COMP 2804 Solutions Assignment 1
|
|
- Amy Mathews
- 5 years ago
- Views:
Transcription
1 COMP 2804 Solutios Assiget 1 Questio 1: O the first page of your assiget, write your ae ad studet uber Solutio: Nae: Jaes Bod Studet uber: 007 Questio 2: I Tic-Tac-Toe, we are give a 3 3 grid, cosistig of uared cells Two players, Xavier ad Olivia, tae turs arig the cells of this grid Whe it is Xavier s tur, he chooses a uared cell ad ars it with the letter X Siilarly, whe it is Olivia s tur, she chooses a uared cell ad ars it with the letter O The first tur is by Xavier The players cotiue aig turs util all cells have bee ared Below, you see a exaple of a copletely ared grid O X X O X X X O O What is the uber of copletely ared grids? Justify your aswer What is the uber of differet ways (ie, ordered sequeces) i which the grid ca be copletely ared, whe followig the rules give above? Justify your aswer Solutio: For the first part, observe that the grid cosists of ie cells, ad that a copletely ared grid cotais five X s ad four O s To copletely ar a iitially epty grid, we have to choose five cells (out of ie); i the chose cells, we write X, whereas we write O i the reaiig four cells Therefore, the uber of copletely ared grids is equal to the uber of ways to choose five eleets out of ie Thus, the aswer is ( ) 9 5 For the secod part, we uber the cells arbitrarily 1, 2, 3,, 9 Ay sequece to ar all cells is othig but a perutatio of these ie itegers For exaple, the perutatio idicates the followig sequece of oves: Xavier writes X i cell 7, 7, 3, 4, 1, 9, 6, 8, 2, 5 1
2 Olivia writes O i cell 3, Xavier writes X i cell 4, Olivia writes O i cell 1, Xavier writes X i cell 9, Olivia writes O i cell 6, Xavier writes X i cell 8, Olivia writes X i cell 2, Xavier writes X i cell 5 Thus the aswer is 9! Questio 3: A password is a strig of 100 characters, where each character is a digit or a lowercase letter A password is called valid if it does ot start with abc, ad it does ot ed with xyz, ad it does ot start with 3456 Deterie the uber of valid passwords Justify your aswer Solutio: We defie the followig sets: U is the set of all strigs of 100 characters, where each character is a digit or a lowercase letter A is the set of all strigs i U that start with abc B is the set of all strigs i U that ed with xyz C is the set of all strigs i U that start with 3456 The questio ass for the value of A B C Usig De Morga ad the Copleet Rule, we get A B C = A B C = U A B C 2
3 To deterie the size of U, we observe that each strig i U cosists of 100 characters, ad for each character, there are = 36 choices By the Product Rule, we have U = To deterie the size of A, we observe that each strig i U cosists of 100 characters The first three characters are fixed, whereas for each of the reaiig 97 characters, there are = 36 choices By the Product Rule, we have A = By siilar reasoig, we have ad B = 36 97, C = 36 96, A B = 36 94, B C = Sice A C = ad A B C =, we have A C = 0 ad A B C = 0 By the Priciple of Iclusio ad Exclusio, we have A B C = A + B + C A B A C B C + A B C We coclude that = = A B C = U A B C = = Questio 4: Let ad be itegers with 0 There are + 1 studets i Carleto s Coputer Sciece progra The Carleto Coputer Sciece Society has a Board of Directors, cosistig of oe presidet ad vice-presidets The presidet caot be vice-presidet Prove that ( ) ( ) + 1 ( + 1) = ( + 1 ), 3
4 by coutig, i two differet ways, the uber of ways to choose a Board of Directors Solutio: First way: First tas: Choose a presidet; there are + 1 ways to do this Secod tas: Choose vice-presidets Sice the presidet has already bee chose, there are are ( ) ways to do this By the Product Rule, the uber of ways to choose a Board of Directors is equal to Secod way: First tas: Choose vice-presidets; there are ( +1 ( ) ( + 1) (1) ) ways to do this Secod tas: Choose a presidet Sice the vice-presidets have already bee chose, there are are + 1 ways to do this By the Product Rule, the uber of ways to choose a Board of Directors is equal to ( ) + 1 ( + 1 ) (2) The values of (1) ad (2) ust be equal, because both of the cout the uber of ways to choose a Board of Directors Questio 5: Let ad be itegers with 2, ad cosider the set S = {1, 2, 3,, 2} A ordered sequece of eleets of S is called valid if this sequece is strictly icreasig, or this sequece is strictly decreasig, or this sequece cotais oly eve ubers (ad duplicate eleets are allowed) Deterie the uber of valid sequeces Justify your aswer Solutio: We defie the followig sets: A is the set of all ordered sequeces of eleets of S that are strictly icreasig B is the set of all ordered sequeces of eleets of S that are strictly decreasig C is the set of all ordered sequeces of eleets of S that cotai oly eve ubers 4
5 The questio ass for the value of A B C What is the size of A? Ay sequece of A correspods to a uique subset of S havig size Coversely, ay subset of S havig size correspods to a uique sequece i A It follows that ( ) 2 A = By the sae reasoig, we have B = ( ) 2 What is the size of C? Each sequece i C is a ordered sequece of leth, ad each eleet is a eve eleet of S Sice S cotais ay eve ubers, it follows that that there are choices for each eleet i ay such sequece It follows that C = We observe that A B =, because we assue that 2 This also iplies that A B C = Thus, A B = A B C = 0 What is the size of A C? Ay sequece of A C correspods to a uique subset of {2, 4, 6,, 2} havig size Coversely, ay subset of {2, 4, 6,, 2} havig size correspods to a uique sequece i A C It follows that ( ) A C = By the sae reasoig, we have B C = By the Priciple of Iclusio ad Exclusio, we have ( ) A B C = A + B + C A B A C B C + A B C ( ) ( ) ( ) ( ) 2 2 = (( ) ( )) 2 = 2 + Questio 6: Let,, ad be itegers with 0, ad let S be a set of size Prove that ( )( ) ( )( ) =, 5
6 by coutig, i two differet ways, the uber of ordered pairs (A, B) with A S, B S, A B, A =, ad B = Solutio: First way: First tas: Choose a subset A of S Sice A has size, there are ( ) ways to do this Secod tas: Choose a subset B of S; this subset has size ad it ust cotai all eleets of A This is the sae as choosig a subset of S \ A (which has size ) of size There are ( ) ways to do this By the Product Rule, the uber of ordered pairs (A, B) with A S, B S, A B, A =, ad B = is equal to ( )( ) (3) Secod way: First tas: Choose a subset B of S Sice B has size, there are ( ) ways to do this Secod tas: Choose a subset A of S; this subset has size ad it ust be a subset of B There are ( ) ways to do this By the Product Rule, the uber of ordered pairs (A, B) with A S, B S, A B, A =, ad B = is equal to ( )( ) (4) The values of (3) ad (4) ust be equal, because both cout the uber of ordered pairs (A, B) with A S, B S, A B, A =, ad B = Questio 7: Let ad be itegers with 0 How ay bitstrigs of legth + 1 have exactly ay 1s? Let be a iteger with 0 What is the uber of bitstrigs of legth + 1 that have exactly ay 1s ad that start with } 1 {{ 1} 0? Use the above two results to prove that =0 ( ) = ( ) + 1 6
7 Solutio: We have see the aswer to the first part i class: ( ) + 1 For the secod part, we wat to cout bitstrigs havig legth + 1, that have exactly ay 1s, that start with } 1 {{ 1} 0 This eas that the first + 1 positios are fixed, ad of these positios cotai 1, i the last ( + 1) ( + 1) = positios, we have to place ay 1s This eas that we have to cout the bitstrigs of legth havig exactly ay 1s We ow fro class that the aswer is ( ) For the third part, we are goig to cout, i two differet ways, the bitstrigs of legth + 1 that have exactly ay 1s The first way is give by the first part of this questio: ( ) + 1 (5) For the secod way, we observe the followig: Each bitstrig of legth + 1 that has exactly ay 1s is of oe of the followig types: it starts with 0, ie, with } 1 {{ 1} 0 Fro the secod part, the uber of these is equal =0 to ( ) it starts with 10, ie, with } 1 {{ 1} 0 Fro the secod part, the uber of these is equal =1 to ( ) 1 1 7
8 it starts with 110, ie, with } 1 {{ 1} 0 Fro the secod part, the uber of these is =2 equal to ( ) 2 2 it starts with 1110, ie, with } 1 {{ 1} 0 Fro the secod part, the uber of these is =3 equal to ( ) 3 3 Etc, etc it starts with } 1 {{ 1} 0 Fro the secod part, the uber of these is equal to = ( ) If we add up the uber of bitstrigs of all these types, we see that the uber of bitstrigs of legth + 1 that have exactly ay 1s is equal to ( ) ( ) ( ) ( ) ( ) ( ) = (6) Sice the values of (5) ad (6) are equal, because both cout the sae bitstrigs, we are doe Questio 8: Let ad be itegers with 0 Use Questios 4, 6, ad 7 to prove that ( ) ) = =0 ( =0 Solutio: Usig Questio 6, we get ( ) ( ) ( = ( =0 ) =0 ) I the latter su, the ter ( ) does ot deped o the suatio variable Thus, we ca tae it out of the suatio ad get ( ) ) = ( 1 ( ) ) =0 ( 8 =0
9 Usig Questio 7, this becoes ( ) ( ) = 1 ( ) + 1 ( =0 ) Fially, usig Questio 4, this becoes ( ) ) = =0 Questio 9: I this exercise, we cosider the sequece of itegers ( 3 0, 3 1, 3 2,, Prove that this sequece cotais two distict eleets whose differece is divisible by 1000 That is, prove that there exist two itegers ad with 0 < 1000, such that 3 3 is divisible by 1000 Hit: Cosider each eleet i the sequece odulo 1000 ad use the Pigeohole Priciple Use the first part to prove that this sequece cotais a eleet whose decial represetatio eds with 001 I other words, the last three digits i the decial represetatio are 001 Solutio: For the first part, cosider the sequece 3 0 od 1000, 3 1 od 1000, 3 2 od 1000,, od 1000 Each uber i this sequece is a iteger belogig to the set {0, 1, 2, 3,, 999}; this set has size 1000 Sice the sequece cosists of 1001 ubers, the Pigeohole Priciple tells us that there ust be at least two ubers i the sequece that are equal I other words, there exist two itegers ad with 0 < 1000, such that This eas that ie, 3 3 is divisible by od 1000 = 3 od 1000 (3 3 ) od 1000 = 0, For the secod part: I the first part, we have foud two itegers ad with 0 < 1000, such that 3 3 is divisible by 1000 Observe that 3 3 = 3 ( 3 1 ) 9
10 Thus, sice 3 3 is divisible by 1000, 3 ( 3 1 ) is divisible by 1000 Sice the greatest coo divisor of 3 ad 1000 is equal to 1, the uber 3 1 is divisible by 1000 Thus, there is a positive iteger such that 3 1 = 1000, ie, 3 = This eas that the last three digits i the decial represetatio of 3 are
distinct distinct n k n k n! n n k k n 1 if k n, identical identical p j (k) p 0 if k > n n (k)
THE TWELVEFOLD WAY FOLLOWING GIAN-CARLO ROTA How ay ways ca we distribute objects to recipiets? Equivaletly, we wat to euerate equivalece classes of fuctios f : X Y where X = ad Y = The fuctios are subject
More informationJacobi symbols. p 1. Note: The Jacobi symbol does not necessarily distinguish between quadratic residues and nonresidues. That is, we could have ( a
Jacobi sybols efiitio Let be a odd positive iteger If 1, the Jacobi sybol : Z C is the costat fuctio 1 1 If > 1, it has a decopositio ( as ) a product of (ot ecessarily distict) pries p 1 p r The Jacobi
More informationSection 5.1 The Basics of Counting
1 Sectio 5.1 The Basics of Coutig Combiatorics, the study of arragemets of objects, is a importat part of discrete mathematics. I this chapter, we will lear basic techiques of coutig which has a lot of
More informationDiscrete Mathematics: Lectures 8 and 9 Principle of Inclusion and Exclusion Instructor: Arijit Bishnu Date: August 11 and 13, 2009
Discrete Matheatics: Lectures 8 ad 9 Priciple of Iclusio ad Exclusio Istructor: Arijit Bishu Date: August ad 3, 009 As you ca observe by ow, we ca cout i various ways. Oe such ethod is the age-old priciple
More information6.4 Binomial Coefficients
64 Bioial Coefficiets Pascal s Forula Pascal s forula, aed after the seveteeth-cetury Frech atheaticia ad philosopher Blaise Pascal, is oe of the ost faous ad useful i cobiatorics (which is the foral ter
More informationMath 4707 Spring 2018 (Darij Grinberg): midterm 2 page 1. Math 4707 Spring 2018 (Darij Grinberg): midterm 2 with solutions [preliminary version]
Math 4707 Sprig 08 Darij Griberg: idter page Math 4707 Sprig 08 Darij Griberg: idter with solutios [preliiary versio] Cotets 0.. Coutig first-eve tuples......................... 3 0.. Coutig legal paths
More informationHomework 3. = k 1. Let S be a set of n elements, and let a, b, c be distinct elements of S. The number of k-subsets of S is
Homewor 3 Chapter 5 pp53: 3 40 45 Chapter 6 p85: 4 6 4 30 Use combiatorial reasoig to prove the idetity 3 3 Proof Let S be a set of elemets ad let a b c be distict elemets of S The umber of -subsets of
More informationCSE 191, Class Note 05: Counting Methods Computer Sci & Eng Dept SUNY Buffalo
Coutig Methods CSE 191, Class Note 05: Coutig Methods Computer Sci & Eg Dept SUNY Buffalo c Xi He (Uiversity at Buffalo CSE 191 Discrete Structures 1 / 48 Need for Coutig The problem of coutig the umber
More informationCS 70 Second Midterm 7 April NAME (1 pt): SID (1 pt): TA (1 pt): Name of Neighbor to your left (1 pt): Name of Neighbor to your right (1 pt):
CS 70 Secod Midter 7 April 2011 NAME (1 pt): SID (1 pt): TA (1 pt): Nae of Neighbor to your left (1 pt): Nae of Neighbor to your right (1 pt): Istructios: This is a closed book, closed calculator, closed
More informationInjections, Surjections, and the Pigeonhole Principle
Ijectios, Surjectios, ad the Pigeohole Priciple 1 (10 poits Here we will come up with a sloppy boud o the umber of parethesisestigs (a (5 poits Describe a ijectio from the set of possible ways to est pairs
More informationf(1), and so, if f is continuous, f(x) = f(1)x.
2.2.35: Let f be a additive fuctio. i Clearly fx = fx ad therefore f x = fx for all Z+ ad x R. Hece, for ay, Z +, f = f, ad so, if f is cotiuous, fx = fx. ii Suppose that f is bouded o soe o-epty ope set.
More information42 Dependence and Bases
42 Depedece ad Bases The spa s(a) of a subset A i vector space V is a subspace of V. This spa ay be the whole vector space V (we say the A spas V). I this paragraph we study subsets A of V which spa V
More information(ii) Two-permutations of {a, b, c}. Answer. (B) P (3, 3) = 3! (C) 3! = 6, and there are 6 items in (A). ... Answer.
SOLUTIONS Homewor 5 Due /6/19 Exercise. (a Cosider the set {a, b, c}. For each of the followig, (A list the objects described, (B give a formula that tells you how may you should have listed, ad (C verify
More informationChapter 2. Asymptotic Notation
Asyptotic Notatio 3 Chapter Asyptotic Notatio Goal : To siplify the aalysis of ruig tie by gettig rid of details which ay be affected by specific ipleetatio ad hardware. [1] The Big Oh (O-Notatio) : It
More information6.3 Testing Series With Positive Terms
6.3. TESTING SERIES WITH POSITIVE TERMS 307 6.3 Testig Series With Positive Terms 6.3. Review of what is kow up to ow I theory, testig a series a i for covergece amouts to fidig the i= sequece of partial
More informationMATH 324 Summer 2006 Elementary Number Theory Solutions to Assignment 2 Due: Thursday July 27, 2006
MATH 34 Summer 006 Elemetary Number Theory Solutios to Assigmet Due: Thursday July 7, 006 Departmet of Mathematical ad Statistical Scieces Uiversity of Alberta Questio [p 74 #6] Show that o iteger of the
More informationPutnam Training Exercise Counting, Probability, Pigeonhole Principle (Answers)
Putam Traiig Exercise Coutig, Probability, Pigeohole Pricile (Aswers) November 24th, 2015 1. Fid the umber of iteger o-egative solutios to the followig Diohatie equatio: x 1 + x 2 + x 3 + x 4 + x 5 = 17.
More informationTeacher s Marking. Guide/Answers
WOLLONGONG COLLEGE AUSRALIA eacher s Markig A College of the Uiversity of Wollogog Guide/Aswers Diploa i Iforatio echology Fial Exaiatio Autu 008 WUC Discrete Matheatics his exa represets 60% of the total
More informationLecture Overview. 2 Permutations and Combinations. n(n 1) (n (k 1)) = n(n 1) (n k + 1) =
COMPSCI 230: Discrete Mathematics for Computer Sciece April 8, 2019 Lecturer: Debmalya Paigrahi Lecture 22 Scribe: Kevi Su 1 Overview I this lecture, we begi studyig the fudametals of coutig discrete objects.
More informationMA541 : Real Analysis. Tutorial and Practice Problems - 1 Hints and Solutions
MA54 : Real Aalysis Tutorial ad Practice Problems - Hits ad Solutios. Suppose that S is a oempty subset of real umbers that is bouded (i.e. bouded above as well as below). Prove that if S sup S. What ca
More informationDouble Derangement Permutations
Ope Joural of iscrete Matheatics, 206, 6, 99-04 Published Olie April 206 i SciRes http://wwwscirporg/joural/ojd http://dxdoiorg/04236/ojd2066200 ouble erageet Perutatios Pooya aeshad, Kayar Mirzavaziri
More informationMATH10040 Chapter 4: Sets, Functions and Counting
MATH10040 Chapter 4: Sets, Fuctios ad Coutig 1. The laguage of sets Iforally, a set is ay collectio of objects. The objects ay be atheatical objects such as ubers, fuctios ad eve sets, or letters or sybols
More informationECE 901 Lecture 4: Estimation of Lipschitz smooth functions
ECE 9 Lecture 4: Estiatio of Lipschitz sooth fuctios R. Nowak 5/7/29 Cosider the followig settig. Let Y f (X) + W, where X is a rado variable (r.v.) o X [, ], W is a r.v. o Y R, idepedet of X ad satisfyig
More informationand each factor on the right is clearly greater than 1. which is a contradiction, so n must be prime.
MATH 324 Summer 200 Elemetary Number Theory Solutios to Assigmet 2 Due: Wedesday July 2, 200 Questio [p 74 #6] Show that o iteger of the form 3 + is a prime, other tha 2 = 3 + Solutio: If 3 + is a prime,
More informationMath 155 (Lecture 3)
Math 55 (Lecture 3) September 8, I this lecture, we ll cosider the aswer to oe of the most basic coutig problems i combiatorics Questio How may ways are there to choose a -elemet subset of the set {,,,
More informationLet us consider the following problem to warm up towards a more general statement.
Lecture 4: Sequeces with repetitios, distributig idetical objects amog distict parties, the biomial theorem, ad some properties of biomial coefficiets Refereces: Relevat parts of chapter 15 of the Math
More informationMath 475, Problem Set #12: Answers
Math 475, Problem Set #12: Aswers A. Chapter 8, problem 12, parts (b) ad (d). (b) S # (, 2) = 2 2, sice, from amog the 2 ways of puttig elemets ito 2 distiguishable boxes, exactly 2 of them result i oe
More informationSequence A sequence is a function whose domain of definition is the set of natural numbers.
Chapter Sequeces Course Title: Real Aalysis Course Code: MTH3 Course istructor: Dr Atiq ur Rehma Class: MSc-I Course URL: wwwmathcityorg/atiq/fa8-mth3 Sequeces form a importat compoet of Mathematical Aalysis
More informationBernoulli Polynomials Talks given at LSBU, October and November 2015 Tony Forbes
Beroulli Polyoials Tals give at LSBU, October ad Noveber 5 Toy Forbes Beroulli Polyoials The Beroulli polyoials B (x) are defied by B (x), Thus B (x) B (x) ad B (x) x, B (x) x x + 6, B (x) dx,. () B 3
More informationIntermediate Math Circles November 4, 2009 Counting II
Uiversity of Waterloo Faculty of Mathematics Cetre for Educatio i Mathematics ad Computig Itermediate Math Circles November 4, 009 Coutig II Last time, after lookig at the product rule ad sum rule, we
More information3sin A 1 2sin B. 3π x is a solution. 1. If A and B are acute positive angles satisfying the equation 3sin A 2sin B 1 and 3sin 2A 2sin 2B 0, then A 2B
1. If A ad B are acute positive agles satisfyig the equatio 3si A si B 1 ad 3si A si B 0, the A B (a) (b) (c) (d) 6. 3 si A + si B = 1 3si A 1 si B 3 si A = cosb Also 3 si A si B = 0 si B = 3 si A Now,
More informationFind a formula for the exponential function whose graph is given , 1 2,16 1, 6
Math 4 Activity (Due by EOC Apr. ) Graph the followig epoetial fuctios by modifyig the graph of f. Fid the rage of each fuctio.. g. g. g 4. g. g 6. g Fid a formula for the epoetial fuctio whose graph is
More information18th Bay Area Mathematical Olympiad. Problems and Solutions. February 23, 2016
18th Bay Area Mathematical Olympiad February 3, 016 Problems ad Solutios BAMO-8 ad BAMO-1 are each 5-questio essay-proof exams, for middle- ad high-school studets, respectively. The problems i each exam
More informationMath 2112 Solutions Assignment 5
Math 2112 Solutios Assigmet 5 5.1.1 Idicate which of the followig relatioships are true ad which are false: a. Z Q b. R Q c. Q Z d. Z Z Z e. Q R Q f. Q Z Q g. Z R Z h. Z Q Z a. True. Every positive iteger
More informationInfinite Sequences and Series
Chapter 6 Ifiite Sequeces ad Series 6.1 Ifiite Sequeces 6.1.1 Elemetary Cocepts Simply speakig, a sequece is a ordered list of umbers writte: {a 1, a 2, a 3,...a, a +1,...} where the elemets a i represet
More information[ 47 ] then T ( m ) is true for all n a. 2. The greatest integer function : [ ] is defined by selling [ x]
[ 47 ] Number System 1. Itroductio Pricile : Let { T ( ) : N} be a set of statemets, oe for each atural umber. If (i), T ( a ) is true for some a N ad (ii) T ( k ) is true imlies T ( k 1) is true for all
More informationX. Perturbation Theory
X. Perturbatio Theory I perturbatio theory, oe deals with a ailtoia that is coposed Ĥ that is typically exactly solvable of two pieces: a referece part ad a perturbatio ( Ĥ ) that is assued to be sall.
More informationChapter 6 Infinite Series
Chapter 6 Ifiite Series I the previous chapter we cosidered itegrals which were improper i the sese that the iterval of itegratio was ubouded. I this chapter we are goig to discuss a topic which is somewhat
More informationExercises 1 Sets and functions
Exercises 1 Sets ad fuctios HU Wei September 6, 018 1 Basics Set theory ca be made much more rigorous ad built upo a set of Axioms. But we will cover oly some heuristic ideas. For those iterested studets,
More information1.2 AXIOMATIC APPROACH TO PROBABILITY AND PROPERTIES OF PROBABILITY MEASURE 1.2 AXIOMATIC APPROACH TO PROBABILITY AND
NTEL- robability ad Distributios MODULE 1 ROBABILITY LECTURE 2 Topics 1.2 AXIOMATIC AROACH TO ROBABILITY AND ROERTIES OF ROBABILITY MEASURE 1.2.1 Iclusio-Exclusio Forula I the followig sectio we will discuss
More informationsin(n) + 2 cos(2n) n 3/2 3 sin(n) 2cos(2n) n 3/2 a n =
60. Ratio ad root tests 60.1. Absolutely coverget series. Defiitio 13. (Absolute covergece) A series a is called absolutely coverget if the series of absolute values a is coverget. The absolute covergece
More informationUNIVERSITY OF NORTHERN COLORADO MATHEMATICS CONTEST. First Round For all Colorado Students Grades 7-12 November 3, 2007
UNIVERSITY OF NORTHERN COLORADO MATHEMATICS CONTEST First Roud For all Colorado Studets Grades 7- November, 7 The positive itegers are,,, 4, 5, 6, 7, 8, 9,,,,. The Pythagorea Theorem says that a + b =
More information18.S34 (FALL, 2007) GREATEST INTEGER PROBLEMS. n + n + 1 = 4n + 2.
18.S34 (FALL, 007) GREATEST INTEGER PROBLEMS Note: We use the otatio x for the greatest iteger x, eve if the origial source used the older otatio [x]. 1. (48P) If is a positive iteger, prove that + + 1
More informationMA131 - Analysis 1. Workbook 2 Sequences I
MA3 - Aalysis Workbook 2 Sequeces I Autum 203 Cotets 2 Sequeces I 2. Itroductio.............................. 2.2 Icreasig ad Decreasig Sequeces................ 2 2.3 Bouded Sequeces..........................
More informationCombinatorics and Newton s theorem
INTRODUCTION TO MATHEMATICAL REASONING Key Ideas Worksheet 5 Combiatorics ad Newto s theorem This week we are goig to explore Newto s biomial expasio theorem. This is a very useful tool i aalysis, but
More information10.5 Positive Term Series: Comparison Tests Contemporary Calculus 1
0. Positive Term Series: Compariso Tests Cotemporary Calculus 0. POSITIVE TERM SERIES: COMPARISON TESTS This sectio discusses how to determie whether some series coverge or diverge by comparig them with
More informationPARTIAL DIFFERENTIAL EQUATIONS SEPARATION OF VARIABLES
Diola Bagayoko (0 PARTAL DFFERENTAL EQUATONS SEPARATON OF ARABLES. troductio As discussed i previous lectures, partial differetial equatios arise whe the depedet variale, i.e., the fuctio, varies with
More informationSolutions to Final Exam
Solutios to Fial Exam 1. Three married couples are seated together at the couter at Moty s Blue Plate Dier, occupyig six cosecutive seats. How may arragemets are there with o wife sittig ext to her ow
More informationObjective Mathematics
. If sum of '' terms of a sequece is give by S Tr ( )( ), the 4 5 67 r (d) 4 9 r is equal to : T. Let a, b, c be distict o-zero real umbers such that a, b, c are i harmoic progressio ad a, b, c are i arithmetic
More informationMA131 - Analysis 1. Workbook 3 Sequences II
MA3 - Aalysis Workbook 3 Sequeces II Autum 2004 Cotets 2.8 Coverget Sequeces........................ 2.9 Algebra of Limits......................... 2 2.0 Further Useful Results........................
More informationSequences I. Chapter Introduction
Chapter 2 Sequeces I 2. Itroductio A sequece is a list of umbers i a defiite order so that we kow which umber is i the first place, which umber is i the secod place ad, for ay atural umber, we kow which
More informationThe Structure of Z p when p is Prime
LECTURE 13 The Structure of Z p whe p is Prime Theorem 131 If p > 1 is a iteger, the the followig properties are equivalet (1) p is prime (2) For ay [0] p i Z p, the equatio X = [1] p has a solutio i Z
More informationUSA Mathematical Talent Search Round 3 Solutions Year 27 Academic Year
/3/27. Fill i each space of the grid with either a or a so that all sixtee strigs of four cosecutive umbers across ad dow are distict. You do ot eed to prove that your aswer is the oly oe possible; you
More informationRandomized Algorithms I, Spring 2018, Department of Computer Science, University of Helsinki Homework 1: Solutions (Discussed January 25, 2018)
Radomized Algorithms I, Sprig 08, Departmet of Computer Sciece, Uiversity of Helsiki Homework : Solutios Discussed Jauary 5, 08). Exercise.: Cosider the followig balls-ad-bi game. We start with oe black
More informationSolutions. tan 2 θ(tan 2 θ + 1) = cot6 θ,
Solutios 99. Let A ad B be two poits o a parabola with vertex V such that V A is perpedicular to V B ad θ is the agle betwee the chord V A ad the axis of the parabola. Prove that V A V B cot3 θ. Commet.
More informationBooks Recommended for Further Reading
Books Recommeded for Further Readig by 8.5..8 o 0//8. For persoal use oly.. K. P. Bogart, Itroductory Combiatorics rd ed., S. I. Harcourt Brace College Publishers, 998.. R. A. Brualdi, Itroductory Combiatorics
More informationGENERALIZATIONS OF ZECKENDORFS THEOREM. TilVIOTHY J. KELLER Student, Harvey Mudd College, Claremont, California
GENERALIZATIONS OF ZECKENDORFS THEOREM TilVIOTHY J. KELLER Studet, Harvey Mudd College, Claremot, Califoria 91711 The Fiboacci umbers F are defied by the recurrece relatio Fi = F 2 = 1, F = F - + F 0 (
More informationAl Lehnen Madison Area Technical College 10/5/2014
The Correlatio of Two Rado Variables Page Preliiary: The Cauchy-Schwarz-Buyakovsky Iequality For ay two sequeces of real ubers { a } ad = { b } =, the followig iequality is always true. Furtherore, equality
More informationPermutations & Combinations. Dr Patrick Chan. Multiplication / Addition Principle Inclusion-Exclusion Principle Permutation / Combination
Discrete Mathematic Chapter 3: C outig 3. The Basics of Coutig 3.3 Permutatios & Combiatios 3.5 Geeralized Permutatios & Combiatios 3.6 Geeratig Permutatios & Combiatios Dr Patrick Cha School of Computer
More informationCS 336. of n 1 objects with order unimportant but repetition allowed.
CS 336. The importat issue is the logic you used to arrive at your aswer.. Use extra paper to determie your solutios the eatly trascribe them oto these sheets. 3. Do ot submit the scratch sheets. However,
More information2.4 Sequences, Sequences of Sets
72 CHAPTER 2. IMPORTANT PROPERTIES OF R 2.4 Sequeces, Sequeces of Sets 2.4.1 Sequeces Defiitio 2.4.1 (sequece Let S R. 1. A sequece i S is a fuctio f : K S where K = { N : 0 for some 0 N}. 2. For each
More informationOn a Smarandache problem concerning the prime gaps
O a Smaradache problem cocerig the prime gaps Felice Russo Via A. Ifate 7 6705 Avezzao (Aq) Italy felice.russo@katamail.com Abstract I this paper, a problem posed i [] by Smaradache cocerig the prime gaps
More informationThe Hypergeometric Coupon Collection Problem and its Dual
Joural of Idustrial ad Systes Egieerig Vol., o., pp -7 Sprig 7 The Hypergeoetric Coupo Collectio Proble ad its Dual Sheldo M. Ross Epstei Departet of Idustrial ad Systes Egieerig, Uiversity of Souther
More informationLangford s Problem. Moti Ben-Ari. Department of Science Teaching. Weizmann Institute of Science.
Lagford s Problem Moti Be-Ari Departmet of Sciece Teachig Weizma Istitute of Sciece http://www.weizma.ac.il/sci-tea/beari/ c 017 by Moti Be-Ari. This work is licesed uder the Creative Commos Attributio-ShareAlike
More informationMath 4707 Spring 2018 (Darij Grinberg): homework set 4 page 1
Math 4707 Sprig 2018 Darij Griberg): homewor set 4 page 1 Math 4707 Sprig 2018 Darij Griberg): homewor set 4 due date: Wedesday 11 April 2018 at the begiig of class, or before that by email or moodle Please
More informationSome Basic Counting Techniques
Some Basic Coutig Techiques Itroductio If A is a oempty subset of a fiite sample space S, the coceptually simplest way to fid the probability of A would be simply to apply the defiitio P (A) = s A p(s);
More information(A sequence also can be thought of as the list of function values attained for a function f :ℵ X, where f (n) = x n for n 1.) x 1 x N +k x N +4 x 3
MATH 337 Sequeces Dr. Neal, WKU Let X be a metric space with distace fuctio d. We shall defie the geeral cocept of sequece ad limit i a metric space, the apply the results i particular to some special
More informationHOMEWORK 2 SOLUTIONS
HOMEWORK SOLUTIONS CSE 55 RANDOMIZED AND APPROXIMATION ALGORITHMS 1. Questio 1. a) The larger the value of k is, the smaller the expected umber of days util we get all the coupos we eed. I fact if = k
More information4.3 Growth Rates of Solutions to Recurrences
4.3. GROWTH RATES OF SOLUTIONS TO RECURRENCES 81 4.3 Growth Rates of Solutios to Recurreces 4.3.1 Divide ad Coquer Algorithms Oe of the most basic ad powerful algorithmic techiques is divide ad coquer.
More informationSeunghee Ye Ma 8: Week 5 Oct 28
Week 5 Summary I Sectio, we go over the Mea Value Theorem ad its applicatios. I Sectio 2, we will recap what we have covered so far this term. Topics Page Mea Value Theorem. Applicatios of the Mea Value
More information1 (12 points) Red-Black trees and Red-Purple trees
CS6 Hoework 3 Due: 29 April 206, 2 oo Subit o Gradescope Haded out: 22 April 206 Istructios: Please aswer the followig questios to the best of your ability. If you are asked to desig a algorith, please
More informationUniversity of Colorado Denver Dept. Math. & Stat. Sciences Applied Analysis Preliminary Exam 13 January 2012, 10:00 am 2:00 pm. Good luck!
Uiversity of Colorado Dever Dept. Math. & Stat. Scieces Applied Aalysis Prelimiary Exam 13 Jauary 01, 10:00 am :00 pm Name: The proctor will let you read the followig coditios before the exam begis, ad
More information19.1 The dictionary problem
CS125 Lecture 19 Fall 2016 19.1 The dictioary proble Cosider the followig data structural proble, usually called the dictioary proble. We have a set of ites. Each ite is a (key, value pair. Keys are i
More information3.1 Counting Principles
3.1 Coutig Priciples Goal: Cout the umber of objects i a set. Notatio: Whe S is a set, S deotes the umber of objects i the set. This is also called S s cardiality. Additio Priciple: Whe you wat to cout
More informationStatistics for Applications Fall Problem Set 7
18.650. Statistics for Applicatios Fall 016. Proble Set 7 Due Friday, Oct. 8 at 1 oo Proble 1 QQ-plots Recall that the Laplace distributio with paraeter λ > 0 is the cotiuous probaλ bility easure with
More informationStream Ciphers (contd.) Debdeep Mukhopadhyay
Strea Ciphers (cotd.) Debdeep Mukhopadhyay Assistat Professor Departet of Coputer Sciece ad Egieerig Idia Istitute of Techology Kharagpur IDIA -7232 Objectives iear Coplexity Berlekap Massey Algorith ow
More informationSummer MA Lesson 13 Section 1.6, Section 1.7 (part 1)
Suer MA 1500 Lesso 1 Sectio 1.6, Sectio 1.7 (part 1) I Solvig Polyoial Equatios Liear equatio ad quadratic equatios of 1 variable are specific types of polyoial equatios. Soe polyoial equatios of a higher
More informationSEQUENCE AND SERIES NCERT
9. Overview By a sequece, we mea a arragemet of umbers i a defiite order accordig to some rule. We deote the terms of a sequece by a, a,..., etc., the subscript deotes the positio of the term. I view of
More informationMetric Dimension of Some Graphs under Join Operation
Global Joural of Pure ad Applied Matheatics ISSN 0973-768 Volue 3, Nuber 7 (07), pp 333-3348 Research Idia Publicatios http://wwwripublicatioco Metric Diesio of Soe Graphs uder Joi Operatio B S Rawat ad
More informationMath 113 Exam 3 Practice
Math Exam Practice Exam will cover.-.9. This sheet has three sectios. The first sectio will remid you about techiques ad formulas that you should kow. The secod gives a umber of practice questios for you
More informationA PROBABILITY PROBLEM
A PROBABILITY PROBLEM A big superarket chai has the followig policy: For every Euros you sped per buy, you ear oe poit (suppose, e.g., that = 3; i this case, if you sped 8.45 Euros, you get two poits,
More informationpage Suppose that S 0, 1 1, 2.
page 10 1. Suppose that S 0, 1 1,. a. What is the set of iterior poits of S? The set of iterior poits of S is 0, 1 1,. b. Give that U is the set of iterior poits of S, evaluate U. 0, 1 1, 0, 1 1, S. The
More informationOptimal Estimator for a Sample Set with Response Error. Ed Stanek
Optial Estiator for a Saple Set wit Respose Error Ed Staek Itroductio We develop a optial estiator siilar to te FP estiator wit respose error tat was cosidered i c08ed63doc Te first 6 pages of tis docuet
More informationBinomial transform of products
Jauary 02 207 Bioial trasfor of products Khristo N Boyadzhiev Departet of Matheatics ad Statistics Ohio Norther Uiversity Ada OH 4580 USA -boyadzhiev@ouedu Abstract Give the bioial trasfors { b } ad {
More informationContents Two Sample t Tests Two Sample t Tests
Cotets 3.5.3 Two Saple t Tests................................... 3.5.3 Two Saple t Tests Setup: Two Saples We ow focus o a sceario where we have two idepedet saples fro possibly differet populatios. Our
More informationSquare-Congruence Modulo n
Square-Cogruece Modulo Abstract This paper is a ivestigatio of a equivalece relatio o the itegers that was itroduced as a exercise i our Discrete Math class. Part I - Itro Defiitio Two itegers are Square-Cogruet
More informationLecture 10: Bounded Linear Operators and Orthogonality in Hilbert Spaces
Lecture : Bouded Liear Operators ad Orthogoality i Hilbert Spaces 34 Bouded Liear Operator Let ( X, ), ( Y, ) i i be ored liear vector spaces ad { } X Y The, T is said to be bouded if a real uber c such
More informationInduction: Solutions
Writig Proofs Misha Lavrov Iductio: Solutios Wester PA ARML Practice March 6, 206. Prove that a 2 2 chessboard with ay oe square removed ca always be covered by shaped tiles. Solutio : We iduct o. For
More information( ) GENERATING FUNCTIONS
GENERATING FUNCTIONS Solve a ifiite umber of related problems i oe swoop. *Code the problems, maipulate the code, the decode the aswer! Really a algebraic cocept but ca be eteded to aalytic basis for iterestig
More informationREVIEW OF CALCULUS Herman J. Bierens Pennsylvania State University (January 28, 2004) x 2., or x 1. x j. ' ' n i'1 x i well.,y 2
REVIEW OF CALCULUS Hera J. Bieres Pesylvaia State Uiversity (Jauary 28, 2004) 1. Suatio Let x 1,x 2,...,x e a sequece of uers. The su of these uers is usually deoted y x 1 % x 2 %...% x ' j x j, or x 1
More informationCounting Well-Formed Parenthesizations Easily
Coutig Well-Formed Parethesizatios Easily Pekka Kilpeläie Uiversity of Easter Filad School of Computig, Kuopio August 20, 2014 Abstract It is well kow that there is a oe-to-oe correspodece betwee ordered
More informationMathematical Preliminaries
Matheatical Preliiaries I this chapter we ll review soe atheatical cocepts that will be used throughout this course. We ll also lear soe ew atheatical otatios ad techiques that are iportat for aalysis
More informationSequences A sequence of numbers is a function whose domain is the positive integers. We can see that the sequence
Sequeces A sequece of umbers is a fuctio whose domai is the positive itegers. We ca see that the sequece 1, 1, 2, 2, 3, 3,... is a fuctio from the positive itegers whe we write the first sequece elemet
More informationARITHMETIC PROGRESSIONS
CHAPTER 5 ARITHMETIC PROGRESSIONS (A) Mai Cocepts ad Results A arithmetic progressio (AP) is a list of umbers i which each term is obtaied by addig a fixed umber d to the precedig term, except the first
More informationIntroduction to Probability. Ariel Yadin. Lecture 2
Itroductio to Probability Ariel Yadi Lecture 2 1. Discrete Probability Spaces Discrete probability spaces are those for which the sample space is coutable. We have already see that i this case we ca take
More informationCIS Spring 2018 (instructor Val Tannen)
CIS 160 - Sprig 2018 (istructor Val Tae) Lecture 5 Thursday, Jauary 25 COUNTING We cotiue studyig how to use combiatios ad what are their properties. Example 5.1 How may 8-letter strigs ca be costructed
More informationDifferent kinds of Mathematical Induction
Differet ids of Mathematical Iductio () Mathematical Iductio Give A N, [ A (a A a A)] A N () (First) Priciple of Mathematical Iductio Let P() be a propositio (ope setece), if we put A { : N p() is true}
More informationSequences and Series of Functions
Chapter 6 Sequeces ad Series of Fuctios 6.1. Covergece of a Sequece of Fuctios Poitwise Covergece. Defiitio 6.1. Let, for each N, fuctio f : A R be defied. If, for each x A, the sequece (f (x)) coverges
More information[ 11 ] z of degree 2 as both degree 2 each. The degree of a polynomial in n variables is the maximum of the degrees of its terms.
[ 11 ] 1 1.1 Polyomial Fuctios 1 Algebra Ay fuctio f ( x) ax a1x... a1x a0 is a polyomial fuctio if ai ( i 0,1,,,..., ) is a costat which belogs to the set of real umbers ad the idices,, 1,...,1 are atural
More informationThe Random Walk For Dummies
The Radom Walk For Dummies Richard A Mote Abstract We look at the priciples goverig the oe-dimesioal discrete radom walk First we review five basic cocepts of probability theory The we cosider the Beroulli
More information