Math 475, Problem Set #6: Solutions


 Quentin Clyde Spencer
 2 years ago
 Views:
Transcription
1 Math 475, Problem Set #6: Solutios A (a) For each poit (a, b) with a, b oegative itegers satisfyig ab 8, cout the paths from (0,0) to (a, b) where the legal steps from (i, j) are to (i 2, j), (i, j 2), ad (i, j ) I ll do this i the first quadrat usig i ad j as x ad ycoordiates (though you could use the i ad j as rowidex ad columidex; you d get differet pictures but the same umbers) (b) Compute the coefficiets of (x 2 xy y 2 ) for 0,, 2, 3, 4 (x 2 xy y 2 ) 0, (x 2 xy y 2 ) x 2 xy y 2, (x 2 xy y 2 ) 2 x 4 2x 3 y 3x 2 y 2 2xy 3 y 4, (x 2 xy y 2 ) 3 x 6 3x 5 y 6x 4 y 2 7x 3 y 3 6x 2 y 4 3xy 5 y 6, ad (x 2 xy y 2 ) 4 x 8 4x 7 y 0x 6 y 2 6x 5 y 3 9x 4 y 4 6x 3 y 5 0x 2 y 6 4xy 7 y 8 (c) Based o parts (a) ad (b), formulate a precise cojecture of the form for all oegative itegers a ad b, the umber of paths from (0, 0) to (a, b) is equal to the coefficiet of i the polyomial The umber of paths from (0, 0) to (a, b) is equal to the coefficiet of x a y b i (x 2 xy y 2 ) (ab)/2 (This assumes that a b is eve; if a b is odd, the umber of paths from (0, 0) to (a, b) is zero) You ca thik of the three terms of x 2 xy y 2 as represetig the three possible steps you re allowed to take: x 2 x 2 y 0 correspods to movig 2 steps to the right, xy x y correspods to movig step to the right ad step up, ad y 2 x 0 y 2 correspods to movig 2 steps upward The terms i the expasio of (x 2 xy y 2 ) m correspod to all the places (a, b)
2 you ca get to usig m steps of the allowed kids, ad the coefficiet of x a y b i this expasio is the umber of ways to get there B Chapter 5, problem 2 Algebraic proof: Rewritig the expressio as ( ), we see that it is the coefficiet of x i the product of x x x 2 ( ) 0 2 ad x 0 x x 2 2 But by applyig the biomial theorem to the two factors, we see that this polyomial is equal to (x ) (x ) (x 2 ) All the terms i this sum are of eve degree, so if is odd, the coefficiet of x i the polyomial vaishes O the other had, if is eve (say 2k), the, settig y x 2, we see that the coefficiet of x i (x 2 ) is the same as the coefficiet of y /2 i (y ), which is ( ) /2 /2 (Remark: Eve if you did t fid this, you could have still gotte partial credit if you oticed that, whe is odd, the terms of the alteratig ( sum of the squares of the biomial coefficiets cacel i pairs: 2 0) cacels ( ), 2 cacels, etc) Combiatorial proof: Imagie me ad wome, from whom we wish to choose idividuals to form a committee Call a committee eve if it cotais a eve umber of wome, ad odd otherwise The is the umber of eve committees, ad 3 3
3 is the umber of odd committees We wish to show that the umber of eve commitees mius the umber of odd commitees is 0 whe is odd ad ( ) /2 /2 whe is eve We will do this by pairig up eve committees with odd committees i such a way that whe is odd, there are o upaired committees, while if is eve, there are exactly /2 upaired committees, all of which are eve if /2 is eve ad odd if /2 is odd To accomplish this, first marry off the me ad wome, ad umber the resultig couples from to If a committee cosists completely of married couples, we do t pair it with aother committee; otherwise, we pair it with the committee obtaied by replacig the lowestumbered committtee member whose spouse is ot o the committee by that committeemember s spouse (Here lowestumbered could mea with the lowest social security umber, or aythig else that lets us break ties) This pairig clearly pairs odd committees with eve committees If is odd, there are o upaired committees If is eve, the there are /2 upaired committees, each of which is either eve or odd accordig to the parity of /2 C Solve Brualdi, Chapter 5, problem 8 i two differet ways: oce usig problem 6 as a model, ad oce usig problem 7 as a model First method: Itegratig k0 x k k ( x) from x 0, we get ) k0 k xk( k (( x) ), ad settig x, we get ) k0 ( ) k( k k (0 ) Multiplyig by, we get ) k0 ( ) k( k k Secod method: ( ) [ ] ( ) 2 3 0
4 D What is the coefficiet of x 3 x 3 2x 3 x 2 4 i the expasio of (x x 2 2x 3 2x 4 ) 9? The relevat term of the multiomial theorem is equal to the product of 9 9! !3!!2! 5040 ad (x ) 3 ( x 2 ) 3 (2x 3 ) ( 2x 4 ) 2 8x 2 x 3 2x 3x 2 4; the total coefficiet is thus (5040) ( 8) E Brualdi, Chapter 5, problem 46 Retai all terms that are greater tha 0 3 ; discard the rest Let x 2 ad y 8 Sice 0 2 8, Theorem 56 applies: (2 8) /3 / /3 /3 2 8 /3 / /3 2 / /3 3 ) /3 4 ()(2)(/3)(2 )( (/3)( 2/3) )(2 3 )( (/3)( 2/3)( 5/3) 2! ( /3 4 ( (/3)( 2/3)( 5/3)( 8/3) 4! )(2 7 ) 2/6 /725/2592 5/5552 These terms decrease i absolute value ad alterate i sig (leavig aside the first two terms), ad the fifth term is less tha 0 3, so we may approximate the ifiite sum to withi 0 3 by takig just the first four terms: 2 /6 /72 5/ / (The true value is 2544 ) F Fix positive itegers, k 3 Cosider a covex go with vertices labelled through Call a covex kgo, whose vertices are a subset of the vertices of the go, a iteral kgo if all of its sides are diagoals of the go (a) How may iteral kgos are there cotaiig the vertex labelled? The iteral kgos cotaiig vertex are i  correspodece with the ktuples (x, x 2,, x k ) satisfyig x x 2 x k k, where each x i is a positive iteger (Thik of x i as the umber of vertices of the go that lie betwee successive vertices of the k go) Lettig y i x i, we see that these i tur correspod to k tuples (y, y 2,, x k ) satisfyig y y 2 y k 2k where each y i is a oegative iteger Applyig oeadstars i the usual 3! )(2 5 )
5 way we see that the umber of such ktuples is ( 2k)(k ) k ) ( k k (b) How may iteral kgos are there all together? (Hit: What do you kow ahead of time about the ratio betwee the aswer to (a) ad the aswer to (b)?) Claim (a) is true about ay particular vertex, ad ot just the vertex labelled So if we multiply k k by (the umber of vertices) ad divide by k (to take ito accout the fact that each kgo has k differet vertices), we get the correct aswer, amely k k k ( k )! k!( 2k)! Alterative aalysis: Let s cout, i two differet ways, the umber of ways to draw a iteral kgo ad pick a vertex of that kgo ad color it blue O the oe had, the aswer is N b times k, where N b is the aswer to problem (b), sice there are N b possible iteral kgos, each of which has k vertices that are available to be colored blue O the other had, you could choose the blue vertex first There are vertices of the origial go that could be colored blue For each choice of the blue vertex, there are N a iteral kgos that cotai that vertex, where N a is the aswer to part (a) (We proved this i the case where the special vertex is vertex, but there s othig special about the vertex that we happeed to call vertex ; the aswer is N a for ay particular vertex) Hece N b k N a, from which it follows that N b (/k)n a
x(x 1)(x 2)... (x k + 1) = [x] k n+m 1
1 Coutig mappigs For every real x ad positive iteger k, let [x] k deote the fallig factorial ad x(x 1)(x 2)... (x k + 1) ( ) x = [x] k k k!, ( ) k = 1. 0 I the sequel, X = {x 1,..., x m }, Y = {y 1,...,
More informationChapter Gaussian Elimination
Chapter 04.06 Gaussia Elimiatio After readig this chapter, you should be able to:. solve a set of simultaeous liear equatios usig Naïve Gauss elimiatio,. lear the pitfalls of the Naïve Gauss elimiatio
More informationMath Discrete Math Combinatorics MULTIPLICATION PRINCIPLE:
Math 355  Discrete Math 4.14.4 Combiatorics Notes MULTIPLICATION PRINCIPLE: If there m ways to do somethig ad ways to do aother thig the there are m ways to do both. I the laguage of set theory: Let
More informationIntro to Sequences / Arithmetic Sequences and Series Levels
Itro to Sequeces / Arithmetic Sequeces ad Series Levels Level : pg. 569: #7, 0, 33 Pg. 575: #, 7, 8 Pg. 584: #8, 9, 34, 36 Levels, 3, ad 4(Fiboacci Sequece Extesio) See Hadout Check for Uderstadig Level
More informationHW 1 Solutions Math 115, Winter 2009, Prof. Yitzhak Katznelson
HW Solutios Math 5, Witer 2009, Prof. Yitzhak Katzelso.: Prove 2 + 2 2 +... + 2 = ( + )(2 + ) for all atural umbers. The proof is by iductio. Call the th propositio P. The basis for iductio P is the statemet
More informationrepresented by 4! different arrangements of boxes, divide by 4! to get ways
Problem Set #6 solutios A juggler colors idetical jugglig balls red, white, ad blue (a I how may ways ca this be doe if each color is used at least oce? Let us preemptively color oe ball i each color,
More informationContinued Fractions continued. 3. Best rational approximations
Cotiued Fractios cotiued 3. Best ratioal approximatios We hear so much about π beig approximated by 22/7 because o other ratioal umber with deomiator < 7 is closer to π. Evetually 22/7 is defeated by 333/06
More informationModule 4: Mathematical Induction
Module 4: Mathematical Iductio Theme 1: Priciple of Mathematical Iductio Mathematical iductio is used to prove statemets about atural umbers. As studets may remember, we ca write such a statemet as a predicate
More information1 The Binomial Theorem: Another Approach
The Biomial Theorem: Aother Approach Pascal s Triagle I class (ad i our text we saw that, for iteger, the biomial theorem ca be stated (a + b = c a + c a b + c a b + + c ab + c b, where the coefficiets
More informationSECTION 1.5 : SUMMATION NOTATION + WORK WITH SEQUENCES
SECTION 1.5 : SUMMATION NOTATION + WORK WITH SEQUENCES Read Sectio 1.5 (pages 5 9) Overview I Sectio 1.5 we lear to work with summatio otatio ad formulas. We will also itroduce a brief overview of sequeces,
More informationORDERS OF GROWTH KEITH CONRAD
ORDERS OF GROWTH KEITH CONRAD Itroductio Gaiig a ituitive feel for the relative growth of fuctios is importat if you really wat to uderstad their behavior It also helps you better grasp topics i calculus
More informationExample 2 Find the square root of 0. The only square root of 0 is 0 (since 0 is not positive or negative, so those choices don t exist here).
BEGINNING ALGEBRA Roots ad Radicals (revised summer, 00 Olso) Packet to Supplemet the Curret Textbook  Part Review of Square Roots & Irratioals (This portio ca be ay time before Part ad should mostly
More informationReview for College Algebra Final Exam
Review for College Algebra Fial Exam (Please remember that half of the fial exam will cover chapters 14. This review sheet covers oly the ew material, from chapters 5 ad 7.) 5.1 Systems of equatios i
More informationTHE LEAST SQUARES REGRESSION LINE and R 2
THE LEAST SQUARES REGRESSION LINE ad R M358K I. Recall from p. 36 that the least squares regressio lie of y o x is the lie that makes the sum of the squares of the vertical distaces of the data poits from
More information1.3 Binomial Coefficients
18 CHAPTER 1. COUNTING 1. Biomial Coefficiets I this sectio, we will explore various properties of biomial coefficiets. Pascal s Triagle Table 1 cotais the values of the biomial coefficiets ( ) for 0to
More informationTo get the next Fibonacci number, you add the previous two. numbers are defined by the recursive formula. F 1 F n+1
Liear Algebra Notes Chapter 6 FIBONACCI NUMBERS The Fiboacci umbers are F, F, F 2, F 3 2, F 4 3, F, F 6 8, To get the ext Fiboacci umber, you add the previous two umbers are defied by the recursive formula
More informationLinear Algebra II. Notes 6 25th November 2010
MTH6140 Liear Algebra II Notes 6 25th November 2010 6 Quadratic forms A lot of applicatios of mathematics ivolve dealig with quadratic forms: you meet them i statistics (aalysis of variace) ad mechaics
More informationUnit 8 Rational Functions
Uit 8 Ratioal Fuctios Algebraic Fractios: Simplifyig Algebraic Fractios: To simplify a algebraic fractio meas to reduce it to lowest terms. This is doe by dividig out the commo factors i the umerator ad
More informationTHE ARITHMETIC OF INTEGERS.  multiplication, exponentiation, division, addition, and subtraction
THE ARITHMETIC OF INTEGERS  multiplicatio, expoetiatio, divisio, additio, ad subtractio What to do ad what ot to do. THE INTEGERS Recall that a iteger is oe of the whole umbers, which may be either positive,
More informationLecture 4: Cauchy sequences, BolzanoWeierstrass, and the Squeeze theorem
Lecture 4: Cauchy sequeces, BolzaoWeierstrass, ad the Squeeze theorem The purpose of this lecture is more modest tha the previous oes. It is to state certai coditios uder which we are guarateed that limits
More informationARITHMETIC AND GEOMETRIC PROGRESSIONS
Arithmetic Ad Geometric Progressios Sequeces Ad ARITHMETIC AND GEOMETRIC PROGRESSIONS Successio of umbers of which oe umber is desigated as the first, other as the secod, aother as the third ad so o gives
More informationTrigonometric Form of a Complex Number. The Complex Plane. axis. ( 2, 1) or 2 i FIGURE 6.44. The absolute value of the complex number z a bi is
0_0605.qxd /5/05 0:45 AM Page 470 470 Chapter 6 Additioal Topics i Trigoometry 6.5 Trigoometric Form of a Complex Number What you should lear Plot complex umbers i the complex plae ad fid absolute values
More information7 b) 0. Guided Notes for lesson P.2 Properties of Exponents. If a, b, x, y and a, b, 0, and m, n Z then the following properties hold: 1 n b
Guided Notes for lesso P. Properties of Expoets If a, b, x, y ad a, b, 0, ad m, Z the the followig properties hold:. Negative Expoet Rule: b ad b b b Aswers must ever cotai egative expoets. Examples: 5
More informationMath 115 HW #4 Solutions
Math 5 HW #4 Solutios From 2.5 8. Does the series coverge or diverge? ( ) 3 + 2 = Aswer: This is a alteratig series, so we eed to check that the terms satisfy the hypotheses of the Alteratig Series Test.
More informationBINOMIAL EXPANSIONS 12.5. In this section. Some Examples. Obtaining the Coefficients
652 (1226) Chapter 12 Sequeces ad Series 12.5 BINOMIAL EXPANSIONS I this sectio Some Examples Otaiig the Coefficiets The Biomial Theorem I Chapter 5 you leared how to square a iomial. I this sectio you
More information.04. This means $1000 is multiplied by 1.02 five times, once for each of the remaining sixmonth
Questio 1: What is a ordiary auity? Let s look at a ordiary auity that is certai ad simple. By this, we mea a auity over a fixed term whose paymet period matches the iterest coversio period. Additioally,
More informationSAMPLE QUESTIONS FOR FINAL EXAM. (1) (2) (3) (4) Find the following using the definition of the Riemann integral: (2x + 1)dx
SAMPLE QUESTIONS FOR FINAL EXAM REAL ANALYSIS I FALL 006 3 4 Fid the followig usig the defiitio of the Riema itegral: a 0 x + dx 3 Cosider the partitio P x 0 3, x 3 +, x 3 +,......, x 3 3 + 3 of the iterval
More information5.3. Generalized Permutations and Combinations
53 GENERALIZED PERMUTATIONS AND COMBINATIONS 73 53 Geeralized Permutatios ad Combiatios 53 Permutatios with Repeated Elemets Assume that we have a alphabet with letters ad we wat to write all possible
More informationHomework 1 Solutions
Homewor 1 Solutios Math 171, Sprig 2010 Please sed correctios to herya@math.staford.edu 2.2. Let h : X Y, g : Y Z, ad f : Z W. Prove that (f g h = f (g h. Solutio. Let x X. Note that ((f g h(x = (f g(h(x
More information8.5 Alternating infinite series
65 8.5 Alteratig ifiite series I the previous two sectios we cosidered oly series with positive terms. I this sectio we cosider series with both positive ad egative terms which alterate: positive, egative,
More information1. MATHEMATICAL INDUCTION
1. MATHEMATICAL INDUCTION EXAMPLE 1: Prove that for ay iteger 1. Proof: 1 + 2 + 3 +... + ( + 1 2 (1.1 STEP 1: For 1 (1.1 is true, sice 1 1(1 + 1. 2 STEP 2: Suppose (1.1 is true for some k 1, that is 1
More informationInfinite Sequences and Series
CHAPTER 4 Ifiite Sequeces ad Series 4.1. Sequeces A sequece is a ifiite ordered list of umbers, for example the sequece of odd positive itegers: 1, 3, 5, 7, 9, 11, 13, 15, 17, 19, 21, 23, 25, 27, 29...
More informationRadicals and Fractional Exponents
Radicals ad Roots Radicals ad Fractioal Expoets I math, may problems will ivolve what is called the radical symbol, X is proouced the th root of X, where is or greater, ad X is a positive umber. What it
More informationLecture 13. Lecturer: Jonathan Kelner Scribe: Jonathan Pines (2009)
18.409 A Algorithmist s Toolkit October 27, 2009 Lecture 13 Lecturer: Joatha Keler Scribe: Joatha Pies (2009) 1 Outlie Last time, we proved the BruMikowski iequality for boxes. Today we ll go over the
More informationThe Euler Totient, the Möbius and the Divisor Functions
The Euler Totiet, the Möbius ad the Divisor Fuctios Rosica Dieva July 29, 2005 Mout Holyoke College South Hadley, MA 01075 1 Ackowledgemets This work was supported by the Mout Holyoke College fellowship
More information{{1}, {2, 4}, {3}} {{1, 3, 4}, {2}} {{1}, {2}, {3, 4}} 5.4 Stirling Numbers
. Stirlig Numbers Whe coutig various types of fuctios from., we quicly discovered that eumeratig the umber of oto fuctios was a difficult problem. For a domai of five elemets ad a rage of four elemets,
More informationSection 8.3 : De Moivre s Theorem and Applications
The Sectio 8 : De Moivre s Theorem ad Applicatios Let z 1 ad z be complex umbers, where z 1 = r 1, z = r, arg(z 1 ) = θ 1, arg(z ) = θ z 1 = r 1 (cos θ 1 + i si θ 1 ) z = r (cos θ + i si θ ) ad z 1 z =
More informationSum and Product Rules. Combinatorics. Some Subtler Examples
Combiatorics Sum ad Product Rules Problem: How to cout without coutig. How do you figure out how may thigs there are with a certai property without actually eumeratig all of them. Sometimes this requires
More informationMocks.ie Maths LC HL Further Calculus mocks.ie Page 1
Maths Leavig Cert Higher Level Further Calculus Questio Paper By Cillia Fahy ad Darro Higgis Mocks.ie Maths LC HL Further Calculus mocks.ie Page Further Calculus ad Series, Paper II Q8 Table of Cotets:.
More information8.3 POLAR FORM AND DEMOIVRE S THEOREM
SECTION 8. POLAR FORM AND DEMOIVRE S THEOREM 48 8. POLAR FORM AND DEMOIVRE S THEOREM Figure 8.6 (a, b) b r a 0 θ Complex Number: a + bi Rectagular Form: (a, b) Polar Form: (r, θ) At this poit you ca add,
More informationSequences II. Chapter 3. 3.1 Convergent Sequences
Chapter 3 Sequeces II 3. Coverget Sequeces Plot a graph of the sequece a ) = 2, 3 2, 4 3, 5 + 4,...,,... To what limit do you thik this sequece teds? What ca you say about the sequece a )? For ǫ = 0.,
More informationAlternatives To Pearson s and Spearman s Correlation Coefficients
Alteratives To Pearso s ad Spearma s Correlatio Coefficiets Floreti Smaradache Chair of Math & Scieces Departmet Uiversity of New Mexico Gallup, NM 8730, USA Abstract. This article presets several alteratives
More informationHere are a couple of warnings to my students who may be here to get a copy of what happened on a day that you missed.
This documet was writte ad copyrighted by Paul Dawkis. Use of this documet ad its olie versio is govered by the Terms ad Coditios of Use located at http://tutorial.math.lamar.edu/terms.asp. The olie versio
More informationIn nite Sequences. Dr. Philippe B. Laval Kennesaw State University. October 9, 2008
I ite Sequeces Dr. Philippe B. Laval Keesaw State Uiversity October 9, 2008 Abstract This had out is a itroductio to i ite sequeces. mai de itios ad presets some elemetary results. It gives the I ite Sequeces
More informationMATH 140A  HW 5 SOLUTIONS
MATH 40A  HW 5 SOLUTIONS Problem WR Ch 3 #8. If a coverges, ad if {b } is mootoic ad bouded, rove that a b coverges. Solutio. Theorem 3.4 states that if a the artial sums of a form a bouded sequece; b
More information1 Notes on Little s Law (l = λw)
Copyright c 29 by Karl Sigma Notes o Little s Law (l λw) We cosider here a famous ad very useful law i queueig theory called Little s Law, also kow as l λw, which asserts that the time average umber of
More informationLesson 17 Pearson s Correlation Coefficient
Outlie Measures of Relatioships Pearso s Correlatio Coefficiet (r) types of data scatter plots measure of directio measure of stregth Computatio covariatio of X ad Y uique variatio i X ad Y measurig
More informationWinter Camp 2012 Sequences Alexander Remorov. Sequences. Alexander Remorov
Witer Camp 202 Sequeces Alexader Remorov Sequeces Alexader Remorov alexaderrem@gmail.com Warmup Problem : Give a positive iteger, cosider a sequece of real umbers a 0, a,..., a defied as a 0 = 2 ad =
More informationTheorems About Power Series
Physics 6A Witer 20 Theorems About Power Series Cosider a power series, f(x) = a x, () where the a are real coefficiets ad x is a real variable. There exists a real oegative umber R, called the radius
More informationSequences and Series
CHAPTER 9 Sequeces ad Series 9.. Covergece: Defiitio ad Examples Sequeces The purpose of this chapter is to itroduce a particular way of geeratig algorithms for fidig the values of fuctios defied by their
More informationNUMBERS COMMON TO TWO POLYGONAL SEQUENCES
NUMBERS COMMON TO TWO POLYGONAL SEQUENCES DIANNE SMITH LUCAS Chia Lake, Califoria a iteger, The polygoal sequece (or sequeces of polygoal umbers) of order r (where r is r > 3) may be defied recursively
More informationConvexity, Inequalities, and Norms
Covexity, Iequalities, ad Norms Covex Fuctios You are probably familiar with the otio of cocavity of fuctios. Give a twicedifferetiable fuctio ϕ: R R, We say that ϕ is covex (or cocave up) if ϕ (x) 0 for
More informationThe second difference is the sequence of differences of the first difference sequence, 2
Differece Equatios I differetial equatios, you look for a fuctio that satisfies ad equatio ivolvig derivatives. I differece equatios, istead of a fuctio of a cotiuous variable (such as time), we look for
More informationRiemann Sums y = f (x)
Riema Sums Recall that we have previously discussed the area problem I its simplest form we ca state it this way: The Area Problem Let f be a cotiuous, oegative fuctio o the closed iterval [a, b] Fid
More informationChapter Suppose you wish to use the Principle of Mathematical Induction to prove that 1 1! + 2 2! + 3 3! n n! = (n + 1)! 1 for all n 1.
Chapter 4. Suppose you wish to prove that the followig is true for all positive itegers by usig the Priciple of Mathematical Iductio: + 3 + 5 +... + ( ) =. (a) Write P() (b) Write P(7) (c) Write P(73)
More informationFIBONACCI NUMBERS: AN APPLICATION OF LINEAR ALGEBRA. 1. Powers of a matrix
FIBONACCI NUMBERS: AN APPLICATION OF LINEAR ALGEBRA. Powers of a matrix We begi with a propositio which illustrates the usefuless of the diagoalizatio. Recall that a square matrix A is diogaalizable if
More information8 The Poisson Distribution
8 The Poisso Distributio Let X biomial, p ). Recall that this meas that X has pmf ) p,p k) p k k p ) k for k 0,,...,. ) Agai, thik of X as the umber of successes i a series of idepedet experimets, each
More informationDivide and Conquer. Maximum/minimum. Integer Multiplication. CS125 Lecture 4 Fall 2015
CS125 Lecture 4 Fall 2015 Divide ad Coquer We have see oe geeral paradigm for fidig algorithms: the greedy approach. We ow cosider aother geeral paradigm, kow as divide ad coquer. We have already see a
More informationwhen n = 1, 2, 3, 4, 5, 6, This list represents the amount of dollars you have after n days. Note: The use of is read as and so on.
Geometric eries Before we defie what is meat by a series, we eed to itroduce a related topic, that of sequeces. Formally, a sequece is a fuctio that computes a ordered list. uppose that o day 1, you have
More informationFourier Series and the Wave Equation Part 2
Fourier Series ad the Wave Equatio Part There are two big ideas i our work this week. The first is the use of liearity to break complicated problems ito simple pieces. The secod is the use of the symmetries
More informationFigure 40.1. Figure 40.2
40 Regular Polygos Covex ad Cocave Shapes A plae figure is said to be covex if every lie segmet draw betwee ay two poits iside the figure lies etirely iside the figure. A figure that is ot covex is called
More informationCS103A Handout 23 Winter 2002 February 22, 2002 Solving Recurrence Relations
CS3A Hadout 3 Witer 00 February, 00 Solvig Recurrece Relatios Itroductio A wide variety of recurrece problems occur i models. Some of these recurrece relatios ca be solved usig iteratio or some other ad
More information9.1 Simplify Radical Expressions
9.1 Simplifyig Radical Expressios (Page 1 of 20) 9.1 Simplify Radical Expressios Radical Notatio for the th Root of a If is a iteger greater tha oe, the the th root of a is the umer whose th power is
More informationFactoring x n 1: cyclotomic and Aurifeuillian polynomials Paul Garrett <garrett@math.umn.edu>
(March 16, 004) Factorig x 1: cyclotomic ad Aurifeuillia polyomials Paul Garrett Polyomials of the form x 1, x 3 1, x 4 1 have at least oe systematic factorizatio x 1 = (x 1)(x 1
More informationDepartment of Computer Science, University of Otago
Departmet of Computer Sciece, Uiversity of Otago Techical Report OUCS200609 Permutatios Cotaiig May Patters Authors: M.H. Albert Departmet of Computer Sciece, Uiversity of Otago Micah Colema, Rya Fly
More informationCS103X: Discrete Structures Homework 4 Solutions
CS103X: Discrete Structures Homewor 4 Solutios Due February 22, 2008 Exercise 1 10 poits. Silico Valley questios: a How may possible sixfigure salaries i whole dollar amouts are there that cotai at least
More informationCounting II 3, 7 3, 2 3, 9 7, 2 7, 9 2, 9
Coutig II Sometimes we will wat to choose objects from a set of objects, ad we wo t be iterested i orderig them. For example, if you are leavig for vacatio ad you wat to pac your suitcase with three of
More information= 2, 3, 4, etc. = { FLC Ch 7. Math 120 Intermediate Algebra Sec 7.1: Radical Expressions and Functions
Math 120 Itermediate Algebra Sec 7.1: Radical Expressios ad Fuctios idex radicad = 2,,, etc. Ex 1 For each umber, fid all of its square roots. 121 2 6 Ex 2 1 Simplify. 1 22 9 81 62 8 27 16 16 0 1 180 22
More information23 The Remainder and Factor Theorems
 The Remaider ad Factor Theorems Factor each polyomial completely usig the give factor ad log divisio 1 x + x x 60; x + So, x + x x 60 = (x + )(x x 15) Factorig the quadratic expressio yields x + x x
More informationEquation of a line. Line in coordinate geometry. Slopeintercept form ( 斜 截 式 ) Intercept form ( 截 距 式 ) Pointslope form ( 點 斜 式 )
Chapter : Liear Equatios Chapter Liear Equatios Lie i coordiate geometr I Cartesia coordiate sstems ( 卡 笛 兒 坐 標 系 統 ), a lie ca be represeted b a liear equatio, i.e., a polomial with degree. But before
More informationWHEN IS THE (CO)SINE OF A RATIONAL ANGLE EQUAL TO A RATIONAL NUMBER?
WHEN IS THE (CO)SINE OF A RATIONAL ANGLE EQUAL TO A RATIONAL NUMBER? JÖRG JAHNEL 1. My Motivatio Some Sort of a Itroductio Last term I tought Topological Groups at the Göttige Georg August Uiversity. This
More information4 n. n 1. You shold think of the Ratio Test as a generalization of the Geometric Series Test. For example, if a n ar n is a geometric sequence then
SECTION 2.6 THE RATIO TEST 79 2.6. THE RATIO TEST We ow kow how to hadle series which we ca itegrate (the Itegral Test), ad series which are similar to geometric or pseries (the Compariso Test), but of
More informationThe Field of Complex Numbers
The Field of Complex Numbers S. F. Ellermeyer The costructio of the system of complex umbers begis by appedig to the system of real umbers a umber which we call i with the property that i = 1. (Note that
More informationMATH 361 Homework 9. Royden Royden Royden
MATH 61 Homework 9 Royde..9 First, we show that for ay subset E of the real umbers, E c + y = E + y) c traslatig the complemet is equivalet to the complemet of the traslated set). Without loss of geerality,
More informationRepeating Decimals are decimal numbers that have number(s) after the decimal point that repeat in a pattern.
5.5 Fractios ad Decimals Steps for Chagig a Fractio to a Decimal. Simplify the fractio, if possible. 2. Divide the umerator by the deomiator. d d Repeatig Decimals Repeatig Decimals are decimal umbers
More information8.1 Arithmetic Sequences
MCR3U Uit 8: Sequeces & Series Page 1 of 1 8.1 Arithmetic Sequeces Defiitio: A sequece is a comma separated list of ordered terms that follow a patter. Examples: 1, 2, 3, 4, 5 : a sequece of the first
More informationSoving Recurrence Relations
Sovig Recurrece Relatios Part 1. Homogeeous liear 2d degree relatios with costat coefficiets. Cosider the recurrece relatio ( ) T () + at ( 1) + bt ( 2) = 0 This is called a homogeeous liear 2d degree
More information2.3. GEOMETRIC SERIES
6 CHAPTER INFINITE SERIES GEOMETRIC SERIES Oe of the most importat types of ifiite series are geometric series A geometric series is simply the sum of a geometric sequece, Fortuately, geometric series
More informationMath 114 Intermediate Algebra Integral Exponents & Fractional Exponents (10 )
Math 4 Math 4 Itermediate Algebra Itegral Epoets & Fractioal Epoets (0 ) Epoetial Fuctios Epoetial Fuctios ad Graphs I. Epoetial Fuctios The fuctio f ( ) a, where is a real umber, a 0, ad a, is called
More informationS. Tanny MAT 344 Spring 1999. be the minimum number of moves required.
S. Tay MAT 344 Sprig 999 Recurrece Relatios Tower of Haoi Let T be the miimum umber of moves required. T 0 = 0, T = 7 Iitial Coditios * T = T + $ T is a sequece (f. o itegers). Solve for T? * is a recurrece,
More informationSection 11.3: The Integral Test
Sectio.3: The Itegral Test Most of the series we have looked at have either diverged or have coverged ad we have bee able to fid what they coverge to. I geeral however, the problem is much more difficult
More information3.2 Introduction to Infinite Series
3.2 Itroductio to Ifiite Series May of our ifiite sequeces, for the remaider of the course, will be defied by sums. For example, the sequece S m := 2. () is defied by a sum. Its terms (partial sums) are
More informationChapter 5: Inner Product Spaces
Chapter 5: Ier Product Spaces Chapter 5: Ier Product Spaces SECION A Itroductio to Ier Product Spaces By the ed of this sectio you will be able to uderstad what is meat by a ier product space give examples
More informationBinet Formulas for Recursive Integer Sequences
Biet Formulas for Recursive Iteger Sequeces Homer W. Austi Jatha W. Austi Abstract May iteger sequeces are recursive sequeces ad ca be defied either recursively or explicitly by use of Biettype formulas.
More informationLecture 17 TwoWay Finite Automata
This is page 9 Priter: Opaque this Lecture 7 TwoWay Fiite Automata Twoway fiite automata are similar to the machies we have bee studyig, except that they ca read the iput strig i either directio. We
More informationOnestep equations. Vocabulary
Review solvig oestep equatios with itegers, fractios, ad decimals. Oestep equatios Vocabulary equatio solve solutio iverse operatio isolate the variable Additio Property of Equality Subtractio Property
More informationMa/CS 6b Class 17: Extremal Graph Theory
//06 Ma/CS 6b Class 7: Extremal Graph Theory Paul Turá By Adam Sheffer Extremal Graph Theory The subfield of extremal graph theory deals with questios of the form: What is the maximum umber of edges that
More informationSum of Exterior Angles of Polygons TEACHER NOTES
Sum of Exterior Agles of Polygos TEACHER NOTES Math Objectives Studets will determie that the iterior agle of a polygo ad a exterior agle of a polygo form a liear pair (i.e., the two agles are supplemetary).
More information1 Correlation and Regression Analysis
1 Correlatio ad Regressio Aalysis I this sectio we will be ivestigatig the relatioship betwee two cotiuous variable, such as height ad weight, the cocetratio of a ijected drug ad heart rate, or the cosumptio
More informationHandout: How to calculate time complexity? CSE 101 Winter 2014
Hadout: How to calculate time complexity? CSE 101 Witer 014 Recipe (a) Kow algorithm If you are usig a modied versio of a kow algorithm, you ca piggyback your aalysis o the complexity of the origial algorithm
More informationLesson 12. Sequences and Series
Retur to List of Lessos Lesso. Sequeces ad Series A ifiite sequece { a, a, a,... a,...} ca be thought of as a list of umbers writte i defiite order ad certai patter. It is usually deoted by { a } =, or
More informationSUMS OF nth POWERS OF ROOTS OF A GIVEN QUADRATIC EQUATION. N.A. Draim, Ventura, Calif., and Marjorie Bicknell Wilcox High School, Santa Clara, Calif.
SUMS OF th OWERS OF ROOTS OF A GIVEN QUADRATIC EQUATION N.A. Draim, Vetura, Calif., ad Marjorie Bickell Wilcox High School, Sata Clara, Calif. The quadratic equatio whose roots a r e the sum or differece
More informationFactors of sums of powers of binomial coefficients
ACTA ARITHMETICA LXXXVI.1 (1998) Factors of sums of powers of biomial coefficiets by Neil J. Cali (Clemso, S.C.) Dedicated to the memory of Paul Erdős 1. Itroductio. It is well ow that if ( ) a f,a = the
More informationAsymptotic Growth of Functions
CMPS Itroductio to Aalysis of Algorithms Fall 3 Asymptotic Growth of Fuctios We itroduce several types of asymptotic otatio which are used to compare the performace ad efficiecy of algorithms As we ll
More informationCHAPTER 7: Central Limit Theorem: CLT for Averages (Means)
CHAPTER 7: Cetral Limit Theorem: CLT for Averages (Meas) X = the umber obtaied whe rollig oe six sided die oce. If we roll a six sided die oce, the mea of the probability distributio is X P(X = x) Simulatio:
More informationSection IV.5: Recurrence Relations from Algorithms
Sectio IV.5: Recurrece Relatios from Algorithms Give a recursive algorithm with iput size, we wish to fid a Θ (best big O) estimate for its ru time T() either by obtaiig a explicit formula for T() or by
More informationarxiv:1012.1336v2 [cs.cc] 8 Dec 2010
Uary SubsetSum is i Logspace arxiv:1012.1336v2 [cs.cc] 8 Dec 2010 1 Itroductio Daiel M. Kae December 9, 2010 I this paper we cosider the Uary SubsetSum problem which is defied as follows: Give itegers
More informationOur aim is to show that under reasonable assumptions a given 2πperiodic function f can be represented as convergent series
8 Fourier Series Our aim is to show that uder reasoable assumptios a give periodic fuctio f ca be represeted as coverget series f(x) = a + (a cos x + b si x). (8.) By defiitio, the covergece of the series
More informationSolving DivideandConquer Recurrences
Solvig DivideadCoquer Recurreces Victor Adamchik A divideadcoquer algorithm cosists of three steps: dividig a problem ito smaller subproblems solvig (recursively) each subproblem the combiig solutios
More information1. C. The formula for the confidence interval for a population mean is: x t, which was
s 1. C. The formula for the cofidece iterval for a populatio mea is: x t, which was based o the sample Mea. So, x is guarateed to be i the iterval you form.. D. Use the rule : pvalue
More information