < ": So fx n g converges to 1: .cos n / n .cos n / n n1 K1 (4) Let yn D . 1/n : Show that f yn g does not converge. Solution. Assume f yn g converges, say to y : Let " D 0:1: Then there exists K 2 N such that n K implies j. 1/n y j < " D 0:1: Taking an odd integer n k ; we get j 1 y j < 0:1; which implies y 2 . 1:1; 0:9/: Taking a even integer n K ; we get j1 y j < 0:1; which implies y 2 .0:9; 1:1/: Since no y is in both . 1:1; 0:9/ and .0:9; 1:1/; we have a contradiction. (5) Let z n D n 1= n : Show that fz n g converges to 1 by checking the definition. Solution. (Let u n D jz n 1j D z n 1: By the binomial theorem, n .n 2 1/ u2 C n 2 C un n n D z n D .1 C u n /n D 1 C nu n C n r n .n 2 1/ u2 n so that u n implies jz n 2 n 1 :) For every " > 0; there exists integer K > 1 C r 1j D u n 2 n 1 r "2 (by the Archimedean principle). Then n K 2 K 1 < ": So fz n g converges to 1. n!1 Theorem (Uniqueness of Limit). If fx n g converges to x and y, then x D y (and so we may write lim x n D x). Proof. For every " > 0, we will show jx y j < ". (By the infinitesimal principle, we will get x D y .) Let "0 D "=2 > 0: By the definition of convergence, there are K 1 , K 2 2 N such that n K 1 ) jx n x j < "0 and n K 2 ) jx n y j < "0 . Let K D max. K 1 ; K 2 /. By the triangle inequality, jx y j D j.x x K / C .x K y /j jx x K jCjx K y j < "0 C "0 D ". Boundedness Theorem. If fx n g converges, then fx n g is bounded. Proof. Let lim x n D x . For " D 1, there is K 2 N such that n K ) jx n x j < 1 ) jx n j D j.x n n!1 Let M D max.jx 1 j, : : :, jx K 1 j, 1 C jx j/; then for every n 2 N; jx n j M (i.e. x n 2 [ M ; M ]). x / C x j < 1 Cjx j. Remarks. The converse is false. The sequence f. 1/n g is bounded, but not convergent by example (4). In general, bounded sequences may or may not converge. Theorem (Computation Formulas for Limits). If fx n g converges to x and f yn g converges to y, then (i) fx n yn g converges to x y , respectively, i.e. lim .x n n!1 n!1 n!1 yn / D lim x n n!1 n!1 lim yn , (ii) fx n yn g converges to x y, i.e. lim .x n yn / D lim x n (iii) fx n = yn g converges to x = y, provided yn 6D 0 for all n and y 6D 0. 26 n!1 lim yn , Proof. (i) For every " > 0, there are K 1 ; K 2 2 N such that n K 1 ) jx n x j < "=2 and n Let K D max. K 1 ; K 2 /: Then n K implies n K 1 and n K 2 : So for these n ’s, j. x n K 2 ) j yn y j < "=2. yn / .x y /j D j.x n x/ . yn y /j jx n x j C j yn y j < "=2 C "=2 D ": (ii) We prove a lemma first. Lemma. If fan g is bounded and lim bn D 0, then lim an bn D 0. n!1 n!1 Proof. Since fan g is bounded, there is M such that jan j < M for all n . For every " > 0, since "= M > 0 and fbn g converges to 0, there is K 2 N such that n K ) jbn 0j < "= M ) jan bn 0j M jbn j < ". To prove (ii), we write x n yn x y D x n yn x n y C x n y x y D x n . yn y / C y .x n x /. Since fx n g converges, fx n g is bounded by the boundedness theorem. So by (i) and the lemma, n!1 lim x n yn D lim .x n yn n!1 x y / C lim x y D lim x n . yn n!1 n!1 y / C lim y .x n n!1 x / C x y D 0 C 0 C x y D x y: (iii) Note 1 j y j > 0: Since f yn g converges to y ; there is K 0 2 N such that n K 0 implies j yn y j < 1 j y j: By 2 2 1 1 j yn y j < 2 j y j ) 2 j y j < j yn j for n K 0 : Then for every n 2 N; the triangle inequality, j y j j yn j j yn j m D min.j y1 j; : : : ; j y K 0 1 j; 1 j y j/ > 0: 2 1 1 xn 1 1 x (then by (ii), lim Next we will show lim D D lim .x n / D x D ). For every " > 0; let n!1 yn n!1 yn n!1 y yn y y "0 D m j y j" > 0: Since lim yn D y 6D 0, there is K 2 N such that n K ) j yn y j < "0: Then n!1 n K) 1 yn 1 jy yn j " D < 0 D ": y j yn jj y j mj yj Remarks. (1) As in the proofs of the uniqueness of limit and part (i) of the computation formulas, when we have n K 1 ) jan a j < "1 and n K 2 ) jbn bj < "2 ; we may as well take K D max. K 1 ; K 2 / to say n K ) jan a j < "1 and jbn bj < "2 from now on. (2) By mathematical induction, we can show that the computation formulas also hold for finitely many sequences. However, the number of sequences must stay constant as the following example shows 1 D lim n!1 1 1 1 C C 6D lim C n!1 n n n | {z } n terms C lim n!1 1 D 0C n C 0 D 0: Sandwich Theorem (or Squeeze Limit Theorem). If x n lim wn D z . n!1 wn yn for all n 2 N and lim x n D lim yn D z ; then n!1 n!1 Proof. For any " > 0, there is K such that n K ) jx n x n wn yn ; so wn 2 .z "; z C "/, i.e. jwn z j < ". p z j < " and j yn z j < ", i.e. x n , yn 2 .z "; z C "/. Since [10n 2] Example. Let wn D 2 Q for every n 2 N: (Note w1 D 1:4; w2 D 1:41; w3 D 1:414; w4 D 1:4142; : : : :) 10n p p p p 10n 2 1 10n 2 p 10n 2 1 p Then < wn D 2: Since lim D 2; by the sandwich theorem, lim wn D 2: n n n n!1 n!1 10 10 10 Theorem (Limit Inequality). If an 0 for all n 2 N and lim an D a, then a n!1 0. a j < " D ja j, which implies Proof. Assume a < 0. Then for " D ja j, there is K 2 N such that n a " < an < a C " D a C . a / D 0, a contradiction. K ) jan Remarks. By the limit inequality above, if x n yn , lim x n D x , lim yn D y , then taking an D yn x n 0; we get n!1 n!1 the limit y x 0; i.e. x y : Also, if a x n b and lim x n D x ; then lim a D a lim x n D x lim b D b; n!1 n!1 n!1 n!1 1 i.e. x n 2 [a ; b] and lim x n D x imply x 2 [a ; b]: (This is false for open intervals as 1 2 .0; 2/; lim D 0 62 .0; 2/:) n n!1 n!1 n 27 Supremum Limit Theorem. Let S be a nonempty set with an upper bound c: There is a sequence fwn g in S converging to c if and only if c D sup S . 1 1 Proof. If c D sup S ; then for n 2 N, by the supremum property, there is wn 2 S such that c n D sup S n < wn 1 sup S D c. Since lim .c / D lim c D c, the sandwich theorem implies lim wn D c D sup S . n!1 n!1 n!1 n Conversely, if a sequence fwn g in S converges to c; then wn sup S implies c D lim wn sup S : Since c is an n!1 upper bound of S ; so sup S c: Therefore c D sup S : Infimum Limit Theorem. Let S be a nonempty set with a lower bound c: There is a sequence fwn g in S converging to c if and only if c D inf S . 1 1 Examples. (1) Let S D f n : n 2 Ng D f1; 1 ; 1 ; : : :g: Since 0 n for all n 2 N; 0 is a lower bound of the set S : Now 23 1 the sequence f n g in S converges to 0: By the infimum limit theorem, inf S D 0: (2) Let S D fx C 1 : x 2 Q \ .0; 1]; y 2 [1; 2]g: Since 1 x C 1 for all x 2 Q \ .0; 1] and y 2 [1; 2]; 1 is a y 2 y 2 lower bound of S : Now the sequence f 1 C 1 g in S converges to 1 : By the infimum limit theorem, inf S D 1 : n 2 2 2 (3) Let A and B be bounded in R Prove that if A 2 B D fa 2b : a 2 A; b 2 B g; then sup. A 2 B / D sup A 2 inf B : : Solution. Since A and B are bounded, sup A and inf B exist by the completeness axiom. For x 2 A 2 B ; we have x D a 2b for some a 2 A and b 2 B : So x D a 2b sup A 2 inf B : Hence sup A 2 inf B is an upper bound for A 2 B : By the supremum limit theorem, there is a sequence an 2 A such that fan g converges to sup A: By the infimum limit theorem, there is a sequence bn 2 B such that fbn g converges to inf B : Then fan 2bn g is a sequence in A 2 B and fan 2bn g converges to sup A 2 inf B by the computation formulas for limits. By the supremum limit theorem, therefore sup. A 2 B / D sup A 2 inf B : Definition. A subsequence of fx n g is a sequence fx n j g, where n j 2 N and n 1 < n 2 < n 3 < : : :. Examples. For the sequence x 1 ; x 2 ; x 3 ; x 4 ; x 5 ; x 6 ; : : : ; if we set n j D j 2; we get the subsequence x 1 ; x 4 ; x 9 ; x 16 ; : : : : If n j D 2 j C 1; then we get the subsequence x 3 ; x 5 ; x 7 ; x 9 ; : : : : If n j is the j -th prime number, then we get the subsequence x 2 ; x 3 ; x 5 ; x 7 ; : : : : Remarks. (1) Taking n j D j ; we see that every sequence is a subsequence of itself. A subsequence can also be thought of as obtained from the original sequence by throwing away possibly some terms. Also, a subsequence of a subsequence of fx n g is a subsequence of fx n g: (2) By mathematical induction, we have n j j for all j 2 N because n 1 1 and n j C1 > n j j implies n j C1 j C 1: Subsequence Theorem. If lim x n D x, then lim x n j D x for every subsequence fx n j g of fx n g. (The converse is trivially true because every sequence is a subsequence of itself.) Proof. For every " > 0, there is K 2 N such that n K ) jx n n!1 j !1 x j < ". Then j K ) nj K ) jx n j x j < ". Question. How can we tell if a sequence converges without knowing the limit (especially if the sequence is given by a recurrence relation)? For certain types of sequences, the question has an easy answer. increasing x2 x3 : : : > > > x1 = < = decreasing x1 x2 x3 : : : monotone Definitions. fx n g is iff ; respectively. fx n g is iff strictly monotone > strictly increasing > > x1 < x2 < x3 < : : : > : ; : ; strictly decreasing x1 > x2 > x3 > : : : increasing or decreasing fx n g is ; respectively. strictly increasing or decreasing Monotone Sequence Theorem. If fx n g is increasing and bounded above, then lim x n D supfx n g. (Similarly, if fx n g is decreasing and bounded below, then lim x n D inffx n g.) n!1 n!1 8 > < 9 8 9 28 Examples. (1) Let 0 < c < 1 and x n D c1= n : Then x n < 1 and cnC1 < cn ) x n D c1= n < c1=.nC1/ D x nC1 : So by 2 the monotone sequence theorem, fx n g has a limit x : Now x 2n D .c1=2n /2 D c1= n D x n : Taking limits and using the 2 subsequence theorem, we get x D x : So x D 0 or 1. Since 0 < c D x 1 x ; the limit x is 1. Similarly, if c 1; then c1= n will decrease to the limit 1. r q Proof. Let M D supfx n g, which exists by the completeness axiom. By the supremum property, for any " > 0, there is M . Then j K ) M " < x K x j M ) jx j M j D M x j < ": x K such that M " < x K Remark. Note the completeness axiom was used to show the limit of x n exists (without giving the value). (2) Does 2C 2C p 2C represent a real number? Here we have a nested radical defined by x 1 D 2 and x nC1 D 2 C x n : The question is whether fx n g converges to a real number x . (Computing a few terms, we suspect that fx n g is increasing. To find an upper bound, p observe that if lim x n D x ; then x D 2 C x implies x D 2:) Now by mathematical induction, we can show that n!1 x n < x nC1 < 2: (If x n < x nC1 < 2; then 2 C x n < 2 C x nC1 < 4; so taking square roots, we get x nC1 < x nC2 < 2:) 2 By the monotone sequence theorem, fx n g has a limit x : We have x 2 D lim x nC1 D lim 2 C x n D 2 C x : Then xD 1 or 2. Since p p p 2 D x1 x ; so x D 2: n!1 n!1 Another common type of sequences is obtained by mixing a decreasing sequence and an increasing sequence into one of the form a1 ; b1 ; a2 ; b2 ; a3 ; b3 ; : : : : In the next example, we will have such a situation and we need two theorems to handle these kind of sequences. Nested Interval Theorem. If In D [an ; bn ] is such that I1 a D lim an n!1 n!1 lim bn D b: If lim .bn n!1 1 an / D 0, then \ In contains exactly one number. nD1 [ a2 [ a3 ... ] b3 ] b2 ] b1 I2 I3 : : : , then 1 \ In nD1 D [a ; b ]; where [ a1 Proof. I1 I2 I3 : : : implies fan g is increasing and bounded above by b1 and fbn g is decreasing and bounded below by a1 . By the monotone sequence theorem, fan g converges to a D supfan g and fbn g converges to b D inffbn g. Since an bn for every n 2 N, taking limits, we have an a b bn . Consequently, x 2 [an ; bn ] (i.e. an x bn ) for all n if and only if lim an D a 1 and \ In nD1 n!1 x b D lim bn : So \ In D [a ; b]. If 0 D lim .bn n!1 n D1 n!1 1 an / D b a ; then a D b D fa g: Remarks. Note in the proof, the monotone sequence theorem was used. So the nested interval theorem also implicitly depended on the completeness axiom. Intertwining Sequence Theorem. If fx 2m g and fx 2m 1 g converge to x ; then fx n g also converges to x : Proof. For every " > 0; since fx 2m g converges to x ; there is K 0 2 N such that m K 0 ) jx 2m x j < ": Since fx 2m 1 g also converges to x ; there is K 1 2 N such that m K 1 ) jx 2m 1 x j < ": Now if n K D max.2 K 0 ; 2 K 1 1/; then either n D 2m 2 K 0 ) jx n x j D jx 2m x j < " or n D 2m 1 2 K 1 1 ) jx n x j D jx 2m 1 x j < ": 1 Example. Does represent a number? 1 1C 1C Here we have a continued fraction defined by x 1 D 1 and x nC1 D 1=.1 C x n /: We have x 1 D 1; x 2 D 1=2; x 3 D 2=3; x 4 D 3=5; : : : : Plotting these on the real line suggests 1=2 x 2n < x 2nC2 < x 2nC1 < x 2n 1 1 for all n 2 N: This can be easily established by mathematical induction. (If 1=2 x 2n < x 2nC2 < x 2nC1 < x 2n 1 1; then 1 C x 2n < 1 C x 2nC2 < 1 C x 2nC1 < 1 C x 2n 1 : Taking reciprocal and applying the recurrence relation, we have x 2nC1 > x 2nC3 > x 2nC2 > x 2n : Repeating these steps once more, we get x 2nC2 < x 2nC4 < x 2nC3 < x 2nC1 :) Let In D [x 2n ; x 2n 1 ]; then I1 jx m I2 1 I3 D : Now jx m 1 x m j .1 C x m 1/.1 C x m / 29 x m C1 j D 1 1 C xm 1 1 C xm < jx m 1 1 2 xm j 1 2 .1 C /.1 C / D 4 jx m 9 1 x m j: Using this, we get jx 2n 1 x 2n j < 4 j x 2n 9 1 2 x 2n 1 j < 44 j x 2n 99 3 x 2n 2 j < < 4 4 jx 1 9 9 | {z } 2n 2 x2 j D 4 9 2n 2 1 2 : By the sandwich theorem, lim .x 2n n!1 n!1 n!1 n!1 x 2n / D 0: So by the nested interval theorem, \ In D fx g for some p x and lim x 2n D x D lim x 2n 1 : By the intertwining sequence theorem, lim x n D lim 1=.1 C x n / D 1=.1 C x /: Then x D . 1 5/=2: Since x 2 I1 ; x D . 1 C n!1 p 1 nD1 x : So x D 5/=2: n!1 lim x nC1 D (Instead of estimating the lengths of the In ’s and squeezing them to 0 to see their intersection is a single point, we can also do the following. Let In D [x 2n ; x 2n 1 ] be as above so that I1 I2 I3 : By the nested interval 1 1 theorem, \ In D [a ; b]; where a D lim x 2n and b D lim x 2n 1 : Taking limits on both sides of x 2nC1 D and n!1 n!1 nD1 1 C x 2n 1 1 1 x 2n D ; we get b D and a D : Then b.1 C a / D 1 D a .1 C b/; which yields b C ab D a C ab; 1 C x 2n 1 1Ca 1Cb so a D b: Hence \ In is a single point.) Back to answering the question above in general, French mathematician Augustine Cauchy (1789–1857) introduced the following condition. Definition. fx n g is a Cauchy sequence iff for every " > 0, there is K 2 N such that m , n K implies jx m x n j < ". 1 nD1 Remark. Roughly, the condition says that the terms of these sequences are getting closer and closer to each other. 1 1 1 1 1 : (Note that if m ; n K ; say m n ; then we have jx m x n j D 2 <2 :) n2 n m2 n K2 1 For every " > 0; we can take an integer K > p (by the Archimedean principle). Then m ; n K implies Example. Let x n D jx m xn j 1 < ": So fx n g is a Cauchy sequence. K2 " Theorem. If fx n g converges, then fx n g is a Cauchy sequence. Proof. For every " > 0; since lim x n D x ; there is K 2 N such that j K ) jx j x j < "=2. For m , n n!1 have jx m x n j jx m x j C jx x n j < "=2 C "=2 D ". So fx n g is a Cauchy sequence. K , we The converse of the previous theorem is true, but it takes some work to prove that. The difficulty lies primarily on how to come up with a limit of the sequence. The strategy of showing every Cauchy sequence in R must converge is first to find a subsequence that converges, then show that the original sequence also converge to the same limit. Theorem. If fx n g is a Cauchy sequence, then fx n g is bounded. Proof. Let " D 1: Since fx n g is a Cauchy sequence, there is K 2 N such that m ; n K ) j x m x n j < " D 1: In particular, for n K ; jx K x n j < 1 ) jx n j D j.x n xK / C xK j jx n x K j C jx K j < 1 C jx K j: Let M D max.jx 1 j; : : : ; jx K 1 j; 1 C jx K j/; then for all n 2 N; jx n j M (i.e. x n 2 [ M ; M ]/: Bolzano-Weierstrass Theorem. If fx n g is bounded, then fx n g has a subsequence fx n j g that converges. Proof. (Bisection Method) Let a1 D inffx n g, b1 D supfx n g and I1 D [a1 ; b1 ]. Let m 1 be the midpoint of I1 . If there are infinitely many terms of fx n g in [a ; m 1 ], then let a2 D a1 , b2 D m 1 and I2 D [a2 ; b2 ]. Otherwise, there will be infinitely many terms of fx n g in [m 1 ; b1 ], then let a2 D m 1 , b2 D b1 and I2 D [a2 ; b2 ]. For k D 2, 3, 4, : : :, repeat this bisection on Ik to get IkC1 . We have I j D [a j ; b j ] and I1 I2 I3 : : :. By the nested interval theorem, since 1 b1 a1 lim .b j a j / D lim D 0, \ In contains exactly one number x . j !1 j !1 2 j 1 nD1 Take n 1 D 1; then x n1 D x 1 2 I1 . Suppose n j is chosen with x n j 2 I j . Since there are infinitely many terms x n in I j C1 , choose n j C1 > n j and x n j C1 2 I j C1 . Then lim jx n j x j lim .b j a j / D 0. Therefore, lim x n j D x . j !1 j !1 j !1 30 Remarks. In the proof, the nested interval theorem was used, so the Bolzano-Weierstrass theorem depended on the completeness axiom. Alternate Proof. We will show every sequence fx n g has a monotone subsequence. (If fx n g is bounded, then the monotone sequence theorem will imply the subsequence converges.) Call x m a peak of fx n g if x m x k for all k > m : If fx n g has infinitely many peaks, then we order the peaks by strictly increasing subscripts m 1 < m 2 < m 3 < : By the definition of a peak, x m 1 x m 2 x m 3 : So fx m j g is a decreasing subsequence of fx n g: On the other hand, if fx n g has only finitely many peaks x m 1 ; ; x m k ; then let n 1 D maxfm 1 ; ; m k g C 1: Since x n1 is not a peak, there is n 2 > n 1 such that x n2 > x n1 : Inductively, if x n j is not a peak, there is n j C1 > n j with x n j C1 > x n j : So fx n j g is a strictly increasing subsequence of fx n g: Remarks. This alternate proof used the monotone sequence theorem, so it also depended on the completeness axiom. Cauchy’s Theorem. fx n g converges if and only if fx n g is a Cauchy sequence. Proof. The ‘only if’ part was proved. For the ‘if’ part, since fx n g is a Cauchy sequence, fx n g is bounded. By the Bolzano-Weierstrass theorem, fx n g has a subsequence fx n j g that converges in R, say lim x n j D x . We will show lim x n D x . For every " > 0, since fx n g is a Cauchy sequence, there is K 1 2 N such that m , n n K 1 ) jx m n!1 j !1 x n j < "=2. Since lim x n j D x , there is K 2 2 N such that j J j !1 K 2 ) jx n j x j < "=2. If J D max. K 1 ; K 2 /, then n J K 1, J K 2 and jx n xj jx n x n J j C jx n J 1 x j < "=2 C "=2 D ". Example. Does the sequence fx n g converge, where x 1 D sin 1 and x k D x k We will check the Cauchy condition. For m > n ; x m 1 1 1 < C m2 n .n C 1 / 1 xn D m X k DnC1 C .x k sin k for k D 2; 3; 4; : : :? k2 m X sin k xk 1 / D and k2 k DnC1 C jx m xn j .n C 1/2 C C C .m 1/m D 1 n 1 C nC1 1 m 1 1 K 1 1 D m n 1 1 <: m n So for " > 0; by the Archimedean principle, there is K > 1=": Then m ; n the Cauchy theorem, fx n g converges. Limits of Functions K ) jx m xn j < < ": Therefore, by Let S be an interval (or more generally a set). Suppose f : S ! R is a function. We would like to define lim f .x /. For this to be a meaningful expression, in the interval case, x 0 must be a point on the interval or an endpoint. In the general case, x 0 must be “approachable” by the points of S . CONVENTION. In discussing the limit of a function f : S ! R at x 0 ; x 0 is always assumed to be the limit of some sequence in S n fx 0 g so that x 0 can be approached by points of S : (We say x 0 is an accumulation (or limit or cluster) point of S iff x 0 is the limit of a sequence fx n g in S n fx 0 g:) x ! x0 y = f (x ) L y = f (x) L Let f : S ! R be a function. To say lim f .x / D L roughly means the distance between f .x / and L is as small as we please when x 2 S is sufficiently close to x 0 .(Again, small and close need to be clarified.) S x ! x0 x0 ( S S )( x0 Example. Let f .x / D .x 3 3x 2 /=.x 3/: If x gets close to 3 and x 6D 3; then f .x / D x 2 should be close to 9. In other words, the distance d f .x /; 9 D j f .x / 9j goes to 0 when the distance d .x ; 3/ D jx 3j gets small. So for 31 every positive "; the distance j f .x / 9j will soon or later be less than " when the distance jx 3j becomes small enough. For " D 0:1; how small d .x ; 3/ be (that is, for what x ) will make d f .x /; 9 D j f .x / 9j < "? Equivalently, we are seeking a positive so that 0 < jx 3j < ) j f .x / 9j < ": Now j f .x / 9j < " , 8:9 < f .x / < 9:1: So 2:9833 0:0166: If we take D 0:016; then 0 < jx 3j < ) j f .x / (For small " > 0; j f .x / p take D min. 9 C " 3; 3 9j p p 8 :9 < x < 9j < ": p p 9 :1 3:0166 and 3< p 0:0167 < x 3< < " , 9 " < f .x / < 9 C " ) 9 9 "/; then 0 < jx 3j < ) j f .x / " 3<x 9j < ":) 9C" 3: So we may Following the example, we are ready to state the precise definition of the limit of a function. Definition. Let f : S ! R be a function. We say f .x / converges to L (or has limit L ) as x tends to x 0 in S iff for every " > 0, there is > 0 such that for every x 2 S ; 0 < jx x 0 j < implies j f .x / L j < ". This is denoted by lim f .x / D L (or lim f .x / D L in short.) x ! x0 x 2S x ! x0 In the definition, depends on " and x 0 : For different " (or different x 0 ), will be different. If a limit value exists, then it is unique. The proof is similar to the sequential case and is left as an exercise for the readers. Examples. (1) For g : [0; 1/ ! R defined by g .x / D the "- definition. p x ; show that lim g .x / D 0 and lim g .x / D 2 by checking x !0 x !4 Solution. For every " > 0; let D "2 : Then for every x 2 [0; 1/; 0 < jx 0j < implies jg .x / 0j D This takes care of the checking for the first limit. For the second limit, note that j x x 2 [0; 1/; 0 < jx 4j < jx 4j jx 4j 2j D p : For every " > 0; let 2 x C2 4j jx implies jg .x / 2j < D ": 2 2 p p x< p D ": D 2": Then for every (2) Let f : R n f0g ! R be defined by f .x / D Solution. (Note that 1 and 1 5x jx 1 2j D 10 10x 1 1 : Show that lim f .x / D by checking the "- definition. x !2 5x 10 jx 2j 10 for x 2 .1; 3/:) For every " > 0; take implies x 2 .1; 3/: So f .x / 1 10 D min.1; 10"/: Then jx 10": For every x 2 R n f0g; 0 < jx 2j < 2j 10 < 10 " and we are done. Notation: We will write x n ! x 0 in S n fx 0 g to mean sequence fx n g in S n fx 0 g converges to x 0 : Sequential Limit Theorem. lim f .x / D L if and only if for every x n ! x 0 in S n fx 0 g, lim f .x n / D L : x ! x0 x 2S n!1 In quantifier symbols (8 D for any, for every, for all, 9 D there is, there exists), x ! x0 x 2S x ! x0 x 2S lim f .x / D L lim f .x / 6D L () () 8" 9" > 0; 9 > 0 such that 8x 2 S; 0 < jx < jx x0j < x0j < ) j f .x / L j < "; Lj > 0 such that 8 > 0; 9x 2 S ; 0 and j f .x / x0j < ": L j < ". Proof. If xlim f .x / D L ; then for every " > 0, there is > 0 such that for x 2 S , 0 < jx ! x0 If lim x n D x 0 and x n 6D x 0 , then there is K 2 N such that n n!1 x 2S ) j f .x / lim f .x n / D L . 32 n!1 K ) 0 < jx n x 0 j < .) j f .x n / L j < "/. So Conversely, if xlim f .x / 6D L ; then there is " > 0 such that for every > 0; there is x 2 S with 0 < jx !x x 2S 0 x0 j < 1 1 and j f .x / L j ". Now, setting D n ; there is x n 2 S with 0 < jx n x 0 j < D n and j f .x n / L j ". By the sandwich theorem, x n ! x 0 in S n fx 0 g: So lim f .x n / D L : Then 0 D lim j f .x n / L j ", a contradiction. n!1 n!1 Remarks. If lim f .x n / exists for every x n ! x 0 in S n fx 0 g; then all the limit values are the same. To see this, suppose n!1 x n ! x 0 and wn ! x 0 in S n fx 0 g: Then the intertwining sequence fz n g D fx 1 ; w1 ; x 2 ; w2 ; x 3 ; w3; : : :g converges to x 0 in S n fx 0 g: Since f f .x n /g and f f .wn /g are subsequences of the convergent sequence f f .z n /g, they have the same limit. Consequences of Sequential Limit Theorem. (1) (Computation Formulas) If f ; g : S ! R are functions, lim f .x / D L 1 and lim g .x / D L 2 ; then x ! x0 x 2S x ! x0 x 2S x ! x0 x 2S lim f .x / 89 >C> <= > : = > ; g .x / D L 1 89 >C > <= > : = > ; L 2 D xlim f .x / !x x 2S 0 89 >C> <= > : = > x ! x0 ; x 2S lim g .x / respectively (in the case of division, provided g .x / 6D 0 for every x 2 S and L 2 6D 0). Proof. Since f .x / and g .x / have limits L 1 and L 2 ; respectively, as x tends to x 0 in S ; by the sequential limit theorem, f .x n / and g .x n / will have limits L 1 and L 2 ; respectively, for every x n ! x 0 in S n fx 0 g: By the computation formulas for sequences, the limit of f .x n / C g .x n / is L 1 C L 2 for every x n ! x 0 in S n fx 0 g: By the sequential limit theorem, f .x / C g .x / has limit L 1 C L 2 as x tends to x 0 in S : Replacing C by ; ; =; we get the proofs for the other parts. Alternate Proof. If xlim f .x / D L 1 and xlim g .x / D L 2 ; then for every " > 0; there are ! x0 ! x0 every x 2 S ; 0 < jx x 2S x 2S 1 > 0 such that for < jx x0j < 2 x0 j < implies jg .x / L 2 j < : Now we take D min. 1 ; 2 / > 0 so that 2 0 < jx x 0 j < implies both 0 < jx x 0 j < 1 and 0 < jx x 0 j < f .x / C g .x / " 1 implies j f .x / L 1j < " 2 and 2 > 0 such that for every x and 2 so that 1 2 2 S; 0 : Then, for every x L 2j < 2 S; . L 1 C L 2/ D j. f .x / L 1 / C .g .x / L 2 /j j f .x / L 1 j C jg .x / " 2 C " 2 D ": The other parts of the computation formulas can be proved by adapting the arguments for the sequential case. (2) (Sandwich Theorem or Squeeze Limit Theorem) If f .x / x ! x0 x 2S g .x / h .x / for every x 2 S and xlim f .x / D L D ! x0 x 2S lim h .x /; then lim g .x / D L : x ! x0 x 2S (3) (Limit Inequality) If f .x / 0 for all x 2 S and lim f .x / D L ; then L x ! x0 x 2S 0: The proofs of (2) and (3) can be done by switching to sequences like the first proof of (1) or by adapting the arguments of the sequential cases and checking the "- definition like the alternate proof of (1). Next we will discuss one-sided limits. Definitions. For f : .a ; b/ ! R and x 0 2 .a ; b/; the left hand limit of f at x 0 is f .x 0 / D lim f .x / D The right hand limit of f at x 0 is f .x 0 C/ D limC f .x / D x ! x0 x ! x0 x 2.a ;b / x ! x0 x 2. x0 ;b / x ! x0 x ! x0 x 2.a ; x0 / lim f .x /. lim f .x /: Theorem. For x 0 2 .a ; b/; lim f .x / D L if and only if f .x 0 / D L D f .x 0 C/. 33 Proof. If lim f .x / D L ; then for every " > 0; there is x ! x0 > 0 such that for x 2 .a ; b/; 0 < jx x 0j < ) L j < ": In particular, for x 2 .a ; x 0 /; 0 < jx x 0 j < ) j f .x / L j < ": So f .x 0 / D L : Similarly, for j f .x / x 2 .x 0 ; b/; 0 < jx x 0 j < ) j f .x / L j < ": So f .x 0 C/ D L : Conversely, if f .x 0 / D L D f .x 0 C/; then for every " > 0; there is 1 > 0 such that for x 2 .a ; x 0 /; 0 < jx x 0 j < 1 ) j f .x / L j < " and there is 2 > 0 such that for x 2 .x 0 ; b/; 0 < jx x 0 j < 2 ) j f .x / L j < ": As D min. 1 ; 2 / x 0 j < ) j f .x / L j < ": So lim f .x / D L : 1 and 2 ; we have for x 2 .a ; b /; 0 < j x x ! x0 increasing > > f .x / = < decreasing f .x / Definitions. A function f : S ! R is on S iff for every x , y 2 S , x < y ) strictly increasing > f .x / < > > : ; : strictly decreasing f .x / > monotone increasing or decreasing Also, f is on S iff f is on S ; respectively. strictly monotone strictly increasing or strictly decreasing bounded above For a nonempty subset S0 of S ; we say f is bounded below on S0 iff f f .x / : x 2 S0 g is bounded respectively. If The set S0 is not mentioned, then it is the domain S : ( ) ( 8 > < 9 8 f . y/ > = f . y/ : f . y/ > ; f . y/ ) 9 bounded above bounded below bounded ; For monotone functions, the following theorem is analogous to the monotone sequence theorem. It will be used dx dy D1 rule later. in the next chapter to prove the continuous inverse theorem, which will be used to prove the dy dx Also, it will be used again in the chapter on integration. Monotone Function Theorem. If f is increasing on .a ; b/, then for every x 0 2 .a ; b/, f .x 0 / D supf f .x / : a < x < x 0 g and f .x 0 C/ D inff f .x / : x 0 < x < bg and f .x 0 / f .x 0 / f .x 0 C/. If f is bounded below, then f .a C/ D inff f .x / : a < x < bg: If f is bounded above, then f .b / D supf f .x / : a < x < bg: Also f has countably many discontinuous points on .a ; b/; i.e. J D fx 0 : x 0 2 .a ; b/; f .x 0 / 6D f .x 0 C/g is countable. (The theorem is similarly true for decreasing functions and all other kinds of intervals.) Proof. If a < x < x 0 < b, then f .x / f .x 0 /. So M D supf f .x / : a < x < x 0 g f .x 0 /: By the supremum property, for every " > 0; there is c 2 .a ; x 0 / such that M " < f .c/ M : If we let D x 0 c; then 8x 2 .a ; x 0 /; 0 x ! x0 < jx x0j < ) c D x0 < x < x 0 ) f .c / C x ! x0 f .x / M ) j f .x / Mj M f .c/ < ": So lim f .x / D M f .x 0 /. Similarly, f .x 0 / lim f .x / D inff f .x / : x 0 < x < bg. In the case f is bounded below or above, the proof of the existence of f .a C/ or f .b / is similar. Next let x 0 2 .a ; b/ with f .x 0 / < f .x 0 C/. By the density of rational numbers, we may choose a q .x 0 / 2 Q between f .x 0 / and f .x 0 C/. The function q : J ! Q is injective because if f is discontinuous at x 0 , x 1 with x 0 < x 1 , then x0 C x1 q .x 0 / < f .x 0 C/ f f .x 1 / < q .x 1 /: 2 By the injection theorem, J is countable, i.e. f has countably many discontinuous points on .a ; b/: The cases of decreasing functions or other kinds of intervals are similar. q ( x1) q (x0 ) x0 x1 Appendix: Infinite Limits and Limit at Infinity We begin with a definition of a sequence having C1 as limit. In this case, the sequence does not have any upper bound in R; i.e. the sequence will pass any fixed r 2 R eventually and keep on going. Sequences with 1 as limit and functions with 1 limit are defined similarly. 34 Definitions. (1) A sequence fx n g diverges to C1 (or has limit C1) iff for every r 2 R; there is K 2 N (depending on r ) such that n K implies x n > r: Similarly, a sequence fx n g diverges to 1 (or has limit 1) iff for every r 2 R; there is K 2 N (depending on r ) such that n K implies x n < r: (2) A function f : S ! R diverges to C1 (or has limit C1) as x tends to x 0 in S iff for every r 2 R; there is 2 S ; 0 < jx x 0 j < implies f .x / > r: Similarly, a function f : S ! R diverges to 1 (or has limit 1) as x tends to x 0 in S iff for every r 2 R; there is > 0 (depending on r and x 0 ) such that for every x 2 S ; 0 < jx x 0 j < implies f .x / < r: > 0 (depending on r and x 0 ) such that for every x Exercise. Prove that if fx n g is an increasing sequence, then either the limit of fx n g exists (as a real number) or the limit is C1: (The decreasing case is similar with C1 replaced by 1:) Limit at infinity for functions are defined similarly as for sequences as follow. Definitions. (1) Let f : S ! R be a function such that C1 is an accumulation point of S (i.e. there is a sequence in S diverges to C1:) We say f .x / converges to L (or has limit L ) as x tends to C1 in S iff for every " > 0; there is K 2 R such that for every x 2 S ; x K implies j f .x / L j < ": Similarly, let f : S ! R be a function such that 1 is an accumulation point of S (i.e. there is a sequence in S diverges to 1:) We say f .x / converges to L (or has limit L ) as x tends to 1 in S iff for every " > 0; there is K 2 R such that for every x 2 S ; x K implies j f .x / L j < ": (2) Let f : S ! R be a function such that C1 is an accumulation point of S (i.e. there is a sequence in S diverges to C1:) We say f .x / diverges to C1 (or has limit C1) as x tends to C1 in S iff for every r 2 R; there is K implies f .x / > r: Similarly, let f : S ! R be a function such that 1 is K 2 R such that for every x 2 S ; x an accumulation point of S (i.e. there is a sequence in S diverges to 1:) We say f .x / diverges to C1 (or has limit C1) as x tends to 1 in S iff for every r 2 R; there is K 2 R such that for every x 2 S ; x K implies f .x / > r: By replacing f .x / > r with f .x / < r; we get the definitions of limit equal to 1 as x tends to 1 in S : It is good exercises for the readers to formulate the computation formulas and properties for these limits, which can be proved by checking these definitions just like the finite cases. 35 Chapter 7. Continuity Definitions. A function f : S ! R is continuous at x 0 2 S iff lim f .x / D f .x 0 /; i.e. for every " > 0, there is > 0 f .x 0 /j < ". For E S ; we say f is continuous on E iff f is such that for all x 2 S , jx x 0 j < ) j f .x / continuous at every element of E : Also, we say f is continuous iff f is continuous on the domain S : Sequential Continuity Theorem. f : S ! R is continuous at x 0 2 S if and only if for every x n ! x 0 in S ; lim f .x n / D f .x 0 / D f . lim x n /: n!1 n!1 x ! x0 x 2S Proof. Just replace L by f .x 0 /, 0 < jx x 0 j < by jx x 0 j < and x n ! x 0 in S n fx 0 g by x n ! x 0 in S (i.e. delete the x n 6D x 0 requirement) in the proof of the sequential limit theorem. Example. It is easy to give examples of continuous functions, such as polynomials. Here is an example of a function 1 if x 2 Q not continuous at any point. Let f .x / D ; then f is discontinuous at every x 2 R! 0 if x 62 Q Reason. For every x 0 2 R; n 2 N, by the density of rational numbers and irrational numbers, there are rn 2 Q, sn 62 Q 1 in .x 0 n ; x 0 /: Now rn ! x 0 , sn ! x 0 , but lim f .rn / D 1 and lim f .sn / D 0. So lim f .x / cannot exist. n!1 n!1 x ! x0 Theorem. If f ; g : S ! R are continuous at x 0 2 S ; then f x ! x0 g ; f g ; f =g (provided g .x 0 / 6D 0) are continuous at x 0 : x ! x0 x ! x0 Proof. Since f ; g are continuous at x 0 ; lim . f C g /.x / D lim f .x / C lim g .x / D f .x 0 / C g .x 0 / D . f C g /.x 0 /: So f C g is continous at x 0 by definition. The subtraction, multiplication and division cases are similar. Theorem. If f : S ! R is continuous at x 0 , f . S / at x 0 . S 0; g : S 0 ! R is continuous at f .x 0 /, then g B f is continuous Proof. By the sequential continuity theorem, all we need to show is that lim .g f /.x n / D .g f /.x 0 / for every n!1 sequence fx n g converging to x 0 : Since f is continuous at x 0 , by the sequential continuity theorem, the limit of f .x n / is f .x 0 /: Since g is continuous at f .x 0 /; so lim g . f .x n // D g . lim f .x n // D g . f .x 0 //: n!1 n!1 In the discussions below, S will denote an interval of positive length. Theorem (Sign Preserving Property). If g : S ! R is continuous and g .x 0 / > 0, then there is an interval I D ; x 0 C / with > 0 such that g .x / > 0 for every x 2 S \ I : (The case g .x 0/ < 0 is similar by considering g .) jg .x / .x 0 Proof. Let " D g .x 0 / > 0: Since g is continuous at x 0 ; there is > 0 such that for x 2 S ; jx g .x 0 /j < ": So x 2 S \ .x 0 ; x 0 C / ) 0 D g .x 0/ " < g .x /: x0 j < ) Intermediate Value Theorem. If f : [a ; b] ! R is continuous and y0 is between f .a / and f .b/ inclusive, then there is (at least one) x 0 2 [a ; b] such that f .x 0 / D y0 . Proof. The cases y0 D f .a / or y0 D f .b/ are trivial as x 0 D a or b will do. We assume f .a / < y0 < f .b/. (The other possibility f .a / > y0 > f .b/ is similar.) The set S D fx 2 [a ; b]: f .x / y0 g is nonempty (a 2 S ) and has b as an upper bound. Then x 0 D sup S 2 [a ; b]. By the supremum limit theorem, there is a sequence fx n g in S such that lim x n D x 0 . Note x n 2 [a ; b] implies n!1 x 0 2 [a ; b]: Then f .x 0 / D lim f .x n / y0 by continuity of f at x 0 . Assume f .x 0 / < y0 . Then x 0 6D b as y0 < f .b/: Define g .x / D y0 f .x / on [a ; b]. Since g .x 0 / > 0, by the sign preserving property, there is x 1 > x 0 such that g .x 1 / > 0: Then f .x 1 / < y0 . So x 0 < x 1 2 S ; which contradicts x 0 D sup S . Therefore, f .x 0 / D y0 . Examples. (1) The equation x 5 C 3x C sin x D cos x C 10 has a solution. To see this, let f .x / D x 5 C 3x C sin x cos x 10; then f .0/ D 11 and 26 f .2/ 30: Since f is continuous, the intermediate value theorem implies f .x / D 0 for some x 2 .0; 2/: 36 n!1 (2) Suppose p.x / D x n C a1 x n 1 C C an with n odd. Let x 0 D 1 C ja1 j C n n n n n C an and p . x 0 / D x 0 C a1 x 0 1 C an : So x 0 p . x 0 / D x 0 C a1 x 0 1 C n1 n and x 0 C p. x 0 / D a1 x 0 C an : By the triangle inequality, n x 0 p.x 0 / n x0 C p. x0 / n ja1 x 0 1 j C C jan j p.x 0 / D 1; then we have n a1 x 0 1 an C jan j n ja 1 j x 0 1 C n C jan jx 0 1 D .ja1 j C n n C jan j/x 0 1 < x 0 : Then p.x 0 / > 0 and p. x 0 / < 0: By the intermediate value theorem, there is a root of p.x / between x 0 and x 0 : Extreme Value Theorem. If f : [a ; b] ! R is continuous, then there are x 0 , w0 2 [a ; b] such that f .w0 / f .x / f .x 0 / for every x 2 [a ; b]: So the range of f is f [a ; b] D [ f .w0 /; f .x 0 /]: In particular, f is bounded on [a ; b] In this case, we write f .x 0 / D supf f .x / : x 2 [a ; b]g D max f .x / and f .w0 / D inf f f .x / : x 2 [a ; b]g D min f .x /: x 2[a ;b ] x 2[a ;b ] a x0 w0 b Proof. We first show f .[a ; b]/ D f f .x / : x 2 [a ; b]g is bounded above. Assume it is not bounded above. Then each n 2 N is not an upper bound. So there is z n 2 [a ; b] such that f .z n / > n : By the Bolzano-Weierstrass theorem, fz n g in [a ; b] has a subsequence fz n j g converging to some z 0 2 [a ; b]: Since f is continuous at z 0 ; lim f .z n j / D f .z 0 /; which implies f f .z n j /g is bounded by the boundedness theorem. However, f .z n j / > n j bounded, a contradiction. j implies f f .z n j /g is not j !1 By the completeness axiom, M D supf f .x / : x 2 [a ; b]g exists. By the supremum limit theorem, there is a sequence fx n g in [a ; b] such that lim f .x n / D M . By the Bolzano-Weierstrass theorem, fx n g has a subsequence fx nk g such that lim x nk D x 0 in [a ; b]. Since f is continuous at x 0 2 [a ; b], M D lim f .x nk / D f .x 0 / by the subsequence theorem and the sequential continuity theorem, respectively. Similarly, inff f .x / : x 2 [a ; b]g D f .w0 / for some w0 2 [a ; b]. The following two theorems are for explaining the dx D1 dy k !1 k !1 n!1 dy rule in the next chapter. dx Continuous Injection Theorem. If f is continuous and injective on [a ; b], then f is strictly monotone on [a ; b] and f [a ; b] D [ f .a /; f .b/] or [ f .b/; f .a /]. (This is true for any nonempty interval in place of [a ; b]: The range is an interval with f .a /; f .b/ as endpoints.) Proof. Since f is injective, either f .a / < f .b/ or f .a / > f .b/. Suppose f (y ) f .a / < f .b/. Let y 2 .a ; b/. Then f . y / cannot be greater than f .b/, otherwise by the intermediate value theorem, there is w 2 .a ; y / such that f .w/ D f .b/, contradicting injectivity. Similarly, f . y / cannot be less than f .a /. Therefore, f ( b) f .a / < f . y / < f .b/. If a x<y b, then similarly f .a / f .x / < f (a ) f . y/ f .b/, i.e. f is strictly increasing on [a ; b] and f [a ; b] D [ f .a /; f .b/] by the intermediate value theorem. a y b The case f .a / > f .b/ leads to f strictly decreasing on [a ; b]. Continuous Inverse Theorem. If f is continuous and injective on [a ; b], then f 1 is continuous on f [a ; b] . Proof. By the last theorem, f is strictly monotone on [a ; b]. We first suppose f is strictly increasing. Then f 1 is also strictly increasing on f [a ; b] D [ f .a /; f .b/]. For y0 2 . f .a /; f .b/], let r D f 1 . y0 / D lim f 1 . y /; which exists by the monotone function theorem. Since f . y / 2 [a ; b]; r 2 [a ; b]: Let f yn g be a sequence in . f .a /; y0 / converging to y0 ; then wn D f 1 . yn / will converge to r by sequential limit theorem. Since f is continuous at r; by the sequential continuity theorem, f .r / D lim f .wn / D lim f . f 1 . yn // D y0 : Then f 1 . y0 / D r D f 1 . y0 /. Similarly, if y0 2 [ f .a /; f .b//, then 1 y ! y0 f 1 n!1 n!1 . y0C/ D f 1. y0 /. So f 1 is continuous on [ f .a /; f .b/] D f [a ; b] . The case f is strictly decreasing is similar. 37 Appendix: Fundamental Theorem of Algebra The extreme value theorem is often used in estimating integrals. More precisely, if f : [a ; b] ! R is continuous, then for m D min f .x / and M D max f .x /; we have m x 2[ a ;b ] x 2[a ;b ] f .x / M and m .b a/ Zb a f .x / d x M .b a /: Here we will mention that there is a version of the extreme value theorem for continuous functions defined on closed disks of finite radius on the plane. As an application of this fact, we can sketch a proof of the fundamental theorem of algebra, which asserts that every nonconstant polynomial with complex coefficients must have a root. Let p.z / D z n C a1 z n jz n j D p.z / n X k D1 1 C C an and m D inf fj p .z /j : z 2 C g: We have n X k D1 ak z n k j p .z /j C jak jjz jn k ) j p .z /j jz jn n X k D1 jak jjz jn k D jz jn 1 | n X k D1 jak jjz j k } : !1 as jzj!1 {z So for jz j large, 1 n X k D1 jak jjz j k p 1 : For a very large K > n 2m ; we have jz j > K implies 2 n X k D1 j p .z /j j z jn 1 jak jjz j k 1n 1 jz j > K n > m : 2 2 Let D K be the closed disk fz : jz j K g: We have m D inffj p.z /j : z 2 C g D inffj p.z /j : z 2 D K g D j p.z 0 /j for some z 0 2 D K by the extreme value theorem, since j p.z /j is continuous on D K : We claim m D j p.z 0 /j D 0: Suppose m 6D 0: Then f .z / D p.z C z 0 /= p.z 0 / is a polynomial of degree n and f .0/ D 1: So f .z / D 1 C b1 z C C bn z n for some b1 ; : : : ; bn with bn 6D 0: Let k be the smallest positive integer such that bk 6D 0: Then f .z / D 1 C bk z k C C bn z n : Since j p .z C z 0 /j m D j p.z 0 /j; we get . / j f .z /j 1 for all z : Introduce the notation ei D cos C i sin ; which describes the points on the unit circle. Since the absolute value of jbk j=bk is 1, so jbk j=bk D ei for some : Let D = k ; then eik bk D ei bk D jbk j: Considering w D rei with r < jbk j 1= k .) 1 r k jbk j > 0/; we have j1 C bk wk j D j1 C bk r k eik j D 1 r k jbk j D 1 r k jbk j and j f .w/j D j1 C bk w C k C bn w j n j1 C bk w j C jbkC1 w k k C1 D1 D1 r k jbk j C jbkC1 jr kC1 C r k jbk j jbkC1 jr | {z !jbk j> 1 jbk j>0 2 jC C jbn w j n C jbn jr n jbn jr n k } : as r !0C Taking w D rei with a very small positive r; we have j f .w/j m D j p.z 0 /j D 0 and z 0 is a root of p.z /: 1 r k . 1 jbk j/ < 1; a contradiction to . /: Therefore 2 38 Chapter 8. Differentiation Definitions. Let S be an interval of positive length. A function f : S ! R is differentiable at x 0 2 S iff f 0 .x 0 / D f .x / f .x 0 / lim exists in R. Also, f is differentiable iff f is differentiable at every element of S : x ! x0 x x0 x 2S Theorem. If f : S ! R is differentiable at x 0 , then f is continuous at x 0 . Proof. By the computation formulas for limits, x ! x0 lim f .x / D lim x ! x0 f .x / x f .x 0 / .x x0 x 0 / C f .x 0 / D f 0 .x 0 / 0 C f .x 0 / D f .x 0 /: Theorem (Differentiation Formulas). If f ; g : S ! R are differentiable at x 0 , then f C g, f g , f g, f =g (when g .x 0 / 6D 0) are differentiable at x 0 . In fact, . f g /0.x 0 / D f 0 .x 0 / g 0.x 0 /, . f g /0 .x 0 / D f 0 .x 0 /g .x 0 / C f .x 0 /g 0 .x 0 / f f 0 .x 0 /g .x 0 / f .x 0 /g 0.x 0 / and . /0 .x 0 / D . g g . x 0 /2 Proof. By the computation formulas for limits, x ! x0 lim .f g /.x / x .f x0 D lim g /.x 0 / D lim x ! x0 f .x / x f .x 0 / x0 g .x / x g .x 0 / D f 0 .x 0 / x0 g 0.x 0 /; x ! x0 lim . f g /.x / x . f g /.x 0 / x0 f . /.x / g lim x ! x0 x x ! x0 f .x / x D D f .x 0 / g .x / g .x / C f .x 0 / x0 x g .x 0 / D f 0 .x 0 /g .x 0 / C f .x 0 /g 0 .x 0 /; x0 g .x / x g .x 0 / f .x 0 / x0 . /.x 0 / x0 f g x ! x0 lim 1 f .x / f .x 0 / g .x 0 / g .x /g .x 0 / x x0 f 0 .x 0 /g .x 0 / f .x 0 /g 0 .x 0 / : g . x 0 /2 Theorem (Chain Rule). If f : S ! R is differentiable at x 0 ; f . S / S 0 and g : S 0 ! R is differentiable at f .x 0 /, then g B f is differentiable at x 0 and .g B f /0 .x 0 / D g 0 . f .x 0 // f 0 .x 0 /. Proof. The function 8 g . f .x 0 // < g . y/ if y 6D f .x 0 / h . y/ D y f .x 0 / :0 g . f .x 0 // if y D f .x 0 / is continuous at f .x 0 / because g B f .x / x y ! f . x0 / lim h . y / D g 0. f .x 0 // D h . f .x 0 //. So g . y / g . f .x 0 // D h . y /. y f .x 0 // holds for every y in the domain of g . Then x ! x0 lim g B f .x 0 / f .x / D lim h . f .x // x ! x0 x0 x f .x 0 / D h . f .x 0 // f 0 .x 0 / D g 0. f .x 0 // f 0 .x 0 /: x0 Remarks. Note f differentiable at x 0 does not imply f 0 is continuous at x 0 : In fact, the function f .x / D x 2 sin 1 for x 1 2 0 .x / D 2x sin 1 cos 1 for x 6D 0 and x 6D 0 and f .0/ D lim x sin D 0 is continuous and differentiable with f x x x !0 x 0 .0/ D lim .x 2 sin 1 0/=x D 0: However, f 0 is not continuous at 0 because lim f 0 .x / D lim 2x sin 1 cos 1 does f x !0 x !0 x !0 x x x 0 .0/: In particular, f is not twice differentiable. (Also, the function g .x / D x 2 sin 1 not exist, hence not equal to f x2 for x 6D 0 and g .0/ D 0 is differentiable on R; but g 0.x / is not continuous at 0 and g 0.x / is unbounded on every open interval containing 0.) 39 Notations. C 0 . S / D C . S / is the set of all continuous functions on S : For n 2 N; C n . S / is the set of all functions f : S ! R such that the n -th derivative f .n/ is continuous on S . C 1 . S / is the set of all functions having n -th derivatives for all n 2 N. Functions in C 1 . S / are said to be continuously differentiable on S : Inverse Function Theorem. If f is continuous and injective on .a ; b/ and f 0 .x 0 / 6D 0 for some x 0 2 .a ; b/, then f 1 dx 1 dy is differentiable at y0 D f .x 0 / and . f 1 /0 . y0 / D 0 , i.e. D1 . f .x 0 / dy dx 8 x x0 > if x 6D x 0 < 1 f .x / f .x 0 / Proof. The function g .x / D D g .x 0 /: is continuous at x 0 because lim g .x / D 0 1 x ! x0 > f .x 0 / : if x D x 0 f 0 .x 0 / Since f is continuous and injective on .a ; b/; f 1 is continuous by the continuous inverse theorem. So lim f 1 . y / D f 1 . y0/ D x 0 and . f 1/0 . y0 / D lim y ! y0 f 1 . y/ y f 1 . y0 / D lim g f y ! y0 y0 y ! y0 1 1 . y / D g .x 0 / D 0 : f .x 0 / x 2.a ;b / x 2.a ;b / Local Extremum Theorem. Let f : .a ; b/ ! R be differentiable. If f .x 0 / D min f .x / or f .x 0 / D max f .x /; then f 0 .x 0 / D 0: Proof. If f .x 0 / D min f .x /; then 0 x 2.a ;b / x ! x0 limC f .x / x f .x 0 / f .x / D f 0 .x 0 / D lim x0 x x ! x0 f .x 0 / x0 0: So f 0 .x 0 / D 0: The other case is similar. Rolle’s Theorem. Let f be continuous on [a ; b] and differentiable on .a ; b/. If f .a / D f .b/; then there is (at least one) z 0 2 .a ; b/ such that f 0 .z 0 / D 0. Proof. This is trivial if f is constant on [a ; b]. Otherwise, by the extreme value theorem, there are x 0 , w0 2 [a ; b] such that f .x 0 / D max f .x / > x 2[a ;b ] w0 is in .a ; b/: By the last theorem, f 0 .x 0 / D 0 or f 0 .w0 / D 0: w0 x0 min f .x / D f .w0 /. Either f .x 0 / 6D f .a / or f .w0 / 6D f .a /. So either x 0 or x 2[a ;b ] a x0 b Mean-Value Theorem. Let f be continuous on [a ; b] and differentiable on .a ; b/. Then there exists x 0 2 .a ; b/ such that f .b/ f .a / D f 0 .x 0 /.b a /. f .b / f .a / Proof. Define F .x / D f .x / .x a / C f .a / . Then F .a / D ba 0 D F .b/. By Rolle’s theorem, there is x 0 2 .a ; b/ such that 0 D F 0 .x 0 / D f .b/ f .a / f 0 .x 0 / , which is equivalent to f .b/ f .a / D f 0 .x 0 /.b a /: ba Examples. (1) For a < b; show j sin b sin a j jb a j: (By the mean-value theorem, there is x 0 2 .a ; b/ such that sin b sin a D .cos x 0 /.b a /; so j sin b sin a j jb a j:) (2) Show .1 C x / 1 C x for x 1 and x > 0 or x 0 2 .x ; 0/ if 1 < x < 0 such that 1: (Let f .x / D .1 C x / f .0/ D f 0 .x 0 /.x 1 x ; then there is x 0 2 .0; x / if 1 .1 C x / (3) Show ln x that x 1 x D f .x / 0/ D ..1 C x 0 / 1/ x 0:/ 1 for x > 0: (Let f .x / D ln x ln x x C 1 D f .x / D f .x / x C 1; then f .1/ D 0: If x > 1; then there is x 0 2 .1; x / such f .1/ D f 0 .x 0 /.x 1/ D 1 x0 1 .x 1/ 0: The case 0 < x < 1 is similar.) 40 (4) To approximate 16:1; we can let f .x / D x : Then f .16:1/ f .16/ D f 0 .c/.16:1 16/ for some c 0 .16/.16:1 16/ D 0:0125; which gives 16 and f .16:1/ f .16/ f between 16 and 16:1. Now c p 16:1 D f .16:1/ 4:0125: Curve Tracing Theorem. If f 0 0 (respectively f 0 > 0; f 0 0; f 0 < 0; f 0 6D 0; f 0 0) everywhere on .a ; b/, then f is increasing (respectively strictly increasing, decreasing, strictly decreasing, injective, constant) on .a ; b/: Proof. If x , y 2 .a ; b/ and x < y , then by the mean value theorem, there is x 0 2 .x ; y / such that f . y / f .x / D f 0.x 0 /. y x / 0 (respectively > 0; 0; < 0; 6D 0; D 0). Therefore, f .x / f . y / (respectively f .x / < f . y /; f .x / f . y /; f .x / > f . y /; f .x / 6D f . y /; f .x / D f . y /). Remarks. For differentiable function f ; the converse of the strictly increasing (respectively strictly decreasing, injective) parts of the curve tracing theorem are false. To see an example, consider f .x / D x 3 ; which is strictly increasing on R; but f 0.0/ D 0: The converse of the increasing (respectively decreasing, constant) parts are true because . f .x / f .x 0 //=.x x 0 / is nonnegative (respectively nonpositive, zero) for x ; x 0 2 .a ; b/ and hence the same for the limit as x tends to x 0 on .a ; b/: Local Tracing Theorem. If f : [a ; b] ! R is continuous and f 0 .c/ > 0 for some c 2 [a ; b]; then there are c0 ; c1 2 R such that c0 < c < c1 and f .x / < f .c/ < f . y / for all x 2 .c0 ; c/ \ [a ; b] and all y 2 .c; c1 / \ [a ; b]: A similar result for the case f 0 .c/ < 0 holds and the inequality becomes f .x / > f .c/ > f . y /: 1 Proof. Let f 0 .c/ > 0: Assume there is no such c0 : Then for every n 2 N; there is x n 2 .c n ; c/ \ [a ; b] such that f .x n / f .c/ f .x n / f .c/: This leads to f 0 .c/ D lim 0; a contradiction. The other parts of the theorem are n!1 xn c similar. 1 Remarks. If f 0 .c/ 0; we may not have f .x / f .c / f . y / similar to above as the function f .x / D x 2 sin with x f .0/ D 0 satisfies f 0 .0/ D 0; but f takes positive and negative values on every open interval about 0. p p Exercise. A function f : [a ; b] ! R is said to have the intermediate value property iff for every y0 in the open interval with endpoints f .a / and f .b/; there exists at least one x 0 2 .a ; b/ such that f .x 0 / D y0 : We have already showed that continuous functions on [a ; b] satisfied this property. Prove that if g is differentiable on [a ; b]; then g 0 has the intermediate value property. In particular, if g 0.x / 6D 0 for all x 2 [a ; b]; then g 0 > 0 or g 0 < 0 everywhere on [a ; b]: Next we will introduce the generalized mean-value theorem, which has two very important applications, namely Taylor’s theorem and L’Hˆ pital’s rule. o Generalized Mean-Value Theorem. Let f , g be continuous on [a ; b] and differentiable on .a ; b/. Then there is x 0 2 .a ; b/ such that g 0.x 0 /. f .b/ f .a // D f 0 .x 0 /.g .b/ g .a //. (Note the case g .x / D x is the mean-value theorem.) Proof. Let F .x / D f .x /.g .b/ g .a // . f .b/ f .a //.g .x / g .a //, then F .a / D f .a /.g .b/ Rolle’s theorem, there is x 0 2 .a ; b/ such that 0 D F 0 .x 0 / D f 0 .x 0 /.g .b/ g .a // g 0.x 0 /. f .b/ g .a // D F .b/. By f .a //: Taylor’s Theorem. Let f : .a ; b/ ! R be n times differentiable on .a ; b/. For every x, c 2 .a ; b/, there is x 0 between x and c such that f . x / D f .c / C f 0 .c/ .x 1! c/ C f 00.c/ .x 2! c/2 C : : : C f .n 1/.c/ .x .n 1/! c /n 1 C f .n / . x 0 / .x n! c/n : (This is called the n-th Taylor expansion of f about c: The term Rn .x / D form of the remainder.) f .n / . x 0 / .x n! c/n is called the Lagrange 1 /! n1 X Proof. Let I be the closed interval with endpoints x and c. For t 2 I , define g .t / D .n where f .0/ D f ; .x x /0 D 1 and define p.t / D .x n 41 t /n : We have g 0.t / f .k / .t / .x k! k D0 t /n 1 t /k ; D f .n/ .t /.x and p0 .t / D .x t /n 1 : By the generalized mean value theorem, there is x 0 between x and c such that p0 .x 0 / [ g .x / g .c/]. Then f .x / D g .c/ f .n / . x 0 / C .x .n 1/! n! c/n D n1 X | {z } | {z } .x x0 /n 1 .n 1/! f .x / f .k/ .c/ .x k! k D0 | {z } | {z } .x c /n = n f .n/ . x0 /. x x0 /n 1 f .n / . x / g 0.x 0 / [ p.x / 0 p.c/] D c/n : c/k C n! .x Lemma. Let h : .a ; b/ ! [0; C1/ be a bounded function, where a may be real or 1 and b may be real or C1: We have lim h .x / D 0 if and only if lim supfh .t / : a < t < x g D 0: Similarly, lim h .x / D 0 if and only if C C Proof. For right hand limits at a ; limC h .x / D 0 is the same as for every " > 0; there exists r 2 R such that x !a x !a x !b lim supfh .t / : x < t < bg D 0: x !a x !a x !b for all a < x < r implies h .x / D jh .x / 0j < ": The later part is the same as for all a < x < r implies supfh .t / : a < t < x g "; which is limC supfh .t / : a < t < x g D 0: The statement for the left hand limits at b is similar. L’Hˆ pital’s Rule ( 0 Version). Let f , g be differentiable on .a ; b/ and g .x /; g 0.x / 6D 0 for all x 2 .a ; b/; where o 0 1 a < b C1: If f 0.x / (a) limC 0 D L ; where x !a g . x / x !a x !a 1 L C1; and (b) limC f .x / D 0 D limC g .x /; then limC x !a Proof. For f 0 =g 0 ! L 2 R; replacing f =g by . f f 0 .x / Since limC 0 x !a g . x / f .x / D L : (Similarly, the rule is also true if x ! b :) g .x / L g /=g D . f =g / L ; we may assume L D 0: 0 .x / f D 0; we may assume 0 is bounded on .a ; c/ .a ; b/: For every x ; y 2 .a ; c/ with y < x ; g .x / by generalized mean-value theorem, there is w 2 . y ; x / such that n f 0 .t / o f .x / f . y / f 0 .w/ D sup :a<t<x : ./ g .x / g . y / g 0 .w/ g 0.t / n f 0 .t / o f .x / Taking limit of y ! a C on both sides, we get sup : a < t < x : By condition (a) and the lemma, g .x / g 0.t / the right side goes to 0. So by the sandwich theorem, we get the left side go to 0, which is the conclusion. For f 0 =g 0 ! L D C1; there is r 2 R such that a < t < r implies f 0 .t /=g 0.t / > 1 so that f 0 and g 0 are both f .x / f . y / f 0 .t / positive or both negative when x 2 .a ; r /: Next for a < y < x < r; we have D0 > 1: As y ! a C; g .x / g . y / g .t / f .x / we see 1: So f and g are both positive or both negative when x 2 .a ; r /: Now g 0= f 0 ! 0; g ! 0 and f ! 0 g .x / as x ! a C: So by above, g = f ! 0: Hence f =g ! C1: The cases L D 1 and x ! b are similar. f 0 .x / f .x / Example. (Even if lim f .x / D 0 D lim g .x / and lim 0 does not exist, lim may still exist as the following x !a x !a x !a g . x / x !a g . x / 1 1 example will show.) Let f .x / D x 2 sin , g .x / D sin x , a D 0, then x 2 x 2 sin x 2 implies lim f .x / D 0 x !0 x x 2x sin 1 cos 1 0 cos 1 f 0 .x / x x x by sandwich theorem, lim g .x / D 0, lim 0 D lim D lim does not exist. However, x !0 x !0 g . x / x !0 x !0 cos x 1 f .x / x 1 lim D lim .x sin / D 1 0 D 0. x !0 g . x / x !0 sin x x L’Hˆ pital’s Rule ( 1 Version). Let f , g be differentiable on .a ; b/ and g .x /; g 0 .x / 6D 0 for all x 2 .a ; b/; where o 1 a < b C1: If 42 f 0.x / (a) limC 0 D L ; where x !a g . x / (b) limC g .x / D C1 x !a 1 L C1; and then limC x !a f .x / D L : (Similarly, the rule is also true if x ! b or g .x / ! g .x / 1). Proof. We will modify the proof above as follow. If L 2 R; then we can reduce to L D 0 as above. In the case f 0 =g 0 ! L D C1; for x ; s 2 .a ; b/; by generalized mean-value theorem, there is t between x and s such that f 0 .t / f .x / D f .s / C 0 .g .x / g .s // ! C1 as x ! a C by (a) and (b). So we may replace f =g by g = f to assume g .t / g .x / g . y / ; we get for a < y < x < c; L D 0: Multiplying the left and right sides of (*) by g . y/ f .x / f . y / g . y/ n sup | o f 0 .t / :a<t <x g 0.t / {z ! L D0 as x ! a C by lemma n }| g .x / g . y / : g . y/ ! 1 as y ! a C {z } o f 0 .t / f .x / " 0.t / : a < t < x < 4 : Also, ylimC g . y / D 0: Then there is !a g f .x / " g .x / g . y / r < x such that for all y 2 .a ; r /; we can get < and < 2. Then g . y/ 2 g . y/ For every " > 0; there is x 2 .a ; b/ such that sup f . y/ g . y/ The conclusion follows. The cases L D f .x / f . y / f .x / "" C < C 2 D ": g . y/ g . y/ 24 1; x ! b and g .x / ! 1 are similar. f 0 .x / f .x / 0.x / does not exist, xlim g .x / may still exist as the folx !a x !a x !a g !a lowing example will show.) Let f .x / D 2x Csin x , g .x / D 2x sin x , a D C1, then lim f .x / D C1 D lim g .x /, Examples. (1) (Even if lim f .x / D C1 D lim g .x / and lim f 0 .x / 2 C cos x f .x / 2 C .sin x /=x lim 0 D lim does not exist. However, lim D lim D 1: x !C1 g . x / x !C1 2 x !C1 g . x / x !C1 2 cos x .sin x /=x xr D 0 for every r 2 R: x !C1 e x jr j (by the Archimedian principle). We have x r x !C1 x !C1 (2) Show that lim Solution. There is an integer n on [1; 1/: Since implies lim x n on [1; C1/: So 0 xr D 0 by the sandwich theorem. x !C1 e x dn n n! xn x D n ! and lim x D 0; applying L’Hˆ pital’s rule n times, we see lim x D 0; which o x !C1 e x !C1 e dxn xr ex xn ex (3) Let f : .a ; C1/ ! R be differentiable. Show that lim f 0.x / C f .x / D 0 if and only if lim f .x / D 0 and x !C1 x !C1 dy then every solution y D f .x / of the equation C y D g .x / will have limit 0 as x ! C1:) dx x !C1 lim f 0 .x / D 0: (This type of result is often needed in the studies of differential equations. Here if lim g .x / D 0; x !C1 Solution. The if part follows by the computation formulas for limit. For the only if part, consider f .x / D the denominator on the right side tend to C1 as x tend to C1: Since lim by L’Hˆ pital’s rule, we get lim f .x / D lim o x !C1 f . x /e x D 0 and lim f 0 .x / D lim f 0 .x / C f .x / x !C1 x !C1 x !C1 ex 43 f .x /e x x !C1 .e x /0 0 f .x /e x : Note ex D lim x !C1 f 0 . x / C f . x / D 0; f . x / D 0: Appendix 1: Convex and Concave Functions. Definitions. Let I be an interval and f : I ! R. We say f is a convex function on I iff for every a , b 2 I , 0 t 1, f t a C .1 f (b) tf (a) +(1 - t ) f ( b) f (a) a ta+ (1- t ) b b y = f (x ) t /b t /b t f .a / C .1 t f .a / C .1 t / f .b/: t / f .b/: We say f is a concave function on I if f t a C .1 convex function Remarks. As t ranges from 0 to 1, the point t a C .1 t /b; f .t a C t /b/ traces the graph of y D f .x / for a x b, while the point t a C .1 t /b; t f .a / C .1 t / f .b/ traces the chord joining .a ; f .a // to .b; f .b//. So f is convex on I if and only if the chord joining two points on the graph always lies above or on the graph. .1 ( b, f ( b)) (a, f ( a )) ( x, f (x)) a x b Theorem. f is convex on I if and only if the slope of the chords always f . x / f .a / f .b / f . x / increase, i.e. a, x, b 2 I , a < x < b ) . xa bx bx 1. Proof. x D t a C .1 t /b for some t 2 [0; 1] () 0 t D ba b b x x f .a / C a b a f .b/ () f t a C .1 t /b a t f .a / C .1 t / f .b/: f .x / x f .a / a f .b/ b f .x / () f .x / x Theorem. For f differentiable on I ; f is convex on I () f 0 is increasing on I . (For f twice differentiable on I ; f is convex on I () f 00 0 on I .) f .x / f .a / f .b / f . x / f .b/ f .a / Proof. ()) If a , b 2 I , a < b, then f 0 .a / D limC limC D D x !a x !a xa bx ba f .x / f .a / f .b/ f .x / lim lim D f 0 .b /. x !b x !b xa bx (() If a , x , b 2 I , a < x < b, then by the mean value theorem, there are r , s such that a < r < x < s < b and f .b / f . x / f .x / f .a / D f 0 .r / f 0 .s / D . xa bx Theorem. If f is convex on .a ; b/, then f is continuous on .a ; b/. f .w/ f .v/ f .x 0 / x0 Proof. For x 0 2 .a ; b/, consider u , v , w 2 .a ; b/ such that u < x 0 < v < w. Then f . x 0 / f .u / f .v/ f .x 0 / x0 u v x0 f .w/ f .v/ f .v/ .v x 0/ C f .x 0 /: f .u / .v x 0 / C f .x 0 / u wv C , we have f .x / Taking limit as v ! x 0 f .x 0 C/ f .x 0 /, i.e. f .x 0 C/ D f .x 0 /. Similarly, we can show 0 f .x 0 / D f .x 0 / by taking u < v < x 0 < w: Therefore, lim f .x / D f .x 0 / for every x 0 2 .a ; b/. w v . Solving for f .v/; we get x ! x0 0 if 0 x < 1 is convex on [0; 1] by checking the slopes of the chords over [0; 1], Example. The function f .x / D 1 if x D 1 but f is not continuous at 1 because lim f .x / D 0 6D 1 D f .1/. x !1 n Remark. The proof of the last theorem uses the fact that on an open interval, at any point x 0 , there are points on its left and points on its right, which is not true for endpoints of a closed interval. Example. Prove that if a ; b 0 and 0 < r < 1; then jar p p p n n have j n a bj ja b j:) br j ja bjr : (In particular, for n D 2; 3; 4; : : : ; we Solution. We may assume a > b; otherwise interchange a and b: Define f : [0; a ] ! Rby f .x / D x r C .a x /r : Since r 1 < 0; so f 00 .x / D r .r 1/ x r 2 C .a x /r 2 0: So f is concave on [0; a ]: Since f .0/ D ar D f .a /; we get r r r f .x / D x C .a x / a for all x 2 [0; a ]: Therefore, if b 2 [0; a ]; we have ja bjr D .a b/r ar br D jar br j: 44 Chapter 9. Riemann Integral For a ; b 2 R; let f be a bounded function on [a ; b]; say j f .x /j K for every x 2 [a ; b]. Let P D fx 0 ; x 1 ; : : : ; x n g be a partition of [a ; b], i.e. a D x 0 < x 1 < : : : < x n 1 < x n D b. The length of [x j 1; x j ] is 1x j D x j x j 1 and the mesh of P is k P k D maxf1x 1 ; : : : ; 1x n g: Also, on [x j 1; x j ]; let m j D inff f .x / : x 2 [x j 1; x j ]g and M j D supf f .x / : x 2 [x j 1; x j ]g: Definitions. For the partition P ; a Riemann sum is a sum of the form S D n X j D1 n X j D1 f .t j /1x j ; where every t j is in [x j 1; x j ]: The lower Riemann sum is L . f ; P / D m j 1x j and the upper Riemann sum is U . f ; P / D Mj K implies n X j D1 M j 1x j : a/ L . f; P / S a = x0 x1 x2 . . . xn - 1 xn = b (Note K mj f .t j / U . f ; P / K .b a /: ) K .b In 1854, George B. Riemann defined the integral of f .x / on [a ; b] to be lim immediate technical problems with such a definition. n X k Pk!0 j D1 f .t j /1x j : There are two (1) As the choices of the t j ’s may vary, it is hard to say how close the Riemann sum is to the actual integral. In particular, it is not clear if the Riemann sum is greater than or less than the integral at any time. (2) The type of limit involved is not the limit of a sequence nor the limit of a function. In fact, there are many variables in a Riemann sum! Instead of dealing with these technicalities, we introduce integral following the approach of J. Gaston Darboux in 1875. Definition. We say P 0 is a refinement of P or P 0 is finer than P iff P and P 0 are partitions of [a ; b] and P Refinement Theorem. If P 0 is a finer partition of [a ; b] than P, then L . f ; P / L . f; P 0 / U . f; P 0 / P0: wj xj - 1 w xj Let m j D inf f f .x / : x 2 [x j 1; x j ]g; m 0j D inff f .x / : x 2 [x j 1; w]g and m 00 D j inf f f .x / : x 2 [w; x j ]g: Since [x j 1 ; w]; [w; x j ] [x j 1; x j ]; we have m j m 0j and m j m 00 : Then m j 1x j D m j .w x j 1 / C m j .x j w/ m 0j .w x j 1/ C m 00 .x j w/: j j So L . f ; P / L . f ; P 0 /. Similarly, U . f ; P 0 / U . f ; P /. U . f ; P /. 0 can be obtained from P by adding one point at a time. It suffices to consider Proof. P the case P 0 D P fwg with w 62 P D fx 0 ; x 1 ; : : : ; x n g. Suppose x j 1 < w < x j : Observe that L . f ; P / “underestimates” the area under the curve; U . f ; P / “overestimates” the area under the curve. Definitions. The lower integral of f on [a ; b] is Zb a .L / f .x / d x D supf L . f ; P /: P is a partition of [a ; b]g (the largest underestimate/ and the upper integral of f on [a ; b] is .U / Zb a f .x / d x D inffU . f ; P /: P is a partition of [a ; b]g Rb Rb (the smallest overestimate): (So for every partition P of [a ; b]; L . f ; P / . L / a f .x / d x .U / a f .x / d x U . f ; P /:) Rb Rb We say f is (Riemann) integrable on [a ; b] iff . L / a f .x / d x D .U / a f .x / d x : In that case, we denote the Rb Rb Ra Ra common value by a f .x / d x . (For b a , define a f .x / d x D b f . x / d x . In particular, a f . x / d x D 0.) 45 U ( f, P) - L ( f, P) = sum of areas of shaded regions Theorem (Integral Criterion). Let f : [a ; b] ! R be a bounded function. The function f is (Riemann) integrable on [a ; b] if and only if for every " > 0, there is a partition P of [a ; b] such that U . f ; P / L . f ; P / < ": Proof. If for every " > 0; there is a partition P of [a ; b] such that U . f ; P / L . f ; P / < ", then 0 .U / a f .x / d x R R R . L / ab f .x / dx U . f; P / L . f; P / < ": By the infinitesimal principle, we get . L / ab f .x / d x D .U / ab f .x / d x ; i.e. f is (Riemann) integrable on [a ; b]: Conversely, if f is (Riemann) integrable on [a ; b]; then for every " > 0; by the supremum property, we have R R . L / a f .x / dx "=2 < L . f; P1 / . L / ab f .x / d x for some partition P1 of [a ; b]: Similarly, .U / ab f .x / d x Rb U . f ; P2 / < .U / a f .x / d x C "=2 for some partition P2 of [a ; b]: Let P D P1 P2 ; then by the refinement Rb theorem, L . f ; P1 / L . f; P / U . f; P / U . f ; P2 /: Since U . f ; P2 / L . f ; P1 / < [.U / a f .x / d x C "=2] Rb [. L / a f .x / d x "=2] D "; so U . f ; P / L . f ; P / U . f ; P2 / L . f ; P1 / < ": Rb Rb Question. Are there integrable functions? Are there non-integrable functions? Below we will show that constant functions and continuous functions are integrable. 1 if x 2 Q is not continuous anywhere. Partitioning an interval [a ; b] into 0 if x 62 Q subintervals [x j 1; x j ], we get by the density of rational numbers and irrational numbers that m j D 0 and M j D 1: Then Example. Recall the function f .x / D L . f ; P / D 0 and U . f ; P / D b a for every partition P of [a ; b]: So . L / Zb a f .x / d x D 0, .U / Zb a f .x / d x D b a. Therefore, f .x / is not integrable on every interval [a ; b] with a < b. Example. If f .x / D c for every x 2 [a ; b]; then L . f ; P / D c.b Zb a .L / f .x / d x D c.b a / D .U / Zb a a / D U . f ; P / for every partition P of [a ; b]: So Zb a f .x / d x : Therefore, f is integrable on [a ; b] and f .x / d x D c.b a /: Uniform Continuity Theorem. If f : [a ; b] ! R is continuous, then f is uniformly continuous (in the sense that for every " > 0; there exists a > 0 such that for all x ; t 2 [a ; b]; jx t j < implies j f .x / f .t /j < ":) 1 Proof. Suppose f is not uniform continuous. Then there is an " > 0 such that for every D n ; n 2 N; there are 1 x n ; tn 2 [a ; b] such that jx n tn j < D n and j f .x n / f .tn /j ": By the Bolzano-Weierstrass theorem, fx n g has 1 a subsequence fx n j g converging to some w 2 [a ; b]: Since jtn j wj jtn j x n j j C jx n j wj C jx n j wj; by nj the sandwich theorem, ftn j g converges to w: Then 0 D j f .w/ f .w/j D lim j f .x n j / f .tn j /j " > 0; which is a contradiction. j !1 Examples. For 2 .0; 1]; a function f : S ! R is H¨ lder continuous of order iff f staisfies the condition that o there exists constant M > 0 such that for all x ; y 2 S ; j f .x / f . y /j M jx y j : In case D 1; f is said to be Lipschitz. The constant M is called the Lipschitz constant or H¨ lder constant for f : All functions with bounded o p derivatives (such as cos x with M D 1) are Lipschitz by the mean-value theorem. The function f .x / D x satisfies p p p 1 jx yj jx y j on [0; C1/ is H¨ lder continuous of order 2 : All H¨ lder functions are uniformly continuous, o o since for every " > 0; we can take D ."= m /1= : Theorem. If f is continuous on [a ; b], then f is integrable on [a ; b]. j f .x / Proof. For " > 0, since f is uniformly continuous on [a ; b], there is > 0 such that x , w 2 [a ; b], jx wj < ) f .w/j < "=.b a /. Let P D fa D x 0 ; x 1 ; : : : ; x n D bg be a partition of [a ; b] with max jx j x j 1 j < . By 1jn 46 the extreme value theorem, M j D f .w j / and m j D f .u j / for some w j ; u j 2 [x j 1; x j ]: Then U . f; P / L . f; P / D n X j D1 f .w j / f .u j / 1 x j < " b n X a j D1 .x j xj 1 /D ".x n b x0/ D ": a By the integral criterion, f is integrable on [a ; b]: Exercises. (1) Let f : [a ; b] ! R be a function and c 2 [a ; b]: Show that f is integrable on [a ; b] if and only if f is integrable on [a ; c] and [c; b]: (2) If f : [a ; b] ! R is bounded and discontinuous only at x 1 ; : : : ; x n 2 [a ; b]; show that f is integrable on [a ; b]: (Hint: This can be done directly or by using (1) to reduce the problem to intervals having only one discontinuity.) Questions. How bad can an integrable function be discontinuous? Which functions are integrable? Definitions.(i) A set S R is of measure 0 (or has zero-length) iff for every " > 0, there are intervals .a1 ; b1 /; .a2 ; b2 /; .a3 ; b3/; : : : such that S (ii) We say a property holds almost everywhere iff the property holds except on a set of measure 0. (It is common to abbreviate almost everywhere by a.e. in advanced courses. In probability, almost surely is used instead of almost everywhere.) Remarks. For example, it is known that monotone functions on intervals are differentiable almost everywhere (see H.L. Royden’s book, Real Analysis, 3rd ed., p. 100). Also, H. Rademacher proved that Lipschitz functions on intervals are differentiable almost everwhere (see L.C. Evans’ book, Partial Differential Equations, p. 281). Lebesgue’s Theorem(1902). For a bounded function f : [a ; b] ! R; f is integrable on [a ; b] if and only if the set S f D fx 2 [a ; b]: f is discontinuous at x g is of measure 0 (i.e. f is continuous almost everywhere). For a proof of Lebesgue’s theorem, please go to appendix 1 of this chapter. Examples. (1) The empty set ; is of measure 0 because ; (2) A countable set fx 1 ; x 2 ; : : :g is of measure 0 because fx 1 ; x 2 ; x 3 ; : : :g 1 X 1 .an ; bn / and jan nD1 n D1 bn j < " . function f : [a ; b] ! R is integrable on [a ; b]; since the set of discontinuities S is ;. 1 X 1 .0; 0/ and j0 nD1 nD1 0j D 0 < ": So every continuous .x 1 " 4 " ; x 1 C / .x 2 4 " 4 ; x2 C 2 " 4 / .x 3 2 " 4 ; x3 C 3 " 4 / 3 and 1 X 2" 2" D < ": n 4 3 nD1 So every monotone function on [a ; b] is integrable on [a ; b]; since the set of discontinuities S is countable by the monotone function theorem, hence of measure 0. (3) There exist uncountable sets, which are of measure 0 (eg. the Cantor set). (4) A countable union of sets of measure 0 is of measure 0. (If S1 ; S2; S3 ; : : : are of measure 0 and S is their union, then for every " > 0; since Sk is of measure 0, there are intervals .ak;n ; bk;n / such that Sk 1 X nD1 jak;n bk;n j < " 4 : Then S k (5) If a set S is of measure 0 and S 0 11 1 XX X" 11 " .ak;n ; bk;n / and jak;n bk;n j < D < ":) k D1 nD1 4k 3 k D1 nD1 k D1 S ; then S 0 is also of measure 0. (For S 0; use the same intervals .an ; bn / for S :) n 1 .ak;n ; bk;n / and nD1 1 if x D r1 or r2 or : : : or rn (6) Arrange Q \ [0; 1] in a sequence r1 ; r2 ; r3 ; : : : and define fn .x / D : Then fn .x / 0 otherwise is integrable on [0; 1]; since the set of discontinuities S f n D fr1 ; r2 ; : : : ; rn g is finite, hence of measure 0. Now 47 1 if x 2 Q is not integrable on [0; 1] since Sf D fx 2 [0; 1] : f is discontinuous at x g D 0 if x 62 Q [0; 1]; which is not of measure 0. So the limit of a sequence of Riemann integrable functions may not be Riemann integrable. n!1 lim fn .x / D f .x / D Remarks. In B. R. Gelbaum and J. M. H. Olmsted’s book Counterexamples in Analysis, p. 106, there is an example of the limit of a sequence of Aontinuous functions on [0; 1]; which is not Riemann integrable. c Theorem. For c 2 [a ; b]; f is integrable on [a ; b] if and only if f is integrable on [a ; c] and [c; b]: Proof. Let S ; S1 ; S2 be the set of discontinuous points of f on [a ; b]; [a ; c]; [c; b]; respectively. If f is integrable on [a ; b]; then by Lebesgue’s theorem, S is of measure 0. Since S1 ; S2 S ; so both S1 ; S2 are of measure 0. By Lebesgue’s theorem, f is integrable on [a ; c] and [c; b]: Conversely, if f is integrable on [a ; c] and [c; b]; then S1; S2 are of measure 0. Since S examples (4) and (5), S is of measure 0. By Lebesgue’s theorem, f is integrable on [a ; b]: Theorem. If f ; g : [a ; b] ! R are integrable on [a ; b]; then f C g ; f Proof. If f and g are integrable on [a ; b]; then f and g are bounded on [a ; b]: So f C g ; f on [a ; b]: S1 S2 fcg; by g and f g are integrable on [a ; b]: g and f g are bounded Observe that if f and g are continuous at x ; then f C g is continuous at x : Taking contrapositive, we see that if f C g is discontinuous at x ; then f or g is discontinuous at x : So if x 2 S f Cg ; then x 2 S f or x 2 Sg ; which implies S f Cg S f Sg : Since f ; g are integrable on [a ; b]; by Lebesgue’s theorem, S f ; Sg are of measure 0. By example (4), S f Sg is of measure 0. By example (5), S f Cg is also of measure 0. Therefore, by Lebesgue’s theorem f C g is integrable on [a ; b]: Using a similar argument, we can see that f g ; f g are integrable on [a ; b]: Remarks. By a similar argument, we can show that if f : [a ; b] ! R is integrable on [a ; b] and g is bounded and continuous on f .[a ; b]/; then Sg f S f : (This is because if f is continuous at x ; then g f is continuous at x : Taking contrapositive, we get if x 2 Sg f ; then x 2 S f : So Sg f S f :) Since S f is of measure 0, by example (5), Sg f is of measure 0. So g f : [a ; b] ! R is integrable on [a ; b]: In particular, if f is integrable on [a ; b]; then taking g .x / D jx j; x 2 ; e x ; cos x ; : : : ; respectively, we get j f j; f 2 ; e f ; cos f ; : : : are integrable on [a ; b]: However, even if f : [a ; b] ! [c; d ] is integrable on [a ; b] and g : [c; d ] ! R is integrable on [c; d ]; g f may not be integrable on [a ; b]: (For example, define f : [0; 1] ! [0; 1] by f .0/ D 1; f .m = n / D 1= n ; where m ; n are positive integers with no common prime factors, and f .x / D 0 for every x 2 [0; 1] n Q: Next, define g : [0; 1] ! [0; 1] by g .0/ D 0 and g .x / D 1 for every x 2 .0; 1]: As an exercise, it can be showed that S f D [0; 1] \ Q and Sg D f0g: So f and g are integrable on [0; 1]: However, g f is the nonintegrable function that is 1 on [0; 1] \ Q and 0 on [0; 1] n Q:) There is even an example of a continuous function f : [0; 1] ! [0; 1] and an integrable function g : [0; 1] ! R such that g f is not integrable (see B. R. Gelbaum and J. M. H. Olmstad’s book Counterexample in Analysis, pp. 106-107). Up to now, we have been trying to determine which functions are integrable. Below we will look at how the integrals of functions can be computed. Theorem (Simple Properties of Riemann Integrals). Let f and g be integrable on [a ; b]: Zb a (1) f .x / g .x / d x D Zb a f .x /d x Zb g . x /d x ; f .x / d x Zb a c f .x / d x D c Zb a f .x / d x for every c 2 R. Zb a (2) If f .x / Zb g .x / for all x 2 [a ; b], then Zc a a Zb a Zb a g .x / d x . Also, f .x / d x Zb a j f .x /j d x . (3) a f .x / d x D Zb a f .x / d x C Zb a Zb c f .x / d x for c 2 [a ; b]: Zb a Proof. To get (1), by the supremum and infimum properties, for every " > 0; there is a partition P of [a ; b] such that f .x / d x C g .x / d x " < L . f ; P / C L .g ; P / 48 L . f C g; P/ f .x / C g .x / d x U . f C g; P/ Letting " ! 0; we get inf. S / D Zb a U . f ; P / C U .g ; P / < Zb Z ab a Zb Zb a a f .x / d x C Zb a g .x / d x C ": f; P / D L . f ; P / and Zb a f .x / C g .x / d x D Zb a f .x / d x C g .x / d x : Next, from U . sup S ; we get g .x / d x D f .x / d x D f .x / d x : Then Zb a f .x / Zb a f .x / C . g .x // d x D f .x / d x C Zb a g .x / d x D Zb a Zb a f .x / d x Zb a g .x / d x : Zb For the second statement, the case c D 0 is clear since a 0 f .x / d x D 0 D 0 f .x / d x : If c > 0; then Zb a c f .x /d x D supfcL . f ; P /: P is a partition of [a ; b]g D c supf L . f ; P /: P is a partition of [a ; b ]g D c Zb a f .x /d x : If c < 0; then Zb a c f .x / d x D Zb a . c/ f . x / d x f Zb D a . c/ f . x / d x f; P / D . c/ Zb a f .x / d x D c Zb a f .x / d x : For (2), observe that g Zb a 0 on [a ; b] implies L .g 0 for every partition P : So Zb g .x / f .x / d x D supf L .g jfj f ; P /: P is a partition of [a ; b]g f j f j on [a ; b ]; we get Zb a 0 implies a f .x / d x f .x / d x Zb a g .x / d x : For the second statement, since For (3), let P be a partition of [a ; b]; P 0 D P fcg; P1 D P 0 \ [a ; c] and P2 D P 0 \ [c; b]: Then P 0 is a finer partition of [a ; b] than P ; P1 is a partition of [a ; c]; P2 is a partition of [c; b] and L . f ; P 0 / D L . f ; P1 / C L . f ; P2 /: Let a a j f . x /j d x Zb Zb j f .x /j d x : A D f L . f ; P /: P is a partition of [a ; b]g and B D f L . f ; P 0 /: P is a partition of [a ; b] and P 0 D P fcgg: Since P 0 is also a partition of [a ; b]; we see B A and so sup B sup A: By the refinement theorem, L . f ; P / L . f ; P 0 / sup B : Hence sup A sup B : Therefore, sup A D sup B and Zb a f .x / d x D sup A D sup B D supf L . f ; P1 / C L . f ; P2 /: P1 ; P2 are partitions of [a ; c]; [c; b ]; respectivelyg D supf L . f ; P1 /: P1 is a partition of [a ; c]g C supf L . f ; P2 /: P2 is a partition of [c; b ]g Zc a D f .x / d x C Zb c f .x / d x : The most important tool for computing an integral is to find an antiderivative or primitive function of the integrable function, which is a function whose derivative is the integrable function. What can we say about such a function? Example. For x 2 [ 1; 1]; define f .x / D is integrable on [ 1; 1]: Now the function F .x / D n 1 1 Zx 0 if x 0 : Since f .x / has only one point of discontinuity at 0, f .x / if x < 0 f .t / dt D n x x if x 0 D jx j if x < 0 49 is continuous on [ 1; 1]; hence uniformly continuous there by the uniform continuity theorem, but F .x / is not differentiable at 0, the same point where f .x / is discontinuous! Theorem. If f is integrable on [a ; b] and c 2 [a ; b]; then F .x / D Proof. Recall there is K > 0 such that j f .t /j implies j F .x / F .x 0 /j D Zx x0 Zx c f .t / dt is uniformly continuous on [a ; b]: x0j < " K for every t 2 [a ; b]: For every " > 0; let D K : Then jx f .t / dt K jx x 0 j < ": The next theorem is the most important theorem in calculus. It not only tells us how to compute an integral, but also the deep fact that differentiation and integration are inverse operations on functions. Roughly, it may be Zx Zx x d d summarized by the formulas h .t / dt D h .t / : h .t / dt D h .x / and c dx c c dt Fundamental Theorem of Calculus. Let c; x 0 2 [a ; b]: (1) If f is integrable on [a ; b], continuous at x 0 2 [a ; b] and F .x / D Zx c f .t / dt , then F 0 .x 0 / D f .x 0 /. g . x / d x D G .b / G .a /. (2) If G is differentiable on [a ; b] with G 0 D g integrable on [a ; b], then Zb a (Note G 0 need not be continuous by an example in the chapter on differentiation.) F .x / F .x 0 / D f .x 0 / using the definition of limit. For every " > 0; since f is x x0 continuous at x 0 ; there exists a > 0 such that for every x 2 [a ; b]; jx x 0 j < ) j f .x / f .x 0 /j < "=2: With the same ; we get 0 < jx x 0 j < implies Proof. (1) We will check lim x ! x0 F .x / x F .x 0 / x0 Rx f .x 0 / D c f .t /dt x R x0 c f .t /dt Rx x0 f .x 0 /dt x0 Rx x0 x D x0 . f .t / x Zb f .x 0 //dt x0 g .x / d x Rx x0 "=2 dt x0 G .a / x G .b / < ": (2) The conclusion will follow from the infinitesimal principle if we can show a <" for every " > 0: Now for every " > 0; since g is integrable on [a ; b]; by the integral criterion, there is a partition P D fa D x 0 ; x 1 ; : : : ; x n D bg such that U .g ; P / L .g ; P / < ": For j D 1; 2; : : : ; n ; by the mean value theorem, there exists t j 2 .x j 1 ; x j / such that G .x j / G .x j 1 / D G 0 .t j /.x j x j 1/ D g .t j /1x j : Then L .g ; P / Zb a n X j D1 g .ti /1x j D n X j D1 .G .x j / G .x j 1 // D G .b/ G .a / G .a / U .g ; P /: Also, L .g ; P / g .x / d x U .g ; P /: Then Zb a g .x / d x G .b/ U .g ; P / L .g ; P / < ": Remarks. In (2) above, the condition G 0 is integrable is important. The function G .x / D x 2 sin 1 for x 6D 0 and x2 0 .x / is unbounded on [0; 1]; hence not integrable there. In Y. Katznelson and G .0/ D 0 is differentiable on R; but G K. Stromberg’s paper Everywhere Differentiable, Nowhere Monotone Functions, American Mathematical Monthly, vol. 81, pp. 349-354, there is even an example of a differentiable function on R such that its derivative is bounded, but not Riemann integrable on any interval [a ; b] with a < b: Theorem (Integration by Parts). If f , g are differentiable on [a ; b] with f 0, g 0 integrable on [a ; b], then Zb a f .x /g 0 .x / d x D f .b/g .b/ f .a / g .a / Zb a f 0 .x /g .x / d x : 50 Proof. By part (2) of the fundamental theorem of calculus, f 0 .x /g .x / C . f g /0 .x / d x D f .b/g .b/ f .a /g .a /: Since . f g /0 .x / D a f .x /g 0 .x /; we can subtract the integral of f 0 .x /g .x / on both sides to get the integration by parts formula. [a ; b] , then Z .x / .a / Z .b / Zb Theorem (Change of Variable Formula). If continuous on .a / f .t / dt D Zb a : [a ; b] ! R is differentiable, 0 is integrable on [a ; b] and f is f .x / 0 .x / d x : Proof. Let g .x / D f f .t / dt : By part (1) of the fundamental theorem of calculus and chain rule, g 0.x / D Zb .x / 0 .x /: So Zb f a .x / 0.x / d x D a g 0 . x / d x D g .b / g .a / D g .b/ D Z .b / .a / f .t / dt : Appendix 1: Proof of Lebesgue’s Theorem We first modify the uniform continuity theorem to obtain a lemma. Lemma. Let f : [a ; b] ! R be a function continuous on a subset K D [a ; b] n there exists a > 0 such that for all x 2 K ; t 2 [a ; b]; jx 1 . j ; j /: Then for every " > 0; j D1 t j < implies j f .x / f .t /j < ": 1 Proof. Suppose the lemma is false. Then there is an " > 0 such that for every D n ; n 2 N; there are x n 2 K ; tn 2 [a ; b] 1 f .tn /j ": By the Bolzano-Weierstrass theorem, fx n g has a subsequence such that jx n tn j < D n and j f .x n / fx n j g converging to some w 2 [a ; b ]: Since jt n j wj jt n j x n j j C jx n j wj 1 C jx n j nj wj; by the sandwich theorem, ftn j g converges to w: / for some i: Since lim x nj D w and w 2 . i ; i /; j !1 1 by the definition of limit, there will be some x n p 2 . i ; i /; contradicting x n p 2 K D [a ; b] n . j ; j /: So w 2 K : j D1 Since f is continuous at w; by the sequential continuity theorem, i Next we will show w 2 K : Suppose w 62 K ; then w 2 . i ; 0 D j f .w/ which is a contradiction. f .w/j D lim j f .x n j / j !1 f .tn j /j " > 0; Proof of Lebesgue’s Theorem. First suppose f : [a ; b] ! R is integrable. Note that if f is discontinuous at x ; then there is an " > 0 such that for every > 0; there is z 2 .x ; x C / \ [a ; b] such that j f .x / f .z /j ": Let Dk be the 1 set of all x 2 [a ; b] such that for every open interval I with x 2 I ; there is z 2 I \ [a ; b] such that j f .x / f .z /j > : k 1 Since every positive " > for some positive integer k by the Archimedian principle and every open interval I with x 2 I k 1 contains an interval .x ; x C / for some > 0; it follows that S f D fx 2 [a ; b] : f is discontinuous at x g D Dk : We will show each Dk is of measure 0, which will imply S f is of measure 0. k D1 For every " > 0; by the integral criterion, there is a partition P D fx 0 ; x 1 ; : : : ; x n g of [a ; b] such that " 1 U . f; P / L . f; P / < : If there is x 2 Dk \ .x j 1; x j /; then M j m j j f .x / f .z /j > for some z 2 2k k .x j 1; x j /: Let J be the set of j such that Dk \ .x j 1; x j / 6D ;: Then Dk n fx 0 ; x 1; : : : ; x n g .x j 1; x j / and X j2J jx j 1 xjj X j2J k . Mj | {z >1= k m j / 1x j } k U . f; P / L . f; P / < " j2J 2 : Next around each x j ; we take an open interval 51 : Then the intervals .x j 1; x j / with j 2 J and I j ’s together contain Dk and the 2.n C 1/ sum of their lengths is less than ": So each Dk (and hence S f ) is of measure 0. I j containing x j with length " ; where N is an upper 1 bound for j f .x /j on [a ; b]: By the definition of measure 0, there are intervals . i ; i / such that S f . i; i/ i D1 1 X 1 and ji . i ; i /: Now take the in the lemma for "0 and i j < "0 : Then f is continuous on K D [a ; b ] n i D1 i D1 a partition P D fx 0 ; x 1 ; : : : ; x n g such that jx j 1 x j j < for j D 1; 2; : : : ; n : If K \ [x j 1 ; x j ] D ;; then 1 [ x j 1; x j ] . i ; i /: If there is x 2 K \ [x j 1; x j ]; then M j m j D supfj f .t0 / f .t1 /j : t0; t1 2 [x j 1; x j ]g i D1 supfj f .t0 / f .x /j C j f .x / f .t1 /j : t0 ; t1 2 [x j 1; x j ]g 2"0 : So 3. N C b a/ U . f; P / L . f; P / D X K \[ x j 1 Conversely, suppose S f is of measure 0. For every " > 0; let "0 D " ;x j ]D; | . Mj {z 2N m j / 1x j C } X K \[ x j 1 ;x j ]6D; | . Mj {z 2 "0 m j / 1x j } 2 N "0 C 2"0 .b a / < ": Therefore, by the integral criterion, f is integrable on [a ; b]: Appendix 2: Riemann’s Definition of the Integral Recall the definition of lim f .x / D L is that for every " > 0; there is we have j f .x / x !0 n X > 0 such that for every x with jx j < ; L j < ": Based on this, Riemann’s approach of integral can be defined as follow. Definition. Let f be a bounded function on [a ; b]: We write lim n X j D1 > 0 such that for every partition P of [a ; b] with k P k < and every choice t j have f .t j /1x j k Pk!0 j D1 f .t j /1x j D I iff for every " > 0; there is a 2 [ x j 1; x j ]; j D 1; 2; : : : ; n ; we I < ": (Such I is unique as in the proof of uniqueness of limit.) Below we will show Darboux’s approach is the same as Riemann’s by establishing the following theorem. Darboux’s Theorem. Let f be a bounded function on [a ; b]: Then the following are equivalent: Zb a (a) f .x / d x D I I n X (b) lim k Pk!0 j D1 f .t j /1x j D I I n X j D1 (c) for every " > 0; there is a partition P of [a ; b] such that for every refinement Q D fx 0 ; x 1 ; : : : ; x n g of P ; every choice t j 2 [x j limit.) Proof. .a / ) .b/ Suppose 1 ; x j ]; we have f .t j /1x j I < ": (Such a number I is unique as in the proof of uniqueness of Zb a f .x / d x D I : For every " > 0; by the proof of the integral criterion, there is a partition P" D fw0 ; w1 ; : : : ; wq g of [a ; b] such that I U . f ; P" / < I C : Let D ; where 2 2 4q K K D supfj f .x /j : x 2 [a ; b]g: If P D fx 0 ; x 1 ; : : : ; x n g is a partition of [a ; b] with k P k < ; then U . f ; P / D S1 C S2 ; where S1 is the sum of the terms from intervals not containing points in P" and S2 the sum of the remaining terms. " Note the rectangles for the terms of S1 are inside the rectangles for U . f ; P" /: So we have S1 < U . f ; P" / < I C : 2 Since P" D fw0 ; w1 ; : : : ; wq g; S2 has at most 2q terms (one for w0 ; wq each and at most two for w1 ; : : : ; wq 1 52 " < L . f; P" / " " each). Then S2 I 2q . K k P k/ n X j D1 2q K D " 2 : Thus U . f; P / D S1 C S2 < I n X j D1 C ": Similarly, L . f ; P / >I ": So " < L . f; P / f .t j /1x j U . f ; P / < I C ": Then f .t j /1x j I < ": .b/ ) .c/ For every " > 0; take as the definition above. Let P be any partition of [a ; b] with k P k < : Then for every refinement Q of P ; we have k Q k k P k < and so (c) follows from the conclusion of (b). " .c/ ) .a / For " > 0; let P be the partition in (c) for : For each j D 1; 2; : : : ; n ; by the supremum limit theorem, there 3 n X " are sequences t j ;k such that lim f .t j ;k / D M j D supf f .x / : x 2 [x j 1; x j ]g: Since f .t j ;k /1x j I < ; so taking k !1 3 j D1 " " 2" : Similarly, we can also get j L . f; P / I j : Then jU . f; P / L . f; P /j < ": limit, we get jU . f ; P / I j 3 Zb a 3 3 By the integral criterion, it follows that f .x / d x exists. By .a / ) .b/ and .b/ ) .c/; it must equal I by uniqueness. Improper Riemann Integral – Integration of Unbounded Functions or on Unbounded Intervals We would like to integrate unbounded functions on unbounded intervals by taking limit of integrals on bounded intervals. Since the functions may or may not be continuous, we have to make sure the functions are integrable on bounded intervals first. Definition. Let I be an interval. We say f : I ! R is locally integrable on I iff f is integrable on every closed and bounded subintervals of I : We denote this by f 2 L loc. I /. Roughly, there are three cases of improper integrals. Case 1: (Unbounded near one endpoint) Let f be locally integrable on [a ; b/; where a is real and b is real or C1: We define the improper (Riemann) integral of f on [a ; b/ to be Zb a f .x / d x D lim d !b Zd a f .x / d x ; provided the limit exists in R; and we say f is improper integrable on [a ; b/ in that case. For f locally integrable on an interval of the form .a ; b]; where a is real or 1 and b is real, the definitions of the improper (Riemann) integral of f on .a ; b] and f is improper (Riemann) integrable on .a ; b] are similar. Case 2: (Unbounded near both endpoints) Let f be locally integrable on .a ; b/; with a ; b real or infinity, and x 0 2 .a ; b/. We define the improper (Riemann) integral of f on .a ; b/ to be Zb a provide both limits exist in R; and say f is improper integrable on .a ; b/ in that case. (Note the integral is the same regardless of the choices of x 0 2 .a ; b/.) Case 3: (Unbounded inside interval) Let f be improper integrable on intervals [a ; c/ and .c; b]: The improper integral of f on [a ; b] is the sum of the improper integrals on the two intervals. The cases where a or b is excluded are similarly defined. In each case, if the improper integral is a number, then we say the improper integral converges to the number, otherwise we say it diverges. Examples.(1) Consider the unbounded function f .x / D ln x on .0; 1]. Observe that f is continuous, hence integrable, Z1 0 f .x / d x D limC c !a Z x0 c f .x / d x C lim d !b Zd x0 f .x / d x on every closed subinterval of .0; 1]: So f 2 L loc .0; 1]: Now limC c !0 Z1 c ln x d x D limC. 1 c !0 c ln c C c/ D 1, so ln x dx D 1 and ln x is improper integrable on .0; 1]: 53 1 on the unbounded interval [2; C1/. Observe that f 2 L loc[2; C1/ because of continuity. x2 Zd Z1 1 1 11 1 1 1 Now lim d x D lim C D , so d x D and 2 is improper integrable on [2; C1/: 2 d !C1 2 x 2 d !C1 d 2 2 x 2 x 2 (3) Consider f .x / D e x on R D . 1; C1/: Since f is continuous everywhere, f 2 L loc. 1; C1/: Now take a (2) Consider f .x / D x 0 2 . 1; C1/; say x 0 D 0: Then lim x c! Z0 c does not exist in R: So e is not improper integrable on . 1; C1/: 1 e x d x D lim .1 c! 1 ec / D 1; but lim Zd d !C1 0 e x d x D lim .ed d !C1 1/ Remarks. By taking limits, many properties of Riemann integrals extend to improper Riemann integrals as well. There are also some helpful tests to determine if a function is improper integrable. p-test. For 0 < a < 1; we have Z a 11 xp d x < 1 if and only if p > 1: Also, Z Za 0 1 d x < 1 if and only if p < 1: xp Z1 a Proof. In the proof of p-test for series, we saw 1 11 xp d x < 1 if and only if p > 1: Adding Z Z 1 1=c 1 1 1 get the first statement. Next, by the change of variable y D ; we have dx D d y : Taking limit as x xp y2 p c 1 Z1 Z1 1 1 c ! 0C; we see that d x < 1 if and only if d y < 1: By the first statement, this is the same as p 2p y 0x 1 Za 1 d x < 1; we get the second statement. 2 p > 1; i.e. p < 1: Again adding p 1x 1 d x < 1; we xp Theorem (Comparison Test). Suppose 0 f .x / g .x / on I and f ; g 2 L loc. I /: If g is improper integrable on I , then f is improper integrable on I . If f is not improper integrable on I , then g is not improper integrable on I . Proof. For the case I D [a ; b/; if 0 function of d and is bounded above by a f Zb g on I and g is improper integrable on I ; then g .x / d x : So Zb a Zd a f .x / d x is an increasing f .x / d x D lim d !b Zd a f .x / d x exists in R by the monotone function theorem. The other cases I D .a ; b] and .a ; b/ are similar. g .x / is a x !a C f . x / Zb Zb g .x / positive number L ; then either (both f .x / d x and g .x / d x converge) or (both diverge). If lim D 0; x !a C f . x / a Zb Za Zb b g .x / then f .x / d x converges implies g .x / d x converges. If lim D 1; then f .x / d x diverges implies x !a C f . x / a a Zb a g .x / g .x / dx diverges. In the case [a ; b/; we take lim and the results are similar. x !b f .x / a Theorem (Limit Comparison Test). Suppose f .x /; g .x / > 0 on .a ; b] and f ; g 2 L loc..a ; b]/: If lim Proof. If lim g .x / L is a positive number L ; then for " D > 0; there is > 0 such that for every x 2 .a ; a C /; we x !a C f . x / 2 Z Z aC Z L g .x / 3L L aC 3 L aC have DL "< < LC"D : Then f .x / d x g .x / d x f .x / d x : So either 2 f .x / 2 2a 2a a Zb a (both f .x / d x and Zb a g .x / d x converge) or (both diverge). If lim x !a C g .x / g .x / D 0; then there exists 0 > 0 such that for every x 2 .a ; a C 0 /; we have 0 < < 1; which f .x / f .x / Z aC 0 a implies 0 < g .x / < f .x /: Then 0 g .x / d x Z aC 0 a f .x / d x : So Zb a f .x / d x < 1 ) Zb a g .x / d x < 1: 54 If lim x !a C g .x / g .x / D 1; then there is 00 > 0 such that for every x 2 .a ; a C 00/; we have > 1; which implies f .x / f .x / Z a C 00 a g .x / > f .x /: Hence g .x / d x Z a C 00 a f .x / d x : So Zb a f .x / d x D 1 ) Zb a g .x / d x D 1: Theorem (Absolute Convergence Test). If f 2 L loc. I / and j f j is improper integrable on I , then f is improper integrable on I . Proof. We have j f j f j f j on I ; which implies 0 f C j f j 2j f j on I : By the comparison test, f C j f j is improper integrable on I : Therefore, f D . f C j f j/ j f j is improper integrable on I : Examples. (4) Is ln x improper integrable on .0; 1]? Observe first that the function is locally integrable on .0; 1] 1 C x2 ln x j ln x j on .0; 1]: Since j ln x j D ln x is improper integrable on .0; 1] (cf. by continuity. Also, 1 C x2 ln x ln x example (1)), so is improper integrable on .0; 1] by the comparison test. Then is improper 2 1Cx 1 C x2 integrable on .0; 1] by the absolute convergence test. Z C1 1 1 dx 1 1 p converge? Observe that on [2; C1/; 0 < p : Both and p 2 are (5) Does 2 2 x x x 1 x 1 x 1 2 Z C1 1 continuous, hence locally integrable, on [2; C1/: Now d x D 1 by p-test. By the comparison test, x 2 Z C1 dx p diverges. x2 1 2 Z C1 sin x sin x (6) Does d x converge? Since is continuous on [1; C1/; it is locally integrable there. Integrating x x 1 by parts, Zc Zc sin x cos c cos x dx D C cos 1 dx: x c x2 1 1 Since j cos cj Z C1 cos x 1; lim cos c cos x D 0: Since c !C1 c x2 1 on [1; C1/ and x2 Z 1 C1 1 x2 d x < 1 by p-test, we have Z x2 1 converges. d x converges by the comparison test and the absolute convergence test. Therefore, 1 C1 sin x x dx Z1 (7) For the improper integral 1 1 0 dx ; the integrand is continuous (hence locally integrable) on [0; 1/: Observe that 1 x3 1 D and the second factor on the right has a positive limit as x ! 1 : More precisely, 3 1x 1 x 1 C x C x2 Z1 Zd 1=.1 x / dx dx 2 since lim D lim .1 C x C x / D 3 and D lim D lim ln.1 d / D 1; 3/ x !1 1=.1 x !1 d !1 d !1 x x x 01 01 Z1 dx by the limit comparison test, diverges. 1 x3 0 (8) For the improper integral ; the integrand is continuous (hence locally integrable) on .0; 5]: Observe 7x C 2x 4 1 1 1 that p Dp p and the second factor on the right has a positive limit as x ! 0 C : More 3 3 4 x 3 7 C 2x 3 7x C 2x p Z5 p 1= 3 x dx 1 3 precisely, since lim p D 7 and p < 1 by p -test as p D < 1; by the limit comparison 3 3 x !0 C 1 = 7 x C 2 x 4 3 x 0 Z5 dx test, p converges. 3 7x C 2x 4 0 0 Z5 p 3 dx 55 Principal Value – Symmetric Integration about Singularities. Definition. Let f 2 L loc.R/; then the principal value of Examples. (1) Because of continuity, f .x / D D lim .2 tan 1 c/ D c !C1 Z 1 1 f .x / d x is P:V: Z 1 2 L loc .R/: Now P:V: 1 C x2 1 Z exists and the improper integral Z0 1 11 dx 1 1 C x2 f .x / d x D lim Zc c c !C1 f .x / d x . 1 1 C x2 Zc D lim c !C1 dx c Z 1 1 1 1 C x2 d x D lim c! 1 1 C x2 Zd 1 d x C lim c d !C1 0 1 d x D lim . tan c! 1 1 C x2 Z 1 c/ C lim .tan d !C1 1 d/ D : (2) Because of continuity, f .x / D x 2 L loc.R/: Now P:V: Z 1 1 Zc x d x D lim Zd 1 1 Z Z0 c !C1 x d x D lim 0 D 0 exists, howc c !C1 ever the improper integral exist. Theorem. If the improper integral Z x d x D lim c! 1 x d x C lim c d !C1 0 c2 d2 x d x D lim . / C lim . / does not c! 1 d !C1 2 2 1 1 1 1 f .x / d x exists, then P:V: Z 1 1 f .x / d x exists and equals the improper integral f .x / dx. (Note the converse is false by example (2).) Z 1 1 Proof. If the improper integral exist. So P:V: Z f .x / d x exists, then it implies that both lim d! Z0 d 1 f .x / d x and lim f .x / d x f .x / d x : Zc c !C1 0 f .x / d x 1 1 f .x / d x D lim d! Zc c Z0 d c !C1 f .x / d x D lim f .x / d x C lim Z0 Z c !C1 f .x / d x C f .x / d x D Zc Z 0 D lim 1 c 1 1 1 c !C1 0 Definition. For a ; b 2 R or 1; let f : [a ; c/ P:V: Zb a .c; b] ! R be locally integrable on [a ; c/ and on .c; b]: Define Zc" a f .x / d x D limC "!0 f .x / d x C Zb c C" f .x / d x : Z1 Example. Consider the improper sense and the principal value sense of 1 1 dx: x Because of continuity, D lim . c !0C 1 x 2 L loc[ 1; 0/ and 1 2 L loc.0; 1]: In the improper sense, x ln c/ does not exist. So the improper integral on [ 1; 0/ P:V: Z1 1 1 d x D limC c !0 x 0 .0; 1] does not exist. However, Z1 Z1 c 1 dx x 1 d x D limC "!0 x Z1 Z 1 " 1 1 1 dx C dx x x " Z D lim .ln j "!0C "j ln j"j/ D 0: 1 d x D ln j 1j ln j1j D 0; which attempts to use part (2) of the fundamental 1x theorem of calculus, but here f .x / D ln jx j is not differentiable on [ 1; 1] failing the required condition. Remarks. (1) It is incorrect to say (2) There is a similar theorem for this type of principal value integrals as the theorem above. 56 Taylor Series of Common Functions Question. How do calculators compute sin x ; cos x ; e x ; x y ; ln x ; : : :? Recall Taylor’s theorem with the Lagrange form of the remainder is the following. Taylor’s Theorem. Let f : .a ; b/ ! R be n times differentiable on .a ; b/. For every x, c 2 .a ; b/, there is x 0 between x and c such that f . x / D f .c / C f 0 .c/ .x 1! c/ C f 00.c/ .x 2! c/2 C : : : C f .n 1/.c/ .x .n 1/! c /n 1 C f .n / . x 0 / .x n! c/n : (This is called the n-th Taylor expansion of f about c: The term Rn .x / D form of the remainder.) There are two other forms of the remainder. f .n / . x 0 / .x n! c/n is called the Lagrange Theorem (Taylor’s Formula with Integral Remainder). Let f be n times differentiable on .a ; b/: For every x ; c 2 .a ; b/; if f .n/ is integrable on the closed interval with endpoints x and c; then f 0 .c/ f .x / D f .c/ C . x c/ C 1! d dt f .n 1/ .c/ C .x c/n .n 1/! 1 C Rn .x /; where Rn .x / D 1 Zx .n 1 /! . x t /n 1 f .n/ .t / dt : c Proof. Note .x t / D 1: Applying integration by parts n Zx c 1 times, we get f .x / f .c/ D D f 0 .t / 1 dt D x Zx Zx c c f 0 .t / .x t / dt 0 f 0 .t /.x t/ C c .x .x t / f 00.t / dt t /2 x D f 0 .c/.x D c/ C c/ C f 0 .c/.x c 2! .n 1/ .c/ f C .x c/n .n 1/! f 00.t / C 1 2! Zx c 1 .x .n t /2 f 000 .t / dt D 1 1 /! Zb a Zx c C .x t /n 1 f .n/ .t / dt : Mean Value Theorem for Integrals. If f is continuous on [a ; b], then x 1 2 [a ; b], (i.e. the mean value or average of f on [a ; b] is integrable and g 0 on [a ; b]; then Zb a f .x / d x D f .x 1 /.b a / for some 1 b Z a Zb a b f .x / d x D f .x 0 /). More generally, if g is a f .x /g .x / d x D f .x 1 / g .x / d x for some x 1 2 [a ; b]. Proof. The first statement is the special case of the second statement when g .x / D 1: So we only need to prove the second statement. Let M and m be the maximum and the minimum of f on [a ; b]; respectively. Since m f .x / M on [a ; b]; we have m Zb Zb a g .x / d x Zb a f .x /g .x / d x ,Z Zb a M g .x / d x : If Zb a g .x / d x D 0; then the last sentence Zb a implies Zb a a f .x /g .x / d x D 0 and so we may take x 1 to be any element of [a ; b]: If Zb a g .x / d x > 0; then dividing by g .x / dx ; we see that f .x /g .x / d x b a g .x / d x is between m and M : By the intermediate value theorem, this expression equals f .x 1 / for some x 1 2 [a ; b]; which gives the second statement. 57 Theorem (Taylor’s Formula with Cauchy Form Remainder). Let f be n times differentiable on .a ; b/: For x ; c 2 .a ; b/; if f .n/ is continuous on the closed interval with endpoints x and c; then there is x 1 between x and c such that f .x / D f .c/ C f 0 .c/ .x 1! c/ C C f .n 1/ .c/ .x .n 1/! 1/n 1 C Rn .x /; where Rn .x / D .x c/.x x 1 /n 1 f .n/ .x 1 / : .n 1/! Proof. This follows by applying the mean value theorem for integrals to the integral remainder of Taylor’s formula above. xk x kC1 k ! converges. This follows from the ratio test because lim k !1 .k C 1/! x k k! k D0 k jx j x D lim D 0 < 1: By term test, we have lim D 0 for every x 2 R: k !1 k C 1 k !1 k ! Remarks. For every x 2 R; the series Examples. (1) For f .x / D sin x , f .n/ .x / D 1 X . 1/k cos x if n D 2k C 1 . So j f .n/ .x /j 1 for every x 2 R: . 1/k sin x if n D 2k .2k C1/.0/ D . 1/k and f .2k / .0/ D 0 for every k 2 N: By Taylor’s theorem, Taking c D 0; we have f n1 X . 1/k x 2k C1 1 sin x D C R2n .x / for every x 2 R. Now j R2n .x /j jx j2n ! 0 as n ! 1 by the re.2k C 1/! .2n /! k D0 marks above. Therefore, sin x D x x3 x5 C 3! 5! x7 C 7! jx j18 D 1 X . 1/k x 2kC1 .2 k C 1 /! k D0 <6 10 13 for 1 < x < C1: C Remarks. For 0 x ; j R18 .x /j 2 compute sin x to 10 decimal places. (2) For f .x / D cos x , f .n/ .x / D . =2/18 18! 18! : So x x3 C 3! x 17 can be used to 17! . 1/k sin x if n D 2k 1 . So j f .n/ .x /j 1 for every x 2 R: . 1/k cos x if n D 2k .2k 1/.0/ D 0 and f .2k / .0/ D . 1/k for every k 2 N: By Taylor’s theorem, Taking c D 0; we have f n1 X . 1/k x 2k 1 sin x D C R2n 1 .x / for every x 2 R. Now j R2n 1 .x /j jx j2n 1 ! 0 as n ! 1 .2 k /! .2 n 1 /! k D0 by the remarks above. Therefore, cos x D 1 x2 x4 C 2! 4! x6 C 6! D 1 X . 1 /k x 2 k .2k /! k D0 for 1 < x < C1: (3) For f .x / D e x , f .n/ .x / D e x : Taking c D 0; we have f .n/ .0/ D 1 for every n 2 N: By Taylor’s theorem, n1 X xk e x0 n max.e0 ; e x / n ex D C Rn .x / for every x 2 R: Now j Rn .x /j D jx j jx j ! 0 as n ! 1 by the k! n! n! k D0 remarks above. So ex D 1 C x C x2 x3 C C 2! 3! D 1 X xk k! k D0 for 1 < x < C1: Remarks. The Taylor series for sin x ; cos x and e x also converge if x is a complex number! In fact, this is how the sine, cosine, and exponential functions are defined for complex numbers. For every x 2 R; we have eix D 1 X k D0 .ix /k k! D 1 X . 1 /n x 2 n .2 n /! nD0 Ci 1 X . 1/n x 2nC1 .2n C 1/! nD0 D cos x C i sin x : 58 In particular, we have ei D cos C i sin D 1 so that ei C 1 D 0: This is known as Euler’s formula. It is the most beautiful formula in mathematics because it connects the five most important constants 1; 0; ; i; e of mathematics in one single equation! (4) (Binomial Theorem) For a 2 R; if 1 < x < 1; then 1/.a 3! 2/ x3 C D 1C .1 C x /a D 1 C ax C a .a 1/ 2 a .a xC 2! 1 X k D1 a .a 1/ .a k! k C 1/ xk: To see this, let f .x / D .1 C x /a ; then f .n/ .x / D a .a integral formula, Rn .x / D 1 Zx 1/ 1/ .a n C 1/.1 C x /a Zx 0 n : By Taylor’s formula with n1 .n 1/! .x t /n 1 f .n/ .t / dt D a .a 0 .a n C 1 / .n 1/! xt 1Ct .1 C t /a 1 dt : xt 1x has derivative g 0.t / D < 0: On [0; x ]; g .t / g .0/ D x : Let 1Ct .1 C t /2 Zx n1 ja .a 1/ .a n C 1/kx j bnC1 ja n jjx j kD .1 C t /a 1 dt ; then j Rn .x /j : Since lim D lim D n!1 bn n!1 .n 1/! n 0 For x 2 [0; 1/; the function g .t / D | {z } jx j < 1; by ratio test, 1 X nD1 call this bn bn converges. By term test, lim bn D 0: Then lim Rn .x / D 0 by the sandwich n!1 n!1 theorem. So the binomial formula is true for x 2 [0; 1/: The case x 2 . 1; 0] is similar. Here are a few more common Taylor series. (Note the series only equal the functions on a small interval.) They 1 are obtained from the cases a D 1; a D 1 (with x replaced by x 2 ) and a D (with x replaced by x 2 ) of the 2 binomial theorem by term-by-term integration. ln.1 C x / D x x2 x3 C 2 3 x3 x5 C 3 5 x4 C 4 x7 C 7 DxC D 1 X k D1 . 1 /k 1 x k k for 1<x 1 tan 1 1 xDx D 1 X k D0 . 1/k x 2kC1 2k C 1 for 1 x 1 sin xDxC 1 x3 1 3 x5 C C 23 2 45 1 X 1 3 5 .2k 1/ x 2kC1 2 4 6 .2 k / 2 k C 1 k D1 for 1 x 1 Remarks. Unfortunately, the Taylor series of a function does not always equal to the function away from the center. 1= x 2 if x 6D 0 can be shown to be infinitely differentiable and f .n/ .0/ D 0 For example, the function f .x / D e 0 if x D 0 for every n 2 N: So the Taylor series of f .x / about c D 0 is 1 X k D0 0x k D 0; i.e. the Taylor series is the zero function. Therefore, the Taylor series of f .x / equals f .x / only at the center c D 0: 59 Practice Exercises The starred problems are difficult and involve more work or deeper thinking. For exercises 1 to 7, negate each of the following expressions or statements. 1. (x > 0 and x < 1) or x D 2. x > 0 and (x < 1 or x D 1 1) 3. For every triangle ABC; \ A C \ B C \C D 180 : 4. There exists a man who does not have any wife. 5. For each x , there is a y such that x C y D 0: 6. 9 8 9 such that j j <: 7. If x and y are positive, then x C y > 0. 8. Give the contrapositive of the following statements. (Since the contrapositives are equivalent to the statements, they say the same thing.) (a) If AB D AC in 4 ABC; then \ B D \C in 4 ABC: (b) If a function is differentiable, then it is continuous. (c) If lim f .x / D a and lim g .x / D b; then lim . f .x / C g .x // D a C b: x !0 2 x !0 (d) If x C bx C c D 0; then x D 9. Compute the following sets. (a) fx ; y ; z g (b) f1; 2g bC p x !0 b2 4c 2 or x D b p b2 2 4c : fw; z g n fu ; v; wg: (Here u ; v; w; x ; y ; z are distinct objects.) f5g: 2 f3; 4g (c) Z\ [0; 10] \ fn C 1 : n 2 Ng: (d) fn 2 N : 5 < n < 9g n f2m : m 2 Ng: (e) .[0; 2] n [1; 3]/ .[1; 3] n [0; 2]/: [2; 3]: Plot the graphs of A BDB A and B B on the plane. A; what can be said about A and B ? 10. (i) Let A D [0; 1] and B D [0; 1] (ii) If A; B are sets that are not the empty set and A 11. (a) If B C; then prove that A B A C: (b) For sets X ; Y; Z ; prove that . X n Y / n Z D . X n Z / n Y: 12. (i) For all sets A; B ; C; is it always true that . A (ii) For all sets A; B ; C; is it always true that if A B/ \ C D A BDA . B \ C /? C; then B D C ? (iii) For all sets A; B ; C; is it always true that A n . B 13. (i) Show that if A B and C D ; then A C B 60 C / D . A n B / \ . A n C /? D: (ii) Is it always true that if A (iii) For a < b; let B and C D ; then A C B D? .a ; b / Q Does D fx : x 2 Q and a < x < bg and [a ; b/ Q D fx : x 2 Q and a x < b g: 11 [ ; 2/ Q D . ; 2/ Q ? n n nD1 nD1 0 if x > 0 and g .x / D 1 1 if x 0 g and g f : 2x : For f and g ; determine if each is 11 14. Define functions f ; g : R ! R by f .x / D injective or surjective. Compute f 15. (i) Let f : A ! B be a function. Show that if there is a function g : B ! A such that g then f is a bijection. (Comment: Such a g is f 1 :) (ii) Show that if f : A ! B and h : B ! C are bijections, then h f D I A and f g D I B ; f : A ! C is a bijection. 16. Let A; B be subsets of R and f : A ! B be a function. If for every b 2 B ; the horizontal line y D b intersects the graph of f at most once, then show that f is injective. If ‘at most once’ is replaced by ‘at least once’, what can be said about f ? For each of the sets in exercises 17 to 26, determine if it is countable or uncountable: 17. intervals .a ; b/ and [a ; b], where we assume a < bI 18. Q .R n Q/I 1 1 C m : n; m 2 Z gI 2n 3 p 19. A D f 20. B D fx C 2 y : x ; y 2 NgI 21. the set C of all lines in R2 passing through the origin; 22. D D fx 2 R : x 5 C x C 2 2 QgI 23. the set E of all circles in R2 with centers at rational coordinate points and positive rational radius. 24. the set F D fa : x 4 C ax 5 D 0 has a rational rootgI 25. the set G D fa 3 C b3 : a 2 X ; b 2 Y g; where X is a nonempty countable subet of R and Y is an uncountable subet of R; 26. the set H D . X n Y / .Y n X /; where X is a countable set and Y is an uncountable set. (Remark: The set . X n Y / .Y n X / is called the symmetric difference of X and Y and is usually denoted by X 4Y: This concept will appear in other algebra and analysis courses later.) 27. Show that the set F of all finite subsets of N is countable. 28. If S is a countable subset of R2 ; show that for any two points x ; y 2 R2 n S ; there is a parallelogram in R2 n S having x ; y as opposite vertices. Here parallelogram means only the 4 edges (including the 4 vertices, but not including any interior point). 61 1 *29. From K 0 D [0; 1]; remove the middle thirds to get K 1 D [0; 1] n . 3 ; 2 / D [0; 1 ] [ 2 ; 1]: Then remove the middle 3 3 3 1 21 thirds of the 2 subintervals of K 1 to get K 2 D [0; 9 ] [ 9 ; 3 ] [ 2 ; 7 ] [ 8 ; 1]: Inductively, remove the middle 39 9 thirds of the 2n subintervals of K n to get K nC1 : The set K D K 0 \ K 1 \ K 2 \ K 3 \ is called the Cantor set. Prove that K is uncountable. (Hint: Consider base 3 representations, i.e. representations of the form .:a1 a2 a3 : : :/3 D a1 a2 a3 C 2C 3C 3 3 3 ; where each ai D 0; 1 or 2. What do the base 3 representations of numbers in K n have in common? Note some numbers have 2 representations, e.g. 1 D .:1000 : : :/3 and .:0222 : : :/3 :) 3 30. For each of the following series, determine if it converges or diverges. (a) 6 cos sin (c) 6 ln 1 C (e) 6 1 k D1 1 k D1 1 k 1 k (term test) (find sum) (b) 6 p (d) 6 (f) 6 1 k D1 1 k .k C 1/.k C 2/ k 111 C k k D1 2 1 ln k k D2 k 1 cos 2k k D1 k2 (g) 6 1 kC2 2 k k D1 k C 1 3 1 k2 (i) 6 ke k D1 1 Arctan k (k) 6 2 k D1 k C 1 1 k C1 (m) 6 tank k D1 k 1 1 (o) 6 k 2 sin p (depends on p) k k D1 a2 a3 1 cos k p k D1 k 1 k (j) 6 k D1 .k C 1/! 11 (l) 6 (compare with a p-series) 1 k D1 k 1 C k 1 1 (n) 6 1 cos (compare with a p-series) k D1 k p p 1 (p) 6 . 1/kC1 . k C 1 k/ k D1 (h) 6 31. Let a1 ::: 0: Prove that 6 ak converges if and only if 6 2k a2k converges. (This is called 1 k D1 Cauchy’s condensation test.) Use this test to determine if 6 32. Show that 6 k 2 3 4 D C 2C 3C 2k 1 22 2 index from 1 to 1:) k D2 1 k D3 1 converges. k ln k ln.ln k / 1 k D1 1 converges and find the sum. (Hint: Compare the same series with (Probabilistic interpretation: the sum is the expected number of births to get babies of both sex. The probability 2 1 of the k -th birth finally resulted in babies of both sex is k D k 1 :) 2 2 *33. Let pn be the n -th prime number, i.e. p1 D 2; p2 D 3; p3 D 5; p4 D 7; 11 : Show that 6 diverges. (Hint: k D1 pk 11 1 Suppose it converges to s : Then the partial sum sn has limit s ; as n ! 1: So for some n ; s sn D 6 < : Let pk 2 k DnC1 N 1 11j 1 Q D p1 p2 pn : Show, by considering the prime factorization of 1 C m Q ; that 6 6 6 m D1 1 C m Q j D1 k DnC1 pk for every positive integer N ; which will lead to a contradiction.) 62 For exercises 34 to 36, use definitions, Archimedean principle, density of rationals, density of irrationals, etc. to support your reasonings. 34. For each of the following sets, if it is bounded above, give an upper bound and find its supremum with proof. If it is bounded below, give a lower bound and find its infimum with proof. p p 1 (b) B D . 1; ] 4 :n2N (a) A D f m C n : m ; n 2 Ng n (c) C D 1 1 C m : m; n 2 N n 2 (d) D D Q \ .0; p 2] 35. Let A and B be nonempty subsets of R; which are bounded above. Let S D f x y : x 2 A; y 2 B g and T D fx y : x 2 A ; y 2 B g: Must S or T be bounded above? Give a proof if you think the answer is ‘yes’ or give a counterexample if you think the answer is ‘no’. 36. Let A and B be nonempty subsets of R; which are bounded above. Let C D f x C y : x 2 A ; y 2 B g: Show that C is bounded above and sup C D sup A C sup B : (Hint: If sup C < sup A C sup B ; then consider " D .sup A C sup B sup C /=2: Apply supremum property to get a contradiction.) 4n C 5 37. Let wn D ; then fwn g should converge to 0. For a given " > 0; show there is a positive integer K such n3 that if n K ; then jwn 0j < ": If " D 0:1; give one such positive integer K : [x ] C [2x ] C C [nx ] 38. Let x be positive and an D : Show that fan g converges to 2 n (Here [ y ] is the greatest integer less than or equal to y :) x 2 by the squeeze limit theorem. 39. Show that if x is a real number, then there is a sequence of rational numbers converging to x : 40. If lim an D A; then show that lim jan j D j Aj: (Hint: Show jx j n!1 n!1 converse true? 41. If fan g converges to A; then 42. Let x 1 D 4 and x nC1 D sequence fx n g converges. n j yj jx y j for x ; y 2 R first.) Is the an C anC1 2 should converge to AC A D A: Prove this by checking the definition. 2 4.1 C x n / for n D 1; 2; 3; : : : : Plot the first 3 terms on the real line. Then prove the 4 C xn 1 no is increasing and bounded above. n 43. Show that the sequence 1C 44. Let fx n g be a bounded sequence in R; Mn D supfx n ; x nC1 ; x nC2 ; : : :g and m n D inffx n ; x nC1 ; x nC2 ; : : :g for n 2 N: (a) Prove that both sequence f Mn g and fm n g converge. (The limit of Mn is called the limit superior of x n and is denoted by limsup x n ; while the limit of m n is called the limit inferior of x n and is denoted by liminf x n :) n!1 n!1 (b) Prove that lim x n D x if and only if lim Mn D x D lim m n (i.e. limsup x n D x D liminf x n ). n!1 n!1 n!1 n!1 n!1 63 45. Let x 1 D 1; x 2 D 2 and x nC1 D the sequence fx n g converges. 46. If 6 jx k xn C xn 2 1 for n D 2; 3; 4; : : : : Plot the first 4 terms on the real line. Then prove Cauchy sequence. 47. If an 0 for all n 2 N and fan g is a Cauchy sequence, then show that f an g is also a Cauchy sequence by checking the definition of Cauchy sequence. 48. Let 0 < k < 1: If jx nC1 xn j k jx n x n 1 j for n D 2; 3; 4; : : : ; then prove that fx n g is a Cauchy sequence. p 1 k D2 x k 1 j converges, then show the sequence fx n g is a Cauchy sequence by checking the definition of a1 C a2 C C an 49. Prove that if fan g converges to A; then f n g converges to A; where n D : Show that the n converse is false. b1 C b2 C C bn (Hint: Let bn D an A and n D ; then the first part of the problem becomes showing f n g n converges to 0. Here use fbn g converges to 0 and so jb K 0 j; jb K 0C1 j; : : : is small when K 0 is large. For n K 0; write b1 C b2 C C b K0 1 b K0 C C bn C :/ nD n n 50. If fx n g is bounded and all its convergent subsequences have the same limit x ; then prove that lim x n D x : n!1 51. Let S D f2 n : n 2 Ng and f : N ! S is injective. Show that lim f .n / D 0: n!1 n!1 *52. Let fan g be a sequence satisfying lim .anC1 *53. Let fx n g be a sequence satisfying lim .x n n!1 an / D 0: Prove that lim an D 0: n!1 2 x n 2 / D 0: Prove that lim 1 n!1 xn xn n 1 D 0: *54. Let fx n g be a sequence and let y1 D 0 and yn D x n also converges. C 2 x n for n D 2; 3; 4; : : : : If f yn g converges, prove that fx n g 55. For a sequence fan g of nonzero numbers, we say the infinite product has value L ) if lim a1 a2 k !1 1 Y nD1 an converges to a nonzero number L (or ak D L : We say it diverges if the limit is 0 or does not exist. Determine if each of the following infinite products converges or diverges. Find the values of the infinite products that converge. (a) 1 Y nD2 1 2 n .n C 1/ (b) 1 Y nD2 1 1 n2 n (c) 1 3C1 n nD2 1 Y n3 (d) 1 Y nD0 .1 C z 2 / for jz j < 1: 0; we have Remarks: In Apostol’s book, the following theorem is proved: In the case every an 1 Y nD1 .1 C an /; 1 Y n D1 .1 an /; 6 an all converge or all diverge. Try to do the above exercises without using this theorem. 1 n D1 56. Prove that every bounded infinite subset of R has an accumulation point. (This is also often called the BolzanoWeierstrass theorem.) 64 57. Let f : .0; C1/ ! R be defined by f .x / D 58. Define f : R ! R by f .x / D x ! x0 : Show that lim f .x / D by checking the definition. x !1 xC1 2 n x 1 8x if x is rational, 2x 2 C 8 if x is irrational. For which x 0 ; does lim f .x / exist? (Hint: Sequential limit theorem.) 59. If f : R ! R is continuous and f .x / D 0 for every rational number x ; show that f .x / D 0 for all x : 60. (a) Find all functions f : Q ! R such that f .x C y / D f .x / C f . y / for all x ; y 2 Q: (b) Find all strictly increasing functions f : R ! R such that f .x C y / D f .x / C f . y / for all x ; y 2 R: 61. For a function f : R ! R; we say f has a local (or relative) maximum at x 0 if there exists an open interval .a ; b/ containing x 0 such that f .x / f .x 0 / for every x 2 .a ; b/: Similarly, we say f has a local (or relative) minimum at x 1 if there exists an open interval .c; d / such that f .x / f .x 1 / for every x 2 .c; d /: If f : R ! R is continuous and has a local maximum or a local minimum at every real number, show that f is a constant function. 62. If f .x / D x 3 ; then f . f .x // D x 9 : Is there a continuous function g : [ 1; 1] ! [ 1; 1] such that g .g .x // D for all x 2 [ 1; 1]? (Hint: If such a function g exists, then it is injective.) x9 63. A fixed point of a function f is a number w such that f .w/ D w: Show that if f : [0; 1] ! [0; 1] is continuous, then f has at least one fixed point. (Hint: Consider g .x / D f .x / x :) 64. Let f : [0; 1] ! [0; 1] be an increasing function (perhaps discontinuous). Suppose 0 < f .0/ and f .1/ < 1; show that f has at least one fixed point. (Hint: Sketch the graph of f and consider the set ft 2 [0; 1] : t < f .t /g. Does it have a supremum?) 65. Let f : R ! R be a continuous function such that j f .x / f . y /j jx y j for all x ; y 2 R: Show that f is bijective. (Hint: Easy to show f is injective. To show f is surjective, let w 2 R and M D jw f .0/j, show that w and f .0/ are between f . M / and f . M /:) *66. Let f : [0; 1] ! .0; C1/ be continuous and M D supf f .x / : x 2 [0; 1]g: Show that Z1 n!1 lim f .x / d x n 1 n DM 0 if the limit exists. (Hint: M D f .x 0 /: For every k 2 N; use the sign preserving property to show that 1 f .x 0 / k f .x / M on an interval containing x 0 . Squeeze the integral.) x2 x if x 6D 0 and g .x / D j cos x j: if x D 0 I2 I3 and \ [an ; bn ] D fcg: nD1 67. Find the derivatives of the functions f .x / D 68. Let f : R ! R be differentiable at c and In D [an ; bn ] be such that I1 f .bn / Prove that if an < bn for all n 2 N; then f 0 .c/ D lim n!1 bn f .a n / : an 1 69. Let f .x / D jx j3 for every x 2 R: Show that f 2 C 2 .R/: However, f 000 .0/ does not exist. 70. Let f : R ! R satisfies j f .a / f .b/j ja bj2 for every a ; b 2 R: Show that f is a constant function. If the exponent 2 in the inequality is replaced by a number greater than 1, must f be a constant function? If 2 is replaced by 1, must f be a constant function? 65 71. Let n be a positive integer and f .x / D .x 2 1/n : Show that the n -th derivative of f has n distinct roots. f .x / for every x > 0; then show that f .x / 0 for 72. Let f : [0; 1/ ! R be continuous and f .0/ D 0: If f 0 .x / every x 2 [0; 1/: (Hint: What if f 0 .x / D f .x /?) 73. If f : .0; C1/ ! R is differentiable and j f 0 .x /j x !0 C 1 n converges. Also, show lim f .x / exists. (Hint: Check fx n g is a Cauchy sequence. For the second part, consider 2 for all x > 0; then show that the sequence x n D f the sequential limit theorem and the remarks following it.) 74. For 0 < x < 2 ; prove that j ln.cos x /j < x tan x : 75. Let f : [0; 1/ ! R be continuous and f .0/ D 0: If j f 0 .x /j j f .x /j for every x > 0; show that f .x / D 0 for every x 2 [0; 1/: (Hint: Let j f j has maximum value M on [0; 1 ]: Apply the mean value theorem to f on [0; 1 ]:) 2 2 76. Let f : R ! R be differentiable. If f 0 is differentiable at x 0 ; show that f .x 0 C h / C f .x 0 h!0 h2 lim 77. Prove that if 0 , then 2 2 4 h/ 2 f .x 0 / D f 00 .x 0 /: 2 2 2 24 (Hint: Apply Taylor’s theorem to the four times differentiable function cos :) 78. Let f : .0; C1/ ! Rbe twice differentiable, M0 D supfj f .x /j : x > 0g < 1; M1 D supfj f 0 .x /j : x > 0g < 1 2 and M2 D supfj f 00.x /j : x > 0g < 1: Show that M1 4 M0 M2 : (Hint: Let h > 0: Apply Taylor’s theorem to 0 .c/:) f .x / with x D c C 2h ; then solve for f 79. (a) If f : .a ; b/ ! Ris differentiable and j f 0 .x /j 2 for all x 2 .a ; b/; then show that f is uniformly continuous. 1 is not uniformly continuous. x 1 cos 1 C : (b) Show that f : .0; C1/ ! R defined by f .x / D sin 80. (a) Prove that if the union of a collection of open intervals contains [a ; b]; then there are finitely many of these intervals, whose union also contains [a ; b]: (Hint: Suppose this is false. Let m 1 D .a C b/=2: Then one of [a ; m 1 ] or [m 1 ; b] is not contained in the union of finitely many of these open intervals. Proceed as in the proof of the Bolzano Weierstrass theorem.) (b) Give a proof of the uniform continuity theorem using part (a). 81. If f is continuous on [a ; b]; f .x / x 2 [a ; b ]: 82. Let f : [a ; d ] be a bounded function and a b c d: 0 for all x 2 [a ; b] and Zb a f .x / d x D 0; then prove that f .x / D 0 for all (i) Use the integral criterion to show that if f is integrable on [a ; b] and [b; c]; then f is integrable on [a ; c]: (ii) Use the integral criterion to show that if f is integrable on [a ; d ]; then f is integrable on [b; c]: 83. Let f : [a ; b] ! R be bounded and fx 2 [a ; b] : f is discontinuous at x g D fx 1 ; x 2 ; : : : ; x n g; where x 1 < x 2 < < x n : Use the integral criterion to show that f is integrable on [a ; b]: (Hint: Divide [a ; b] into subintervals where f is continuous except at one endpoint and note part (i) of problem above.) 66 84. (i) Let f ; g : [a ; b] ! R be bounded and P is a partition of [a ; b]: Show that L . f ; P / C L .g ; P / L . f C g; P/ U . f C g; P/ U . f ; P / C U .g ; P /: (ii) If f and g are integrable on [a ; b]; show that Zb a . f .x / C g .x // d x Zb D a f .x / d x C Zb a g .x / d x : (Hint: First show RHS Z " < L . f ; P / C L .g ; P / L . f C g ; P / for some P using supremum property.) Z 85. Determine if each of the following improper integrals exists or not. (a) 0 1 dx p 1 ex (b) 0 sin x d x Z1 Z1 (c) 0 dx x 2 C 5x dx x . x 1/ Z (d) 1 p 3 dx x dx Z1 Z (e) 0 (f) 0 1 cos x 1 C x2 Z2 86. Find the principal value of each of the following integrals if it exists. (a) P.V. C1 x 2 dx 1 ex (b) P.V. 0 dx x2 1 t x 1e t 87. Prove that for 0 < x < 1; the improper integral 0.x / D Z 0 1 dt exists. This is called the gamma function. (Hint: Consider the cases x 2 .0; 1/ and x 2 [1; 1/ separately.) Past Exam Problems 88. Define the following terms: (a) S is a countably infinite set (b) S is a countable set (c) a series 6 ak converges to a number S (d) (e) (f) (g) (h) (i) (j) a nonempty subset S of R is bounded above the supremum of a subset S of R that is bounded above a sequence fx n g converges to a number x a sequence fx n g is a Cauchy sequence x is an accumulation point of a set S f : S ! R has a limit L at x 0 f : S ! R is continuous at x 0 2 S (the "- definition) k D1 1 89. For each of the following sets S , determine if it is countable or uncountable. Be sure to give reasons to support your answer. (a) S is the set of all intersection points .x ; y / of the line y D x with the graphs of all equations y D x 3 C x C m ; where m 2 Z : (b) S is the set of all intersection points .x ; y / of the graph of y D x 3 C x C 1 with all lines y D mx ; where m2Z : 1 (c) S is the set of all intersection points .x ; y / of the circle x 2 C y 2 D 1 with all hyperbolas x y D ; where m m 2 N: 67 (d) S D fa C b 2 R : ja j 2 M ; b 2 Qg; where M is a uncountable subset of the positive real numbers. j2 (e) S D fa C b 2 R : jap M ; b 2 Qg; where M is a countable subset of the positive real numbers. na C b 2 o p p (f) S D Q. 2/ D p : a ; b ; c; d 2 Q; c C d 2 6D 0 : cCd 2 (g) S D fx 2 C y 2 C z 2 : x 2 A \ B ; y 2 Q \ A; z 2 B \ Qg; where A is a nonempty countable subset of R and B is an uncountable subset of R: (h) S D fx y : x ; y 2 Ag; where A is an uncountable subset of R: (i) S D fx 2 C y 2 : x ; y 2 Ag; where A is a nonempty countable subset of R: (j) S D fx C sin y : x 2 R n A; y 2 Z; where A is a nonempty countable subset of R: g (k) S D f.x ; y /p R2 : x 2 A; y 2 R n Ag; where A is a nonempty countable subset of R: 2 (l) S D fx C y 2 p 2 Z y 2 Ag; where A is a nonempty countable subset of R: :x ; p (m) S D R n fa C b 2 c 3 : a ; b; c 2 T g; where T D fr : r 2 Qg: p p (n) S D T \ U ; where T D R n Q and U D R n f m C n : m ; n 2 N g: (Hint: Consider R n .T \ U /:) (o) S is the set of all squares on the plane that can be circumscribed by circles with rational radii and centers with rational coordinates. p (p) S is the set of all nonconstant polynomials with coefficients in G; where i D 1 and G D fa C bi : a ; b 2 Z: g 90. Determine if each of the following series converges or diverges. Be sure to give reasons to support your answer. p 1 cos k 1ek and 6 p (a) 6 2 k D1 k C 2k k D1 k 1 .2k /! 1 .cos k /.sin 2k / (b) 6 k 4 and 6 2k k D1 3 k k D1 11 1 1 1 1 1 (c) 6 cos C sin and 6 sin k k k kC1 k D1 2 k D1 1 2k C 3k 1 1 (d) 6 k and 6 cos sin k k D1 1 C 4k k D1 1 21 = k C 31 = k 1 1 (e) 6 1= k and 6 .cos k / sin C 41 = k k k D1 1 k D1 1 .k !/2 1 1 1 1 (f) 6 2 and 6 cos sin tan k D1 .k /! k D1 k k k 1 2k cos k 1 sin. 1 / k (g) 6 and 6 1/! k D2 .k k D2 ln k 1 k C cos k 1 k cos k (h) 6 and 6 4 3Ck 3k 4 k D1 k D1 1 1 .2 k /! 1 (i) 6 and 6 k cos 2 1/! k k D2 .k C 1/!.k k D1 1 .3 k /! 1 cos.1= k / (j) 6 and 6 1 k D1 k !.2k /! k D2 k 2 1 1 cos k k! (k) 6 and 6 p 1/! k D1 .2k k D1 k C1 1 2k k 2 11 1 (l) 6 and 6 p sin p k D1 k ! k D1 k k 1 1 k 2 sin.1= k / 1 (m) 6 and 6 k D1 k cos k k D1 .2k C 1/! 1 1 cos.sin.1= k // 1 k (n) 6 cos .1 C / and 6 k k D1 k D1 sin.cos.1= k // 91. Determine if each of the following set has an infimum and a supremum. If they exists, find them and give reasons to support your answers. n1 o n2 o 1 (a) S D C : m; n 2 N n :k2N m n k 68 (b) S D x C y : x ; y 2 n n (c) S D x nx (d) S D xC nx o (e) S D : x 2 Q \ [0; 1/ xC (f) S D fx 3p y 3 : x 2 Q \ [0; 1]; y 2 . 1; 0]g C n o 2 1 (g) S D C p : m; n; k 2 N m Cn k2 1 1 1 (h) S D [1 ;1 / 2k 1 2k k D1 (i) S D (j) S D np n1 o 1 :n2N no 1 1 : x 2 [0; 1] \ Q; n 2 N n 1; n 2 o : x 2 .R n Q/ \ [ ; 1/ on 1 ;1 n 2 2 x C y 2 : x ; y 2 .0; 1] \ Q o n (k) S D fx C y : x 2 [0; 1] \ Q; y 2 [0; 1] \ .R n Q/g (l) S D fx 2 R : x .x C 1/ pand x 2 R n Qg 0 k k (m) S D f n! : k ; n 2 N; n! < 2g 1 nQ n n2 nD1 (o) S D fx 2 C y 3 C z 4 : x 2 . 1; 0/ n Q; y 2 .0; 1/ \ Q; z 2 . 1; 1/g (n) S D p 10 C x : x 2 [0; 1] \ Q; n D 1; 2; 3; : : : o 1 ;2 92. For each of the following sequences fx n g; show it converges and find its limit. xn p C x n for n D 1; 2; 3; : : : : (a) x 1 D 1 and x nC1 D 2 p p (b) x 1 D 1; x 2 D 2 and x nC1 D x n C x n 1 for n D 2; 3; 4; : 2 xn (c) x 1 D 1 and x nC1 D for n D 1; 2; 3; : : : : 3 C xn nx o p nC1 (d) x 1 D 15 and x nC1 D 1 1 x n for n D 1; 2; 3; : Also, do the sequence : 16 xn anC1 (e) x n D ; for n D 1; 2; 3; : : : ; where a1 D 1; a2 D 2 and anC1 D an C an 1 for n D 2; 3; 4; : an 1 (f) x 1 D 1 and x nC1 D 1 for n D 1; 2; 3; : : : : 4xn 1 4 4 (g) x 1 D 4 and x nC1 D xn C for n D 1; 2; 3; : : : : (Hint: Sketch the graph of f .x / D x C for x 2 xn x 4 (h) x 1 D 5 and x nC1 D 3 C for n D 1; 2; 3; : xn 1 (i) x 1 D 2 and x nC1 D 2 for n D 1; 2; 3; : xn 2 x C4 (j) x 1 D 0 and x nC1 D n for n D 1; 2; 3; : : : : 5 p 1 (k) x 1 D 1 and x nC1 D x n for n D 1; 2; 3; : : : : 4 s 1 (l) x 1 D 3 and x nC1 D 1 for n D 1; 2; 3; : : : : xn C 1 (m) 0 < x 1 < 1 and 7x nC1 Dpn C 6 for n D 1; 2; 3; : : : : (Hint: x 3 7x C 6 D .x 1/.x 2/.x C 3/:) x3 (n) x 1 2 [1; 1/ and x nC1 D 3x n 2 for n D 1; 2; 3; : : : : 3 (o) x 1 D 0; x 2 2 [0; 1 ] and x nC1 D 1 .1 C x n C x n 1 / for n D 2; 3; 4; : : : : 2 3 2:) 93. If fan g is a decreasing sequence (of nonnegative real numbers) converging to 0, show that the sequence fx n g 69 converges, where x n D 6 . 1/kC1 ak D a1 n k D1 a2 C a3 r C . 1/nC1 an : an C bn 94. Let 0 < a < b and a1 D a ; b1 D b; anC1 D ; bnC1 D 2 and fbn g both converge and their limits are the same. 95. (i) If a (ii) Let x 1 D 1; x 2 D 2 and x nC1 D 1 2 xn C xn 3 3 b and 0 < t < 1, then show that a t a C .1 t /b b: 2 2 an C bn for n D 1; 2; 3 : : : : Show that fan g 2 1 for n D 2; 3; 4; : Show that the sequence fx n g converges. 96. Show that the sequence fx n g given by r x 0 D 0; converges and find its limit. x 1 D 1 and x nC1 D 12 32 xC x 4n 4n 1 for n 2 N 97. Let S be a set of real numbers such that every sequence in S has a convergent subsequence, show that S is bounded. 98. Let A be a nonempty subset of the nonnegative real numbers. If A is bounded above and B D fx 2 C y 2 : x ; y 2 Ag; show B is bounded above and sup B D 2.sup A/2 : 99. Suppose An . 1; 2/ and x n D sup An for n D 1; 2; 3; : : : ; 10: Prove that 10 sup k D1 An D max.x 1 ; x 2 ; : : : ; x 10/: 100. Use the definitions of infimum and supremum to explain the statement: Let f : R R ! [0; 1] be a function, g .x / D supf f .x ; y / : y 2 Rg and h . y / D inff f .x ; y / : x 2 Rg: Show that supfh . y / : y 2 Rg inffg .x / : x 2 Rg: 101. Show that for every x 2 R; there is a strictly increasing sequence of irrational numbers fx n g converging to x : 1 102. Prove that lim 2 n!1 n 103. Prove that lim n!1 p 2 D 0 by checking the definition of limit. n3 1 D 0 by checking the definition of limit. n2 n!1 2 nC1 104. Given lim x n D 0: Prove that lim x n C n!1 1 D 0 by checking the definition of limit. n 105. Given lim x n D n!1 n!1 1 n : Prove that lim x n D 0 by checking the definition of limit. n!1 2 n!1 106. Given lim x n D 8: Prove that lim p 3 x n D 2 by checking the definition of limit. 70 107. Given sequences fx n g and f yn g both converge to A: For n 2 N; let z n D max.x n ; yn /: Show that fz n g converges to A by checking the definition. 108. If fx n g is a sequence such that jx kC1 xk j < 1 for k D 1; 2; 3; : : : ; then show that fx n g is a Cauchy sequence. 2k 109. (a) Let S be an open interval and x 0 2 S : For a function f : S ! R; state the definition of f .x / converges to L (or has limit L ) as x tends to x 0 in S : 1 9 (b) Let f : .1; 3/ ! R be defined by f .x / D x 2 C : Prove that lim f .x / D by checking the definition. x !2 x 2 (c) Let f : .1; 4/ ! R be defined by f .x / D jx 2 9j: Prove that lim f .x / D 5 by checking the definition. x !2 110. If f ; g : .0; 1/ ! R are increasing functions, show that h .x / D max. f .x /; g .x // has countably many jumps on the interval .0; 1/: 111. Give an example of a function f : R ! R that is continuous at x 2 Z but discontinuous at every x 62 Z Be sure ; : to give reasons to support your example. 112. (a) State the intermediate value theorem. (b) Let f : [0; 2] ! Rbe continuous and f .0/ D f .2/: Show that there exists c 2 [0; 1] such that f .c/ D f .cC1/: (c) Show that there is a nonzero continuous function g : R ! R such that g .t / C g .2t / C g .3t / D g .4t / C g .5t / for every t 2 R: (Hint: Try g .t / D jt jr for some constant r:) 113. (a) Show that the set T D fx : sin x 2 Qg is countable. (Hint: For a fixed rational r; how many solutions of sin x D r are there in the interval [k ; k C 2 /?) (b) If f : [0; 1] ! R is continuous and sin f .x / 2 Q for every x 2 [0; 1]; then show that f is a constant function. 114. Show that there is no continuous function f : R ! R such that for every c 2 R, f .x / D c has exactly 2 solutions. 115. Suppose f : R ! R is a function such that f .x C y / D f .x / C f . y / for every x ; y 2 R and j f .x /j every x 6D 0: (a) Show that f is continuous at some x 2 R: (b) Show that f is continuous at every x 2 R: (c) Give an example of such a function. x 4 =jx j for 116. Suppose f ; g : [1; 2] ! [3; 4] are continuous functions and also fg .x / : x 2 [1; 2]g D [3; 4]: Show that there is c 2 [1; 2] such that f .c/ D g .c/: 117. Let f : R ! R be continuous such that j f .x / f . y /j 1 jx y j for every x ; y 2 R: 2 (a) Let w 2 R: Define x 1 D w and x nC1 D f .x n / for n 2 N: Show that fx n g is a Cauchy sequence. (Hint: How is jx kC1 x k j compared to jx 2 x 1 j?) (b) Show that there is x 2 R such that f .x / D x : 118. Give an example of a function f : .0; 2/ ! R that is differentiable for every x 2 .0; 2/; but f 0 .x / is not continuous at x D 1: Be sure to give reasons to support your example. 119. Let f : .0; / ! R satisfies function. p j f .a / f .b /j sin ja bj for every a ; b 2 .0; /: Show that f is a constant 120. (a) State the mean value theorem. (b) For the function f .x / D sin x ; find the smallest constant K such that j f .b/ every a ; b 2 R: x !0 f .a /j K jb a j holds for 121. If f : R ! R is differentiable and lim f 0 .x / exists, then show that f 0 is continuous at 0. 71 122. Use the mean value theorem to prove that f .x / D sin 5x is uniformly continuous. 123. Prove that if f : R ! .0; C1/ is uniformly continuous, then the function g .x / D continuous. p f .x / is also uniformly 124. (a) State Lebesgue’s theorem. (b) Let f ; g : [0; 2] ! [0; 1] be Riemann integrable. Prove that the function h : [0; 2] ! [0; 1] defined by f .x / if x 2 [0; 1/ h .x / D is Riemann integrable. g .x / if x 2 [1; 2] 125. If f ; g : [0; 1] ! R are Riemann integrable, show that the function h .x / D min. f .x /; g .x // is Riemann integrable on [0; 1]: 126. For every positive integer n ; give an example of a Riemann integrable function fn : [0; 1] ! [0; 1] such that lim fn .x / exists for every x 2 [0; 1]; but the function f .x / D lim fn .x / is not Riemann integrable on [0; 1]: n!C1 Be sure to give reasons to support your example. Z n!C1 127. (a) Determine if the improper integral (b) Determine if the principal value P.V. cos 3x d x exists or not. 1 C x2 Z1 1 1 (c) Determine if the improper integral p d x exists or not. 3 x 1 Z1 1 (d) Determine if the principal value P.V. p d x exists or not. 3 x 1 Z 1 cos 3x d x exists or not. 1 1 C x2 Z1 1 (e) Determine if the improper integral (f) Determine if the principal value P.V. 11 Z sin x d x exists or not. 1 sin x d x exists or not. 128. (2002 L1 Midterm) Let f .x / D x 2 C 3: Determine if the set S D f f .w C z / : w 2 [0; 1] \ Q; z 2 [2; 3] n Qg has an infimum and a supremum. If they exist, find them and give reasons to support your answers. 129. (2002 L1 Midterm) Determine if each of the series 6 1 2k k 1 1 and 6 .cos k / sin 2 converges or diverges. k k D1 .2k /! k D1 p 130. (2002 L1 Midterm) Let S be the set of all lines L on the coordinate plane such that L passes through two distinct points in Q Q and T be the set of all points, each of which is the intersection of a pair of distinct lines in S : Determine if T is a countable set or not. 131. (2002 L1 Midterm) Given x n 6D the definition of limit. 1 for all n 2 N: If lim x n D 0; then show that lim n!1 n!1 xn D 0 by checking 1 C xn 132. (2002 L1 Midterm) Given fx n g converges to w 2 R and x n < w for all n 2 N: For every n 2 N; let yn D sup x 2k : k 2 N; k Show that f yn g converges to w: n o 1 133. (2002 L2 Midterm) Let f .x / D sin x : Determine if the set S D f .w/ :w2 ; n Q; n 2 N has an n 43 infimum and a supremum. If they exist, find them and give reasons to support your answers. 72 n n C 1o : 2 134. (2002 L2 Midterm) Determine if each of the series 6 1 .2 k C 1 /5 1 cos k and 6 4 converges or diverges. k! k D1 k D1 k C k C 1 135. (2002 L2 Midterm) Let S be the set of all circles on the coordinate plane that pass through .1; 1/ and another p p point .x 2; x 2/ for some x 2 Q: Determine if S is a countable set or not. 136. (2002 L2 Midterm) Given x n 6D 1 for all n 2 N: If lim x n D 0; then show that lim definition of limit. n!1 n!1 xn xn 1 D 0 by checking the 137. (2002 L2 Midterm) Given fx n g converges to w 2 R and x n > w for all n 2 N: For every n 2 N; let yn D inf x 2k : k 2 N; k Show that f yn g converges to w: 138. (2002 L3 Midterm) Let f .x ; y / D x 2 C y 2 : Determine if S D f 1 C . 1/n ; w : n 2 N; w 2 [1; 2/ n Q has an infimum and a supremum. If they exist, find them and give reasons to support your answers. 139. (2002 L3 Midterm) Determine if each of the series 6 q n n C 1o : 2 p 1 2k 11 and 6 3 .3k /! k D1 k k D1 e C 1 k k sin k converges or diverges. 12 140. (2002 L3 Midterm) Let x 1 D 1 and x nC1 D x C 4x n for n D 1; 2; 3; : : : : Show that fx n g converges. Find the 2n limit of fx n g: 141. (2002 L3 Midterm) Let S be the set of all ordered pairs . p; C /; where p D .x ; y / 2 Q with center p and radius jx y j C 1: Determine if S is a countable set or not. 2 142. (2002 L3 Midterm) If lim x n D 1; then show that lim .x n n!1 n!1 Q and C is the circle 1/ D 0 by checking the definition of limit. 143. (2002 L3 Midterm) Let fx n g be a bounded sequence of real numbers. Let S D fw : there exist infinitely many n 2 N such that w < x n g: (a) Show that S is nonempty and bounded above. (b) Show that fx n g has at least one subsequence fx nk g converging to s D sup S : 144. (2004 L1 Midterm) Find (with proof) all positive numbers b such that the series 6 b C 145. (2004 L1 Midterm) Let A be a nonempty countable subset of R: Let SDf : 2 R; sin 2 Ag 1 k D1 1 k k converges. and T Df : 2 R; sin 62 Ag: Determine (with proof) if each of the sets S and T is countable or uncountable. 146. (2004 L1 Midterm) Show that the sequence fx n g given by x1 D 6 converges and find its limit. 73 and x nC1 D 2 xn C 4 for n D 1; 2; 3; : : : 2xn C 3 147. (2004 L1 Midterm) If x n 6D definition of limit. 1 for all n and lim n!1 xn xn C 1 D 1 ; then prove that lim x n D 1 by checking the n!1 2 1 148. (2004 L1 Midterm) Let S [0; ]; T D fcos2 a C cos2 b : a ; b 2 S g and U D fsin c : c 2 S g: If sup T D ; 2 2 then find the infimum of U with proof. 149. (2004 L2 Midterm) Let S be the set of all intersection points .x ; y / 2 R2 of the graphs of the equations x 2 C my 2 D 1 and mx 2 C y 2 D 1; where m 2 Zn f 1; 1g: Determine if S is countable or uncountable. Provide a proof of your answer. p 150. (2004 L2 Midterm) Let x 1 D 9 and x nC1 D limit of fx n g: xn C 2xn for n D 1; 2; 3; : : : : Prove that fx n g converges. Find the 3 151. (2004 L2 Midterm) Let ak > 0 for k D 1; 2; 3; : : : and 6 ak converges. Determine all positive real number b such that the series 6 1 .b C ak /k converges. Be sure to give a proof of your answers. k D1 k 1 k D1 n 152. (2004 L2 Midterm) Prove that if lim x 2k D 0:5 and lim x 2kC1 D 0:6; then the sequence lim x n D 0 by checking n!1 k !1 k !1 the definition of limit. 153. (2004 L2 Midterm) Let I be a nonempty set. For every t 2 I ; let At be a nonempty subset of [0; 1]: Let x t D sup At : Prove that if A D At ; then sup A D supfx t : t 2 I g: t 2I Z1 154. (2002 Final) Determine if the improper integral 0 cos 3x p d x exists or not. x 155. (2002 Final) Prove that there does not exist any continuous function f : R ! R such that f f .x / C x D 0 for every x 2 R: 156. (2002 Final) Let fx n g be a Cauchy sequence of real numbers. Prove that fsin 5x n g is also a Cauchy sequence by checking the definition of a Cauchy sequence. 157. (2002 Final) Let f ; g : [0; 2] ! R be Riemann integrable. Prove that h : [0; 2] ! R defined by h .x / D is also Riemann integrable on [0; 2]: 158. (2003 Final) Determine (with proof) if each of 6 sink 1 C max f .x /; g .x / min f .x /; g .x / if x 2 [0; 1] if x 2 .1; 2] 1 k D1 11 1 and 6 k k D1 cos.1= k / converges or not. 1= k 2 159. (2003 Final) Let A be a nonempty subset of R such that inf A D 0 and sup A D 1: Determine the infimum and supremum of S D fa 3 4a C 1 : a 2 Ag: 160. (2003 Final) Let P be a countable set of points in R2 : Prove that there exists a circle C with the origin as center and positive radius such that every point of the circle C is not in P : (Note points inside the circle do not belong to the circle.) 161. (2003 Final) Let f : R ! [u ; v ] be a function and w 2 R: For every r > 0; let M .r / D supf f .t / : 0 < jt wj < r g and m .r / 74 D inf f f .t / : 0 < jt wj < r g: (a) Prove that lim m .r / and lim M .r / exist. r !0 C x !w r !0C (b) Prove that lim f .x / D L if and only if lim m .r / D L D lim M .r /: r !0C r !0C 162. (2003 Final) Let f : R ! R be uniformly continuous. Prove that g : R ! R defined by g .x / D also uniformly continuous. 163. (2003 Final) Let f : [0; 1] ! [ 1; 1] be Riemann integrable. n f .x / if 0 < x < 1 g .x / D is also Riemann integrable on [0; 1]: 0 if x D 0 or 1 Z 1 is 1 C f .x /2 Using the integral criterion, prove that 164. (2004 Final) Determine whether the improper integral 0 1 sin x x 3=2 d x converge or not. 165. (2004 Final) Let f : R ! R be an increasing function and g : R ! R be a decreasing function. Prove that f .x /g .x / is discontinuous for only countably many x 2 R: 0 166. (2004 Final) Let f .x / D 1 1= n ( if x 2 [0; 1] n Q if x D 0 if x D m = n for m ; n 2 N with no common prime factor. f: Prove that there exists a Riemann integrable function g : [0; 1] ! [0; 1] such that the composition function g [0; 1] ! [0; 1] is not Riemann integrable on [0; 1]: 167. (2004 Final) Let f : .0; C1/ ! R satisfy j f .x / f . y /j j sin.x 2 / sequence x 1 ; x 2 ; x 3 ; : : : given by x n D f .1= n / is a Cauchy sequence. 168. (2005 Spring Final) Determine the supremum of S D sin. y 2 /j for all x ; y > 0: Prove that the o 1 n1 nD1 x C n2 p 1 : x 2 .2; 3] n Q : 2 C 3x is continuous at 2 by checking x2 C 4 xn 169. (2005 Spring Final) Prove that the function f : R ! R defined by f .x / D the "- definition of a function continuous at a point. 170. (2005 Spring Final) Let x n > 0 for n D 1; 2; 3; : : : : If fx n g is a Cauchy sequence, then prove that fe Cauchy sequence by checking the definition of a Cauchy sequence. 171. (2005 Spring Final) (a) Let S measure 0. g is a [0; 1]: If S is a set of measure 0, then prove that T D fx 2 : x 2 S g is a set of p (b) Let f : [0; 1] ! [0; 1] be integrable. Prove that h : [0; 1] ! [0; 1] defined by h .x / D f . x / is integrable. (Caution: In general it is false that f integrable and g continuous on [a ; b] imply h D f g integrable on [a ; b]:) 172. (2005 Fall Exam, Version 1) (a) Determine with proof if 6 (b) Let ak 6D 1 3k converges or not. k D1 .2k /!k ! 1 1 ak 1 for k D 1; 2; 3; : : : and 6 jak j converges. Determine with proof if 6 converges or not. k D1 k D1 1 C ak x nC1 D 4 4 xn for n D 1; 2; 3; : : : : Prove that x 1 ; x 2 ; x 3 ; : : : 173. (2005 Fall Exam, Version 1) Define x 1 D 4 and converges and find its limit. 174. (2005 Fall Exam, Version 1) Let A be a nonempty bounded subset of R such that sup A D 1 and inf A D 0: Let S D fx y 2 : x 2 A; y 2 [ 2; 2/g: 75 p Determine the supremum and infimum of S with proof. 175. (2005 Fall Exam, Version 1) If lim an D 2; lim bn D 3 and all bn 6D checking the definition of limit. n!1 n!1 2; then prove that lim an C 3 D 1 by n!1 bn C 2 176. (2005 Fall Exam, Version 2) (a) Determine with proof if 6 (b) Let cos ak 6D 0 for k D 1; 2; 3; : : : and 6 jak j converges. Determine with proof if 6 177. (2005 Fall Exam, Version 2) Define x 1 D 1 and converges and find its limit. x nC1 D 2 x n C 15 8 1 k D1 k D1 1 .3k /! k !2k converges or not. 1 ak converges or not. k D1 cos ak for n D 1; 2; 3; : : : : Prove that x 1 ; x 2 ; x 3 ; : : : 178. (2005 Fall Exam, Version 2) Let A be a nonempty bounded subset of R such that sup A D 6 and inf A D 2: Let SDf x y x y : x 2 A; y 2 1 ; 1 g: 2 Determine the supremum and infimum of S with proof. 179. (2005 Fall Exam, Version 2) If lim an D 1 and all an 6D n ; then prove that lim definition of limit. n!1 2 an C n D 1 by checking the n!1 n an 180. (2006 Spring Exam)(a) Let f : S ! R be a function and x 0 be an accumulation point of S : State the definition of f .x / converges to a real number L as x tends to x 0 : r (b) Let f : .0:5; C1/ ! Rbe defined by f .x / D 181. (2006 Spring Exam) Define a1 D 1 and anC1 D exists in R: xC n nC1 p 1 : Prove that lim f .x / D 2 by checking the definition. x !1 x an C cos n for n D 1; 2; 3; : : : : Prove that lim nan n!1 .n C 1/3 182. (2006 Spring Exam) Let a ; b 2 R with a < b and f : [a ; b] ! R be continuous. Also, let f .x / be differentiable for all x 2 .a ; b/: Prove that if the graph of f is not a line segment, then there exist numbers x 1 and x 2 in the open interval .a ; b/ such that f .b/ f .a / f 0 .x 1 / < < f 0 .x 2 /: ba 183. (2006 Spring Exam) Let f ; g : [0; 1] ! R be continuous. If there exists a sequence of numbers x 1 ; x 2 ; x 3 ; : : : 2 [0; 1] such that g .x n / D f .x nC1 / for n D 1; 2; 3; : : : ; then prove that there exists w 2 [0; 1] such that g .w/ D f .w/: Caution Be careful, x ni converges does not imply x ni C1 converges !!! 76 ...
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This note was uploaded on 03/17/2010 for the course STATISTIC 472 taught by Professor Amjad during the Spring '08 term at Yarmouk University.

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