Lecture Notes for CS 313H, Fall 011 August 5. We start by examiig triagular umbers: T () = 1 + + + ( = 0, 1,,...). Triagular umbers ca be also defied recursively: T (0) = 0, T ( + 1) = T () + + 1, or usig sigma-otatio: Topics for discussio: T () = i. i=1 A closed-form expressio for T () ad various ways of provig it. The rate of growth of T (). Combiatorial problems related to triagular umbers. A geeralizatio of triagular umbers: A related cocept factorials: S k () = 1 k + k + + k.! = 1.... Here is a closed-form expressio for T (): This formula ca be also writte as T () = +. T () = ( + 1). (1) We will see that it ca be proved i may differet ways. 1
Problem 1. By S 3 () we deote the sum 1 3 + 3 + + 3. Calculate S 3 () for several values of ad guess what a geeral formula for S 3 () may look like. August 30. Oe possible approach to provig formula (1) is based o iductio. Whe we wat to prove by iductio that some statemet cotaiig a variable is true for all oegative itegers, we do two thigs. First we prove the statemet whe = 0; this part of the proof is called the basis. The we prove the statemet for + 1 assumig that it is true for ; this part of the proof is called the iductio step. (The assumptio that the statemet is true for, which is used i the iductio step, is called the iductio hypothesis.) Oce we have completed both the basis ad the iductio step, we ca coclude that the statemet holds for all oegative itegers. Ideed, accordig to the basis, i holds for = 0. From this fact, accordig to the iductio step, we ca coclude that it holds for = 1. From this fact, accordig to the iductio step, we ca coclude that it holds for =. Ad so o. Here is a proof of formula (1) by iductio. Basis. Whe = 0, the formula turs ito 0(0 + 1) 0 =, which is correct. Iductio step. Assume that ( + 1) 1 + +... + =. We eed to show that ( + 1)( + ) 1 + +... + + ( + 1) =. Usig the iductio hypothesis, we calculate: 1 + +... + + ( + 1) = ( + 1) + ( + 1) This is the ed of the proof. = = ( + 1) + ( + 1) ( + 1)( + ).
Problem. Prove that for ay oegative iteger, 7 + 5 is a multiple of 6. Our work o Problem 1 has led us to the cojecture that for ay oegative iteger 1 3 + 3 + + 3 = ( + 1). () 4 We have also cojectured that 1 + + + = ( + 1)( + 1). (3) 6 Problem 3. Fid a closed-form expressio for 1 + + + geometrically. Problem 4. Use iductio to prove formula (3). Problem 5. Use iductio to prove formula (). Problem 6. Use iductio to solve Problem. September 1. Here is aother proof of formula (1). Cosider the formulas (1 + 1) = 1 + 1 1 + 1 ( + 1) = + 1 + 1 (3 + 1) = 3 + 3 1 + 1... ( + 1) = + 1 + 1. By addig the left-had sides ad the right-had sides, we get: + 3 + 4 + + ( + 1) = 1 + + 3 + + + T () +, or, after subtractig + 3 + + from both sides, ( + 1) = 1 + T () +. Solvig this equatio for T () gives formula (1). Problem 7. Use this method to prove formula (3). So far we used iductio to prove assertios about oegative itegers. More geerally, iductio ca be used to prove a assertio about all itegers that are 0 for some iitial value 0. First we prove the statemet whe = 0. The we prove the statemet for + 1 assumig that it is true for, for a arbitrary 0. Oce this is doe, we ca coclude that the statemet holds for all 0. Problem 8. Prove that for every 4,! >. 3
September 6. Problem 9. Use calculus to fid the smallest value of x x. Problem 10. Fid a closed-form expressio for ( 1 1 ) ( 1 1 ) ( 1 1 )... (1 1 ) 4 9 16. September 8. By P () we deote the umber of parts ito which a plae is divided by lies, provided that these lies are i a geeral positio. Similarly, P 3 () is the umber of parts ito which space is divided by plaes i a geeral positio. Problem 11. Fid P 3 (4). Problem 1. Fid a geeral formula for P (). September 13. Problem 13. Fid a closed-form expressio for P 3 (). The followig approximate formula is ofte useful: for large values of, f(1) + f() + + f() It shows, for istace, that for large values of S k () 1 k + 1 k+1. The -th harmoic umber is defied by the formula H() = 1 1 + 1 + + 1. 0 f(x)dx. (4) I applicatio to harmoic umbers, formula (4) tells us that for large values of H() l. Problem 14. Use formula (4) to estimate!. 4
September 15. Problem 15. A slow but persistet worm, W, starts at oe ed of a footlog rubber bad ad crawls oe ich per miute toward the other ed. At the ed of each miute a equally persistet keeper of the bad, K, whose sole purpose i life is to frustrate W, stretches it oe foot. Thus after oe miute of crawlig, W is 1 ich from the start ad 11 from the fiish; the K stretches the bad oe foot. Durig the stretchig operatio W maitais his relative positio, so that W is ow iches from the startig poit ad iches from the goal. After W crawls for aother miute the score is 3 iches traveled ad 1 to go; but K stretches, ad the distaces become 4.5 ad 31.5. Ad so o. Does the worm ever reach the fiish? (We are assumig a ifiite logevity for W ad K, a ifiite elasticity of the bad, ad a ifiitely tiy worm.) As we have see, the claim that some statemet is true for all oegative itegers ca be sometime proved by iductio. I some cases, strog iductio is more useful. I a proof by strog iductio we show that the statemet is true for a oegative iteger assumig that it is true for each oegative iteger that is <. As i the case of simple ( weak ) iductio, we ca talk about the itegers begiig with some iitial value 0, istead of oegative itegers. 5
September 0. Fiboacci umbers are defied by the formulas Problem 16. Prove that Problem 17. Prove that Fib(0) = 0, Fib(1) = 1, Fib( + ) = Fib() + Fib( + 1) ( 0). Fib(1) + Fib() +... + Fib() = Fib( + ) 1. Fib(1) + Fib() +... + Fib() = Fib()Fib( + 1). 6
September. Problem 18. Fid all values of such that Fib() < 1.1. About a fuctio f defied o oegative itegers we say that it is a geeralized Fiboacci fuctio if it satisfies the equatio f( + ) = f() + f( + 1). Problem 19. A geeralized Fiboacci fuctio f satisfies the coditios f(0) = 1, f(6) = 3. Fid f(3). Problem 0. Fid a formula that expresses f() for ay geeralized Fiboacci fuctio f i terms of f(0), f(1), ad. The formula may use the stadard Fiboacci fuctio Fib. Problem 1. Fid a umber c such that the sequece of its powers f() = c is a geeralized Fiboacci fuctio. 7
September 7. Problem. Fid umbers a, b such that ( ) ( ) 1 + 5 1 5 Fib() = a + b. October 4. Problem 3. Fuctio f is defied by the formula { 10, if > 100, f() = f(f( + 11)), otherwise. Fid f(98). For ay oegative itegers m,, the biomial coefficiet ( m) (read choose m ) is the umber ways to choose a m-elemet subset from a -elemet set. Biomial coefficiets ca be calculated usig the formula ( m ) = ( 1) ( m + 1). m! Whe m, this formula ca be rewritte as ( )! = m m!( m)!. Problem 4. Prove that ( ) + 1 = m + 1 ( ) ( ) +. m m + 1 This formula, combied with the iitial coditios ( ) ( ) 0 = 1, = 0, 0 m + 1 provides a recursive defiitio of biomial coefficiets. Problem 5. Prove the biomial theorem: ( ) ( ) ( ) (a + b) = a + a 1 b + 0 1 a b + + ( ) b. 8
October 6. Problem 6. Prove by iductio: 0 ( ) + 1 0 ( ) + 1 ( ) + + ( ) = 1. Stirlig umbers (of the first kid) are coefficiets i the expasio of the polyomial x(x + 1)(x + ) (x + 1): x(x + 1)(x + ) (x + 1) = For istace, + 0 x + 1 x + x(x + 1)(x + )(x + 3) = 6x + 11x + 6x 3 + x 4 ; x. cosequetly, 4 = 0, 0 4 = 6, 1 4 = 11, 4 = 6, 3 4 = 1. 4 Problem 7. Fid a formula expressig [ +1 [ m+1] i terms of ] [ m ad m+1]. 9
October 11. Problem 8. Prove that for all > 0 = ( 1)!. 1 Problem 9. Prove that for all > 0 = T ( 1). 1 Problem 30. Prove that for all > 0 = ( 1)! H( 1). October 13. Problem 31. Prove that for all 0 0 + 1 + 0 1 + + =! H(). 10
October 18. A set is a collectio of objects. We write x A if object x is a elemet of set A, ad x A otherwise. The set whose elemets are x 1,..., x is deoted by {x 1,..., x }. The set { } is called empty ad deoted also by. The set of oegative itegers is deoted by N: N = {0, 1,,...}. Whe we specify which objects belog to a set, this defies the set completely; there is o such thig as the order of elemets i a set or the umber of repetitios of a elemet i a set. For istace, {, 3} = {3, } = {,, 3}. If C is a coditio, the by {x : C} we deote the set of all objects x satisfyig this coditio. For istace, {x : x = or x = 3} is the same set as {, 3}. If A is a set ad C is a coditio, the by {x A : C} we deote the set of all elemets of A satisfyig coditio C. For istace, {, 3} ca be also writte as {x N : 1 < x < 4}. If a set A is fiite the the umber of elemets of A is also called the cardiality of A ad deoted by A. For istace, = 0, {, 3} =. We say that a set A is a subset of a set B, ad write A B, if every elemet of A is a elemet of B. For istace, N, {, 3} N, N N. For ay sets A ad B, by A B we deote the set {x : x A or x B}, called the uio of A ad B. By A B we deote the set {x : x A ad x B}, 11
called the itersectio of A ad B. For istace, {, 3} {3, 5} = {, 3, 5}, {, 3} {3, 5} = {3}. Problem 3. For ay sets A, B, C, if the A B C. True or false? A B, A C, B C Problem 33. Let A = 3, B = 4. What ca you say about the cardialities of A B ad A B? By A \ B we deote the set {x : x A ad x B}, called the differece of A ad B. For istace, {, 3} \ {3, 5} = {}. Problem 34. Let A = 3, B = 4. What ca you say about the cardialities of A \ B ad B \ A? The Cartesia product of sets A ad B is the set of ordered pairs x, y such that x A ad y B: For istace, A B = { x, y x A ad y B}. {1, } {, 3, 4, 5, 6} = { 1,, 1, 3, 1, 4, 1, 5, 1, 6,,,, 3,, 4,, 5,, 6 }. Problem 35. Determie whether the followig assertios are true. (a) For ay sets A ad B, A B = B A. (b) For ay sets A ad B, if A is ifiite the A B is ifiite too. (c) For ay sets A ad B, if A B = 91 the oe of the sets A, B is a sigleto. 1
October 0. The cardiality of the uio of two sets ca be determied if we kow their cardialities ad the cardiality of their itersectio: A B = A + B A B. This formula is called the iclusio-exclusio priciple. Similar formulas ca be writte for the uio of several sets. Problem 36. Use a set diagram to simplify the expressio (A \ B) (B \ C) (C \ A) (A B C). October 5. By P(A) we deote the power set of a set A, that is, the set of all subsets of A. The cardiality of P(A) is A. Problem 37. For ay sets A, B, True or false? P(A B) = P(A) P(B). Problem 38. For ay pair of differet sets A ad B, P(A) is differet from P(B). True or false? Set-theoretic otatio ca be used to represet some Eglish phrases by mathematical formulas. For istace, the assertio ca be writte as Istead of A is the set of roots of the polyomial x + px + q A = {x R : x + px + q = 0}. All elemets of A are roots of the polyomial x + px + q we ca write A {x R : x + px + q = 0}. 13
November 3. A biary relatio o a set A is a subset of A A. If R is a biary relatio the we write xry istead of x, y R. A relatio R o a set A is said to be reflexive if, for all x A, xrx; symmetric if, for all x, y A, xry implies yrx; trasitive if, for all x, y, z A, xry ad yrz imply xrz. Problem 39. There are 16 biary relatios o the set {1,, 3, 4}. How may of them are reflexive? Symmetric? Both reflexive ad symmetric? Problem 40. (a) Cosider the relatio R o the set N of oegative itegers defied by the coditio: xry if x y + 5. Is it reflexive? Is it symmetric? Is it trasitive? (b) Aswer the same questios for the relatio: xry if x y < 5. 14
November 8. A biary relatio R o a set A ca be visualized as the directed graph with the set of vertices A that has a edge from a vertex x to a vertex y wheever xry. A biary relatio o a fiite set {a 1,..., a } ca be represeted by its adjacecy matrix a matrix such that the etry at the itersectio of colum i ad row j is 1 if a i Ra j, 0 otherwise. If a relatio is reflexive, symmetric, ad trasitive, the we say that it is a equivalece relatio. If R is a equivalece relatio o a set A, ad x is a elemet of A, the the set {y A : xry} is called the equivalece class of x ad is deoted by [x]. The set of all equivalece classes of a equivalece relatio R is called the quotiet set of R. Problem 41. Prove that the equivalece classes of ay equivalece relatio are pairwise disjoit. A partitio of a set A is a set P of o-empty subsets of A such that each elemet of A belogs to a uique elemet of P. The quotiet set of ay equivalece relatio is a partitio. The other way aroud, ay partitio is the quotiet set of some equivalece relatio. Problem 4. Fid all partitios of the set {1,, 3}. For each of them, represet the correspodig equivalece relatio as a graph ad as a adjacecy matrix. Problem 43. Does there exist a partitio of N with ifiitely may equivalece classes such that all of them are ifiite? A relatio R o a set A is atisymmetric if, for all x, y A, xry ad yrx imply x = y. A partial order is a relatio that is reflexive, atisymmetric, ad trasitive. A total order o a set A is a partial order R o A such that for all x, y A, xry or yrx. 15
November 10. A fuctio from a set A to a set B is a subset f of A B satisfyig the followig coditio: for every elemet x of A there exists a uique elemet y of B such that x, y f. This elemet y is called the value of f at x ad is deoted by f(x). Set A is called the domai of f, ad B is called the co-domai of f. The subset of the co-domai B cosistig of the values f(x) for all x A is called the rage of f. Problem 44. Let A = {1,, 3}, B = {, 3, 4, 5}. Fid the umber of fuctios from A to B ad the umber of fuctios from B to A. I the followig examples, the domai of each fuctio is the set S of all bit strigs: S = {ɛ, 0, 1, 00, 01, 10, 11, 000, 001,...}. 1. Fuctio l from S to N: l(x) is the legth of x. For istace, l(00110) = 5.. Fuctio z from S to N: z(x) is the umber zeroes i x. For istace, z(00110) = 3. 3. Fuctio from S to N: (x) is the umber represeted by x i biary otatio. For istace, (00110) = 6. 4. Fuctio e from S to S: e(x) is the strig 1x. For istace, t(00110) = 100110. 5. Fuctio r from S to S: r(x) is the strig x reversed. For istace, r(00110) = 01100. 6. Fuctio p from S to P(S): p(x) is the set of prefixes of x. For istace, p(00110) = {ɛ, 0, 00, 001, 0011, 00110}. If the rage of f is the whole co-domai B the we say that f is a fuctio oto B. A fuctio f from A to B is called oe-to-oe if, for ay pair of differet elemets x, y of A, f(x) is differet from f(y). If a fuctio f is both oto ad oe-to-oe the we say that f is a bijectio. 16
November 17. Problem 45. oe-to-oe? Which of the fuctios l, z,, e, r, p defied above are If f is a fuctio from A to B, ad g is a fuctio from B to C, the the compositio of these fuctios is the fuctio h from A to C defied by the formula h(x) = g(f(x)). This fuctio is deoted by g f. Problem 46. For the fuctios l, z,, e ad r defied above, which of the followig formulas are true? l r = l, z r = z, r =, e r = r e. If f is a bijectio from A to B the the iverse of f is the fuctio g from B to A such that, for every x A, g(f(x)) = x. This fuctio is deoted by f 1. Problem 47. f 1 f. Fid all bijectios f from {1,, 3} to {1,, 3} such that Now we tur to the study of logical otatio, the last topic i this course. Truth values are the symbols F ( false ) ad T ( true ). Propositioal coectives are fuctios that are applied to truth values ad retur truth values. We will use these propositioal coectives: ot (egatio) ad (cojuctio) or (disjuctio) implies (implicatio) is equivalet to (equivalece) Negatio is a uary coective (takes oe argumet); the others are biary (take two argumets). Here are the tables of values for these coectives: p p F T T F 17
p q p q p q p q p q F F F F T T F T F T T F T F F T F F T T T T T T Propositioal variables are variables for truth values, such as p ad q i the tables above. Propositioal formulas are built from truth values ad propositioal variables usig propositioal coectives. For istace, is a propositioal formula. (p q) (r F) 18
November. The truth table of a propositioal formula shows its truth values for all possible combiatios of values of its variables. For istace: p q r (p q) (r F) F F F T F F T F F T F T F T T T T F F T T F T T T T F T T T T T Problem 48. Fid a propositioal formula that gets the value T i two cases: whe p = T, q = F, r = T ad whe p = T, q = T, r = F. About propositioal formulas F ad G we say that they are equivalet to each other if, for each combiatio of the values of their variables, the truth value of F equals the truth value of G. For example, p q is equivalet to p q, p q is equivalet to (p q) ( p q), F is equivalet to p p, T is equivalet to p p. Usig these four facts, we ca elimiate from ay propositioal formula the coectives ad ad the costats F ad T. I other words, ay propositioal formula ca be equivaletly rewritte as a formula formed from variables usig the coectives, ad. Here are two other useful facts about the equivalece of propositioal formulas: (p q) is equivalet to p q, (p q) is equivalet to p q. These facts are called De Morga s laws. Propositioal variables are also called atoms. Atoms ad egated atoms, such as p ad p, are called literals. A propositioal formula is said to be i egatio ormal form if it is formed from literals usig cojuctios ( ) ad disjuctios ( ). Ay propositioal formula ca be equivaletly rewritte i egatio ormal form i two steps: 19
elimiate,, F, T, as described above; use De Morga s laws to push the egatios iside. A propositioal formula is said to be i cojuctive ormal form (CNF) if it is a cojuctio of disjuctios of literals. A propositioal formula is said to be i disjuctive ormal form (DNF) if it is a disjuctio of cojuctios of literals. For istace, the formula p (p q) ( q r s) is i disjuctive ormal form. A formula i egatio ormal form ca be tured ito CNF by distributig disjuctio over cojuctio, ad to DNF by distributig cojuctio over disjuctio. Problem 49. Covert the formulas ad to CNF ad DNF. (p q) r p (q r) A propositioal formula is called a tautology if each of its truth values is T. Problem 50. Determie which of these formulas are tautologies: (p q) (q p), ((p q) p) p, (p (q r)) ((p q) (p r)). 0
November. I logical formulas, we use the propositioal coectives itroduced above ad the quatifiers ( for all ) ad ( exists ). Problem 51. Express i logical otatio: 1. A B C. A = B C 3. A = {a, b} Problem 5. Express i logical otatio usig variables for oegative itegers: 1. Number ca be represeted as the sum of two complete squares.. Number is composite. 1