CSE 21 Mathematics for
|
|
- Percival O’Neal’
- 6 years ago
- Views:
Transcription
1 CSE 2 Mathematics for Algorithm ad System Aalysis Summer, 2005 Outlie What a geeratig fuctio is How to create a geeratig fuctio to model a problem Fidig the desired coefficiet Partitios Expoetial geeratig fuctios Day 8 Geeratig Fuctios Istructor: Neil Rhodes 2 (Ordiary Geeratig Fuctios A differet way to solve problems ivolvig selectio ad arragemet problems with repetitio Ca easily support special costraits ot easy to do otherwise Fid the umber of ways to distribute 70 idetical objects ito 9 boxes with a odd umber betwee 3 ad 7 i the first box, a eve umber betwee 4 ad 20 i the secod box, ad at most 3 i the other boxes. Idea Create a geeratig fuctio g(x which models the problem of selectig r objects. The coefficiet of x r i g(x couts the umber of ways to select r objects g(x = a0 + ax + a2x arx r ax g(x = (+x models the umber of ways to select r objects from objects g(x = ( + x = x x + + x r + + r x Let s look at the coefficiets of x 5 i (+x+x 2 4 Ways to create x 5 : x 0 x 2 x 2 x x x x x 2 etc. I geeral, x e x e 2 x e 3 x e 4 where 0 e, e2, e3, e4 2 Ca be looked at as the umber of iteger solutios to: e + e2 + e3 + e4 = 5 (where 0 e, e2, e3, e4 2 Which ca also be looked at as the umber of ways to place 5 balls i four boxes with at most 2 balls i each box Which ca also be looked at as the umber of ways to select 5 objects from a collectio of 4 differet types of objects (with two idetical objects of each type. g(x = (+x+x 2 4 is a geeratig fuctio for the above problems Actually, g(x is a geeratig fuctio for the problem with ay umber, r, of {balls, objects}, ot just 5 We always write a geeral geeratig fuctio for ay r 3 4
2 Fid a geeratig fuctio for ar, the umber of ways to select r balls from a pile of three gree, three white, three blue, ad three gold balls Same as umber of iteger solutios to: Fid a geeratig fuctio to model coutig all selectios of six objects chose from three types of objects with: repetitio of up to four objects of each type e + e2 + e3 + e4 = r 0 ei 3 ulimited repetitio 5 6 Fid a geeratig fuctio to model: the umber of ways to distribute r idetical objects ito five boxes with a eve umber of objects ot exceedig 0 i the first two boxes ad betwee 3 ad 5 i the other boxes. Fid a geeratig fuctio for ar, the umber of ways a roll of six distict dice ca show a sum of r if The first three dice are odd ad the secod three eve The ith die does ot show a value of i 7 8
3 Fidig the Desired Coefficiet Reduce complicated geeratig fuctio to simple fuctio or product of simple fuctios The simple fuctios: x m+ x = + x + x x m x = + x + x2 + ( + x = 0 x 0 + x + + ( x m = x m + x r + + r x ( ( x 2m + + ( k 2 k x km + + x m If Product of Geeratig Fuctios f(x = a0 + ax + a2x 2 + g(x = b0 + bx + b2x 2 + ad h(x = f(xg(x the h(x = a0b0 + (a0b + ab0x + (a0b2 + ab + a2b0x (a0br + abr- + + ar-b + arb0x r + ( x = + ( + x + ( x ( r + r x r Fid the coefficiet of x 6 i (x 2 + x 3 + x Rewrite as product of simpler fuctios Fid the umber of ways to collect $5 from 20 distict people if each of the first 9 people ca give $0 or $ ad the 20th perso ca give either $0 or $ or $5 : g(x = Fial simplificatio: 2
4 How may ways to distribute 25 idetical balls ito seve distict boxes if the first box ca have o more tha 0 balls? How may ways are there to select 25 toys from seve types of toys with betwee two ad six of each type? 3 4 Defiitio Partitios A partitio of r idetical objects divides the group ito a collectio (uordered subsets of various sizes (Or, a group of positive itegers whose sum is r Partitios of 5 We ll be lookig at creatig geeratig fuctios for partitios Warig No easy way to calculate the coefficiets Geeratig Fuctio for Partitio for ar Wat to kow e: how may s e2: how may 2 s e + 2e2 + + rer = r Choose oes: ( + x + x 2 + x 3 + x + = /-x Choose twos ( + x 2 + x x 2 + = /(-x 2 Choose threes ( + x 3 + x x 3 + = /(-x 3 ad so o. g(x = product of all those = /(-x(-x 2 (-x 3 (-x r 5 6
5 Fid a geeratig fuctio for the umber of ways to express r as a sum of distict itegers Show with geeratig fuctios that every iteger ca be expressed as a uique sum of distict powers of 2 To show: g(x = + x + x 2 + x Ferrers Diagram Graphically displays a partitio of r dots I a set of rows of decreasig size Show that the umber of partitios of a iteger r as a sum of m positive itegers is equal to the umber of partitios of r as a sum of positive itegers, the largest of which is m Cojugate: Rotate 90 degrees couter-clockwise 9 20
6 Expoetial Geeratig Fuctios Used to model problems ivolvig arragemets ad distributios of distict objects problem fid umber of differet arragemets of four letters, choosig from a s, b s, ad c s with at least two a s. Solutio Four letter combiatios: {a, a, a, a}: 4/400 {a,a,a,b}:4/30 {a,a,a,c}:4/30 {a,a,b,b}:4/220 {a,a,b,c}:4/2 {a,a,c,c}:4/20 Total is sum, so we wat that sum as the coefficiet of x 4 Ordiary geeratig fuctio ca solve umber of iteger solutios to: e + e2 + e3 = 4 e"2 e2, e3 "0 But, we wat each solutio to above problem to cotribute: (e+e2+e3/ee2e3 Expoetial Geeratig Fuctio Fuctio of form: g(x = a0 + ax + a2x 2 /2 + a3x 3 /3 + + arx r /r + g(x = (x 2 /2 + x 3 /3 + x 4 /4 + (+x+x 2 /2 + x 3 /3 + 2 Coefficiet of xr i g(x is equal to: e+e2+e3=r Coefficiet of xr/r i g(x is equal to: e+e2+e3=r e e 2 e 3 r e e 2 e Fid the expoetial geeratig fuctio for ar, the umber of differet arragemets of r objects chose from four differet type of objects with each type of object appearig at least 2 ad o more tha 5 times Fid the umber of ways to place r (distict people ito three rooms with at least oe perso i each room 23 24
7 Useful idetities e x = + x + 2 x 2 Fidig Desired Coefficiets e x = + x + x2 2 + x3 xr r x r x r + r 2 (ex + e x = + x2 2 + x4 4 + x Fid the umber of differet r arragemets of objects chose from ulimited supplies of types of objects 2 (ex e x = x + x3 3 + x5 5 + x Fid the umber of ways to place 25 people ito 3 rooms with at least oe perso i each room Fid the umber of r-digit quaterary sequeces (digits: 0,, 2, 3 with a eve umber of 0 s ad a odd umber of s 27 28
( ) 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 informationGenerating Functions. 1 Operations on generating functions
Geeratig Fuctios The geeratig fuctio for a sequece a 0, a,..., a,... is defied to be the power series fx a x. 0 We say that a 0, a,... is the sequece geerated by fx ad a is the coefficiet of x. Example
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 information1 Generating functions for balls in boxes
Math 566 Fall 05 Some otes o geeratig fuctios Give a sequece a 0, a, a,..., a,..., a geeratig fuctio some way of represetig the sequece as a fuctio. There are may ways to do this, with the most commo ways
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 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 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 informationChapter 6. Advanced Counting Techniques
Chapter 6 Advaced Coutig Techiques 6.: Recurrece Relatios Defiitio: A recurrece relatio for the sequece {a } is a equatio expressig a i terms of oe or more of the previous terms of the sequece: a,a2,a3,,a
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 informationNICK DUFRESNE. 1 1 p(x). To determine some formulas for the generating function of the Schröder numbers, r(x) = a(x) =
AN INTRODUCTION TO SCHRÖDER AND UNKNOWN NUMBERS NICK DUFRESNE Abstract. I this article we will itroduce two types of lattice paths, Schröder paths ad Ukow paths. We will examie differet properties of each,
More informationIntroduction To Discrete Mathematics
Itroductio To Discrete Mathematics Review If you put + pigeos i pigeoholes the at least oe hole would have more tha oe pigeo. If (r + objects are put ito boxes, the at least oe of the boxes cotais r or
More informationLecture 6 April 10. We now give two identities which are useful for simplifying sums of binomial coefficients.
Lecture 6 April 0 As aother applicatio of the biomial theorem we have the followig. For 0. 0 I other words startig with the secod row ad goig dow if we sum alog the rows alteratig sig as we go the result
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 informationWorksheet on Generating Functions
Worksheet o Geeratig Fuctios October 26, 205 This worksheet is adapted from otes/exercises by Nat Thiem. Derivatives of Geeratig Fuctios. If the sequece a 0, a, a 2,... has ordiary geeratig fuctio A(x,
More informationDavid Vella, Skidmore College.
David Vella, Skidmore College dvella@skidmore.edu Geeratig Fuctios ad Expoetial Geeratig Fuctios Give a sequece {a } we ca associate to it two fuctios determied by power series: Its (ordiary) geeratig
More informationx c the remainder is Pc ().
Algebra, Polyomial ad Ratioal Fuctios Page 1 K.Paulk Notes Chapter 3, Sectio 3.1 to 3.4 Summary Sectio Theorem Notes 3.1 Zeros of a Fuctio Set the fuctio to zero ad solve for x. The fuctio is zero at these
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 informationPermutations, Combinations, and the Binomial Theorem
Permutatios, ombiatios, ad the Biomial Theorem Sectio Permutatios outig methods are used to determie the umber of members of a specific set as well as outcomes of a evet. There are may differet ways to
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 informationZeros of Polynomials
Math 160 www.timetodare.com 4.5 4.6 Zeros of Polyomials I these sectios we will study polyomials algebraically. Most of our work will be cocered with fidig the solutios of polyomial equatios of ay degree
More informationsubcaptionfont+=small,labelformat=parens,labelsep=space,skip=6pt,list=0,hypcap=0 subcaption ALGEBRAIC COMBINATORICS LECTURE 8 TUESDAY, 2/16/2016
subcaptiofot+=small,labelformat=pares,labelsep=space,skip=6pt,list=0,hypcap=0 subcaptio ALGEBRAIC COMBINATORICS LECTURE 8 TUESDAY, /6/06. Self-cojugate Partitios Recall that, give a partitio λ, we may
More informationGenerating Functions II
Geeratig Fuctios II Misha Lavrov ARML Practice 5/4/2014 Warm-up problems 1. Solve the recursio a +1 = 2a, a 0 = 1 by usig commo sese. 2. Solve the recursio b +1 = 2b + 1, b 0 = 1 by usig commo sese ad
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 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 information4.1 Sigma Notation and Riemann Sums
0 the itegral. Sigma Notatio ad Riema Sums Oe strategy for calculatig the area of a regio is to cut the regio ito simple shapes, calculate the area of each simple shape, ad the add these smaller areas
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 informationLecture 7: Properties of Random Samples
Lecture 7: Properties of Radom Samples 1 Cotiued From Last Class Theorem 1.1. Let X 1, X,...X be a radom sample from a populatio with mea µ ad variace σ
More informationIt is always the case that unions, intersections, complements, and set differences are preserved by the inverse image of a function.
MATH 532 Measurable Fuctios Dr. Neal, WKU Throughout, let ( X, F, µ) be a measure space ad let (!, F, P ) deote the special case of a probability space. We shall ow begi to study real-valued fuctios defied
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 informationBertrand s Postulate
Bertrad s Postulate Lola Thompso Ross Program July 3, 2009 Lola Thompso (Ross Program Bertrad s Postulate July 3, 2009 1 / 33 Bertrad s Postulate I ve said it oce ad I ll say it agai: There s always a
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 informationMATH 304: MIDTERM EXAM SOLUTIONS
MATH 304: MIDTERM EXAM SOLUTIONS [The problems are each worth five poits, except for problem 8, which is worth 8 poits. Thus there are 43 possible poits.] 1. Use the Euclidea algorithm to fid the greatest
More informationReview Problems 1. ICME and MS&E Refresher Course September 19, 2011 B = C = AB = A = A 2 = A 3... C 2 = C 3 = =
Review Problems ICME ad MS&E Refresher Course September 9, 0 Warm-up problems. For the followig matrices A = 0 B = C = AB = 0 fid all powers A,A 3,(which is A times A),... ad B,B 3,... ad C,C 3,... Solutio:
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 information4.1 SIGMA NOTATION AND RIEMANN SUMS
.1 Sigma Notatio ad Riema Sums Cotemporary Calculus 1.1 SIGMA NOTATION AND RIEMANN SUMS Oe strategy for calculatig the area of a regio is to cut the regio ito simple shapes, calculate the area of each
More informationChapter 7 COMBINATIONS AND PERMUTATIONS. where we have the specific formula for the binomial coefficients:
Chapter 7 COMBINATIONS AND PERMUTATIONS We have see i the previous chapter that (a + b) ca be writte as 0 a % a & b%þ% a & b %þ% b where we have the specific formula for the biomial coefficiets: '!!(&)!
More informationIP Reference guide for integer programming formulations.
IP Referece guide for iteger programmig formulatios. by James B. Orli for 15.053 ad 15.058 This documet is iteded as a compact (or relatively compact) guide to the formulatio of iteger programs. For more
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 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 informationEnd-of-Year Contest. ERHS Math Club. May 5, 2009
Ed-of-Year Cotest ERHS Math Club May 5, 009 Problem 1: There are 9 cois. Oe is fake ad weighs a little less tha the others. Fid the fake coi by weighigs. Solutio: Separate the 9 cois ito 3 groups (A, B,
More informationChapter 1 : Combinatorial Analysis
STAT/MATH 394 A - PROBABILITY I UW Autum Quarter 205 Néhémy Lim Chapter : Combiatorial Aalysis A major brach of combiatorial aalysis called eumerative combiatorics cosists of studyig methods for coutig
More informationM A T H F A L L CORRECTION. Algebra I 1 4 / 1 0 / U N I V E R S I T Y O F T O R O N T O
M A T H 2 4 0 F A L L 2 0 1 4 HOMEWORK ASSIGNMENT #4 CORRECTION Algebra I 1 4 / 1 0 / 2 0 1 4 U N I V E R S I T Y O F T O R O N T O P r o f e s s o r : D r o r B a r - N a t a Correctio Homework Assigmet
More information1 Summary: Binary and Logic
1 Summary: Biary ad Logic Biary Usiged Represetatio : each 1-bit is a power of two, the right-most is for 2 0 : 0110101 2 = 2 5 + 2 4 + 2 2 + 2 0 = 32 + 16 + 4 + 1 = 53 10 Usiged Rage o bits is [0...2
More information,... are the terms of the sequence. If the domain consists of the first n positive integers only, the sequence is a finite sequence.
Chpter 9 & 0 FITZGERALD MAT 50/5 SECTION 9. Sequece Defiitio A ifiite sequece is fuctio whose domi is the set of positive itegers. The fuctio vlues,,, 4,...,,... re the terms of the sequece. If the domi
More informationComplex Numbers Solutions
Complex Numbers Solutios Joseph Zoller February 7, 06 Solutios. (009 AIME I Problem ) There is a complex umber with imagiary part 64 ad a positive iteger such that Fid. [Solutio: 697] 4i + + 4i. 4i 4i
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 informationThe Binomial Theorem
The Biomial Theorem Lecture 47 Sectio 9.7 Robb T. Koether Hampde-Sydey College Fri, Apr 8, 204 Robb T. Koether (Hampde-Sydey College The Biomial Theorem Fri, Apr 8, 204 / 25 Combiatios 2 Pascal s Triagle
More informationA 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,, 2, 2, 3, 3,... is a fuctio from the positive itegers whe we write the first sequece elemet as
More informationMath 2784 (or 2794W) University of Connecticut
ORDERS OF GROWTH PAT SMITH Math 2784 (or 2794W) Uiversity of Coecticut Date: Mar. 2, 22. ORDERS OF GROWTH. Itroductio Gaiig a ituitive feel for the relative growth of fuctios is importat if you really
More informationMath 61CM - Solutions to homework 3
Math 6CM - Solutios to homework 3 Cédric De Groote October 2 th, 208 Problem : Let F be a field, m 0 a fixed oegative iteger ad let V = {a 0 + a x + + a m x m a 0,, a m F} be the vector space cosistig
More information6. Uniform distribution mod 1
6. Uiform distributio mod 1 6.1 Uiform distributio ad Weyl s criterio Let x be a seuece of real umbers. We may decompose x as the sum of its iteger part [x ] = sup{m Z m x } (i.e. the largest iteger which
More informationACCESS TO SCIENCE, ENGINEERING AND AGRICULTURE: MATHEMATICS 1 MATH00030 SEMESTER / Statistics
ACCESS TO SCIENCE, ENGINEERING AND AGRICULTURE: MATHEMATICS 1 MATH00030 SEMESTER 1 018/019 DR. ANTHONY BROWN 8. Statistics 8.1. Measures of Cetre: Mea, Media ad Mode. If we have a series of umbers the
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 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 informationA Combinatoric Proof and Generalization of Ferguson s Formula for k-generalized Fibonacci Numbers
Jue 5 00 A Combiatoric Proof ad Geeralizatio of Ferguso s Formula for k-geeralized Fiboacci Numbers David Kessler 1 ad Jeremy Schiff 1 Departmet of Physics Departmet of Mathematics Bar-Ila Uiversity, Ramat
More informationUnit 4: Polynomial and Rational Functions
48 Uit 4: Polyomial ad Ratioal Fuctios Polyomial Fuctios A polyomial fuctio y px ( ) is a fuctio of the form p( x) ax + a x + a x +... + ax + ax+ a 1 1 1 0 where a, a 1,..., a, a1, a0are real costats ad
More informationLecture 10: Mathematical Preliminaries
Lecture : Mathematical Prelimiaries Obective: Reviewig mathematical cocepts ad tools that are frequetly used i the aalysis of algorithms. Lecture # Slide # I this
More informationHOMEWORK #10 SOLUTIONS
Math 33 - Aalysis I Sprig 29 HOMEWORK # SOLUTIONS () Prove that the fuctio f(x) = x 3 is (Riema) itegrable o [, ] ad show that x 3 dx = 4. (Without usig formulae for itegratio that you leart i previous
More informationProperties and Tests of Zeros of Polynomial Functions
Properties ad Tests of Zeros of Polyomial Fuctios The Remaider ad Factor Theorems: Sythetic divisio ca be used to fid the values of polyomials i a sometimes easier way tha substitutio. This is show by
More informationCALCULATION OF FIBONACCI VECTORS
CALCULATION OF FIBONACCI VECTORS Stuart D. Aderso Departmet of Physics, Ithaca College 953 Daby Road, Ithaca NY 14850, USA email: saderso@ithaca.edu ad Dai Novak Departmet of Mathematics, Ithaca College
More informationTrial division, Pollard s p 1, Pollard s ρ, and Fermat s method. Christopher Koch 1. April 8, 2014
Iteger Divisio Algorithm ad Cogruece Iteger Trial divisio,,, ad with itegers mod Iverses mod Multiplicatio ad GCD Iteger Christopher Koch 1 1 Departmet of Computer Sciece ad Egieerig CSE489/589 Algorithms
More informationLecture 6 Chi Square Distribution (χ 2 ) and Least Squares Fitting
Lecture 6 Chi Square Distributio (χ ) ad Least Squares Fittig Chi Square Distributio (χ ) Suppose: We have a set of measuremets {x 1, x, x }. We kow the true value of each x i (x t1, x t, x t ). We would
More informationARRANGEMENTS IN A CIRCLE
ARRANGEMENTS IN A CIRCLE Whe objects are arraged i a circle, the total umber of arragemets is reduced. The arragemet of (say) four people i a lie is easy ad o problem (if they liste of course!!). With
More informationSequences, Mathematical Induction, and Recursion. CSE 2353 Discrete Computational Structures Spring 2018
CSE 353 Discrete Computatioal Structures Sprig 08 Sequeces, Mathematical Iductio, ad Recursio (Chapter 5, Epp) Note: some course slides adopted from publisher-provided material Overview May mathematical
More informationLecture 6 Chi Square Distribution (χ 2 ) and Least Squares Fitting
Lecture 6 Chi Square Distributio (χ ) ad Least Squares Fittig Chi Square Distributio (χ ) Suppose: We have a set of measuremets {x 1, x, x }. We kow the true value of each x i (x t1, x t, x t ). We would
More information14 Classic Counting Problems in Combinatorics
14 Classic Coutig Problems i Combiatorics (Herbert E. Müller, May 2017, herbert-mueller.ifo) Combiatorics is about coutig objects, or the umber of ways of doig somethig. I this article I preset some classic
More informationMT5821 Advanced Combinatorics
MT5821 Advaced Combiatorics 1 Coutig subsets I this sectio, we cout the subsets of a -elemet set. The coutig umbers are the biomial coefficiets, familiar objects but there are some ew thigs to say about
More informationGoodness-of-Fit Tests and Categorical Data Analysis (Devore Chapter Fourteen)
Goodess-of-Fit Tests ad Categorical Data Aalysis (Devore Chapter Fourtee) MATH-252-01: Probability ad Statistics II Sprig 2019 Cotets 1 Chi-Squared Tests with Kow Probabilities 1 1.1 Chi-Squared Testig................
More informationIntroductions to PartitionsP
Itroductios to PartitiosP Itroductio to partitios Geeral Iterest i partitios appeared i the 7th cetury whe G. W. Leibiz (669) ivestigated the umber of ways a give positive iteger ca be decomposed ito a
More informationPERMUTATIONS AND COMBINATIONS
5. PERMUTATIONS AND COMBINATIONS 1. INTRODUCTION The mai subject of this chapter is coutig. Give a set of objects the problem is to arrage some or all of them accordig to some order or to select some or
More informationThe Ratio Test. THEOREM 9.17 Ratio Test Let a n be a series with nonzero terms. 1. a. n converges absolutely if lim. n 1
460_0906.qxd //04 :8 PM Page 69 SECTION 9.6 The Ratio ad Root Tests 69 Sectio 9.6 EXPLORATION Writig a Series Oe of the followig coditios guaratees that a series will diverge, two coditios guaratee that
More informationLinear Regression Analysis. Analysis of paired data and using a given value of one variable to predict the value of the other
Liear Regressio Aalysis Aalysis of paired data ad usig a give value of oe variable to predict the value of the other 5 5 15 15 1 1 5 5 1 3 4 5 6 7 8 1 3 4 5 6 7 8 Liear Regressio Aalysis E: The chirp rate
More information1. Hilbert s Grand Hotel. The Hilbert s Grand Hotel has infinite many rooms numbered 1, 2, 3, 4
. Hilbert s Grad Hotel The Hilbert s Grad Hotel has ifiite may rooms umbered,,,.. Situatio. The Hotel is full ad a ew guest arrives. Ca the mager accommodate the ew guest? - Yes, he ca. There is a simple
More information05 - PERMUTATIONS AND COMBINATIONS Page 1 ( Answers at the end of all questions )
05 - PERMUTATIONS AND COMBINATIONS Page 1 ( Aswers at the ed of all questios ) ( 1 ) If the letters of the word SACHIN are arraged i all possible ways ad these words are writte out as i dictioary, the
More informationLecture 2 Clustering Part II
COMS 4995: Usupervised Learig (Summer 8) May 24, 208 Lecture 2 Clusterig Part II Istructor: Nakul Verma Scribes: Jie Li, Yadi Rozov Today, we will be talkig about the hardess results for k-meas. More specifically,
More informationCombinatorics II. Combinatorics. Product Rule. Sum Rule II. Theorem (Product Rule) Theorem (Sum Rule)
Combiatorics Combiatorics I Slides by Christopher M. Bourke Istructor: Berthe Y. Choueiry Fall 27 Computer Sciece & Egieerig 235 to Discrete Mathematics Sectios 5.-5.6 & 7.5-7.6 of Rose cse235@cse.ul.edu
More informationFinal Review for MATH 3510
Fial Review for MATH 50 Calculatio 5 Give a fairly simple probability mass fuctio or probability desity fuctio of a radom variable, you should be able to compute the expected value ad variace of the variable
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 information3.2 Properties of Division 3.3 Zeros of Polynomials 3.4 Complex and Rational Zeros of Polynomials
Math 60 www.timetodare.com 3. Properties of Divisio 3.3 Zeros of Polyomials 3.4 Complex ad Ratioal Zeros of Polyomials I these sectios we will study polyomials algebraically. Most of our work will be cocered
More informationTHE UNIVERSITY OF TORONTO UNDERGRADUATE MATHEMATICS COMPETITION. In Memory of Robert Barrington Leigh. March 9, 2014
THE UNIVERSITY OF TORONTO UNDERGRADUATE MATHEMATICS COMPETITION I Memory of Robert Barrigto Leigh March 9, 4 Time: 3 hours No aids or calculators permitted. The gradig is desiged to ecourage oly the stroger
More informationMT5821 Advanced Combinatorics
MT5821 Advaced Combiatorics 9 Set partitios ad permutatios It could be said that the mai objects of iterest i combiatorics are subsets, partitios ad permutatios of a fiite set. We have spet some time coutig
More informationFLC Ch 8 & 9. Evaluate. Check work. a) b) c) d) e) f) g) h) i) j) k) l) m) n) o) 3. p) q) r) s) t) 3.
Math 100 Elemetary Algebra Sec 8.1: Radical Expressios List perfect squares ad evaluate their square root. Kow these perfect squares for test. Def The positive (pricipal) square root of x, writte x, is
More informationMath 172 Spring 2010 Haiman Notes on ordinary generating functions
Math 72 Sprig 200 Haima Notes o ordiary geeratig fuctios How do we cout with geeratig fuctios? May eumeratio problems which are ot so easy to hadle by elemetary meas ca be solved usig geeratig fuctios
More informationCS / MCS 401 Homework 3 grader solutions
CS / MCS 401 Homework 3 grader solutios assigmet due July 6, 016 writte by Jāis Lazovskis maximum poits: 33 Some questios from CLRS. Questios marked with a asterisk were ot graded. 1 Use the defiitio of
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 informationPROBLEMS ON ABSTRACT ALGEBRA
PROBLEMS ON ABSTRACT ALGEBRA 1 (Putam 197 A). Let S be a set ad let be a biary operatio o S satisfyig the laws x (x y) = y for all x, y i S, (y x) x = y for all x, y i S. Show that is commutative but ot
More informationMath 525: Lecture 5. January 18, 2018
Math 525: Lecture 5 Jauary 18, 2018 1 Series (review) Defiitio 1.1. A sequece (a ) R coverges to a poit L R (writte a L or lim a = L) if for each ǫ > 0, we ca fid N such that a L < ǫ for all N. If the
More informationAs stated by Laplace, Probability is common sense reduced to calculation.
Note: Hadouts DO NOT replace the book. I most cases, they oly provide a guidelie o topics ad a ituitive feel. The math details will be covered i class, so it is importat to atted class ad also you MUST
More informationCS 171 Lecture Outline October 09, 2008
CS 171 Lecture Outlie October 09, 2008 The followig theorem comes very hady whe calculatig the expectatio of a radom variable that takes o o-egative iteger values. Theorem: Let Y be a radom variable that
More informationApply change-of-basis formula to rewrite x as a linear combination of eigenvectors v j.
Eigevalue-Eigevector Istructor: Nam Su Wag eigemcd Ay vector i real Euclidea space of dimesio ca be uiquely epressed as a liear combiatio of liearly idepedet vectors (ie, basis) g j, j,,, α g α g α g α
More informationBasic Combinatorics. Math 40210, Section 01 Spring Homework 7 due Monday, March 26
Basic Combiatorics Math 40210, Sectio 01 Sprig 2012 Homewor 7 due Moday, March 26 Geeral iformatio: I ecourage you to tal with your colleagues about homewor problems, but your fial write-up must be your
More informationAN ALMOST LINEAR RECURRENCE. Donald E. Knuth Calif. Institute of Technology, Pasadena, Calif.
AN ALMOST LINEAR RECURRENCE Doald E. Kuth Calif. Istitute of Techology, Pasadea, Calif. form A geeral liear recurrece with costat coefficiets has the U 0 = a l* U l = a 2 " ' " U r - l = a r ; u = b, u,
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 information+ au n+1 + bu n = 0.)
Lecture 6 Recurreces - kth order: u +k + a u +k +... a k u k 0 where a... a k are give costats, u 0... u k are startig coditios. (Simple case: u + au + + bu 0.) How to solve explicitly - first, write characteristic
More informationOptimally Sparse SVMs
A. Proof of Lemma 3. We here prove a lower boud o the umber of support vectors to achieve geeralizatio bouds of the form which we cosider. Importatly, this result holds ot oly for liear classifiers, but
More information3. One pencil costs 25 cents, and we have 5 pencils, so the cost is 25 5 = 125 cents. 60 =
JHMMC 0 Grade Solutios October, 0. By coutig, there are 7 words i this questio.. + 4 + + 8 + 6 + 6.. Oe pecil costs cets, ad we have pecils, so the cost is cets. 4. A cube has edges.. + + 4 + 0 60 + 0
More informationThe picture in figure 1.1 helps us to see that the area represents the distance traveled. Figure 1: Area represents distance travelled
1 Lecture : Area Area ad distace traveled Approximatig area by rectagles Summatio The area uder a parabola 1.1 Area ad distace Suppose we have the followig iformatio about the velocity of a particle, how
More informationThis section is optional.
4 Momet Geeratig Fuctios* This sectio is optioal. The momet geeratig fuctio g : R R of a radom variable X is defied as g(t) = E[e tx ]. Propositio 1. We have g () (0) = E[X ] for = 1, 2,... Proof. Therefore
More informationStat 198 for 134 Instructor: Mike Leong
Chapter 2: Repeated Trials ad Samplig Sectio 2.1 Biomial Distributio 2.2 Normal Approximatio: Method 2.3 Normal Approximatios: Derivatio (Skip) 2.4 Poisso Approximatio 2.5 Radom Samplig Chapter 2 Table
More informationCourse : Algebraic Combinatorics
Course 8.32: Algebraic Combiatorics Lecture Notes # Addedum by Gregg Musier February 4th - 6th, 2009 Recurrece Relatios ad Geeratig Fuctios Give a ifiite sequece of umbers, a geeratig fuctio is a compact
More information