Soving Recurrence Relations


 Jesse Cunningham
 2 years ago
 Views:
Transcription
1 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 recurrece relatio with costat coefficiets: 2d degree because it gives T () i terms of T ( 1) ad T ( 2), liear ad costat coefficiets because of the form of the left side, ad homogeeous because of the zero o the right had side. The idea for solvig this relatio is to guess a solutio of the form T () = x for some umber x, ad the to simply substitute this expressio ito the equatio to determie the value(s) of x that work. Sice T ( 1) = x 1 ad T ( 2) = x 2 we get the equatio x + ax 1 + bx 2 = 0 Sice x is clearly ot zero, we ca divide by x 2 to get x 2 + ax + b = 0 which is called the characteristic equatio for the recurrece relatio ( ). Case 1: If this equatio factors as (x r 1 )(x r 2 ) = 0 with r 1 r 2 (so that the characteristic equatio has two distict roots), the the geeral solutio to ( ) is T () = c 1 (r 1 ) + c 2 (r 2 ) where c 1 ad c 2 are costats. This is called a geeral solutio because every fuctio T () that is a solutio to the relatio ( ) has this form. If we also have iitial coditios T (0) = t 0 ad T (1) = t 1 the we ca determie the values of c 1 ad c 2 ad thus get the exact formula for T (). We ll illustrate the process with a typical example. Example 1. Solve the recurrece relatio T () 4T ( 1) + 3T ( 2) = 0, T (0) = 0, T (1) = 2 Note: This could also be writte as 0 = 0 T () = 2 = 1 4T ( 1) 3T ( 2) > 1 Solutio: The characteristic equatio is x 2 4x + 3 = 0, or (x 3)(x 1) = 0, so the geeral solutio is T () = c c 2 1 = c c 2. To fid c 1 ad c 2 we plug i the iitial coditios to get two equatios i those two variables: 0 = T (0) = c c 2 = c 1 + c 2 2 = T (1) = c c 2 = 3c 1 + c 2
2 It s easy to solve these equatios for the solutio c 1 = 1, c 2 = 1, so the fial aswer is T () = 3 1 Check: We ca check our aswer quickly ad easily. The recurrece formula gives us T (2) = 4T (1) 3T (0) = 4(2) 0 = 8 T (3) = 4T (2) 3T (1) = 4(8) 3(2) = 26 T (4) = 4T (3) 3T (2) = 4(26) 3(8) = 80 It appears that the sequece is ideed givig us umbers that are oe less tha the powers of 3 (1, 3, 9, 27, 81,... ), so the formula T () = 3 1 seems to be correct. Case 2: If the characteristic equatio factors as (x r) 2 = 0 (with a sigle root), the we use as our geeral solutio the formula T () = c 1 r + c 2 r As before, iitial coditios ca be used to solve for c 1 ad c 2. Here is a typical example. Example 2. Solve the recurrece relatio 3 = 0 T () = 17 = 1 10T ( 1) 25T ( 2) > 1 Solutio: The characteristic equatio is x 2 10x + 25 = 0, or (x 5) 2 = 0, so the geeral solutio is T () = c c 2 5. As before, we fid c 1 ad c 2 by pluggig i the iitial coditios: 3 = T (0) = c c 2 (0) = c 1 17 = T (1) = c c 2 (1)5 1 = 5c 1 + 5c 2 The solutio here is c 1 = 3 ad c 2 = 2/5, so the exact solutio is T () = (3)5 + (2/5)5 = (15 + 2)5 1 Check: The recurrece formula gives us values T (2) = 10T (1) 25T (0) = 10(17) 25(3) = 95 T (3) = 10T (2) 25T (1) = 525 These are ideed the values give by our formula as well: T (2) = (15 + 4)5 1 = 95, ad T (3) = (15 + 6)5 2 = 525. Case 3: It ca happe i geeral recurrece relatios that the characteristic equatio x 2 + ax + b = 0 has o real roots, but istead has two complex umber roots. There is a method of solutio for such recurreces, but we will ot cocer ourselves with this case sice it does ot typically arise i recurreces that come from studyig recursive algorithms. 2
3 Part 2. Nohomogeeous liear 2d degree relatios with costat coefficiets. Now cosider what happes whe the right side of equatio ( ) (from page 1) is ot zero. We get a equatio of the form ( ) T () + at ( 1) + bt ( 2) = f() We ll lear how to solve ( ) i the special case that f() =(polyomial i )r. Actually, this requires oly a slight modificatio of the method for the homogeeous case. We simply multiply the characteristic equatio by (x r) k+1 where k is the degree of the polyomial part of f(). The solutio method the proceeds as before. This is best illustrated by examples. Example 3. Solve the recurrece relatio { 1 = 0 T () = 3T ( 1) + 2 > 0 Note: This is actually a 1st degree relatio (sice T ( 2) does ot appear), but the same method applies. Solutio: If the ohomogeeous term 2 were ot preset, the characteristic equatio would be simply x 3 = 0. I the presece of the ohomogeeous term, however, we must multiply this by (x 2) 0+1 (I this case f() = 2 is a 0degree polyomial [the costat 1] times a 2 term.) So, the characteristic equatio is actually (x 3)(x 2) = 0, so the geeral solutio is T () = c c 2 2. Note that sice the relatio is oly 1st degree, we oly have oe iitial coditio. Yet we ll eed two equatios to fid the costats c 1 ad c 2. We ca get a secod value to use by simply applyig the recurrece formula for = 1: T (1) = 3T (0) = 3(1) + 2 = 5. We ow proceed as usual: 1 = T (0) = c c = c 1 + c 2 5 = T (1) = c c = 3c 1 + 2c 2 These equatios have solutio c 1 = 3 ad c 2 = 2, so the exact solutio is T () = (3)3 (2)2 = Check: The recurrece formula gives us values T (2) = 3T (1) = 3(5) + 4 = 19 T (3) = 3T (2) = 3(19) + 8 = 65 These are boure out by our solutio: T (2) = = 27 8 = 19, ad T (3) = = = 65. 3
4 Example 4. Solve the recurrece relatio { + 1 = 0, 1 T () = 5T ( 1) 6T ( 2) > 1 Solutio: Here agai, the ohomogeeous term ivolves a zerodegree polyomial, so the modified characteristic equatio will be (x 2 5x + 6)(x 2) 0+1 = 0, or (x 3)(x 2) 2 = 0. The geeral solutio must ow ivolve three terms: a 3 term, a 2 term, ad (because the (x 2) factor appears twice i the characteristic equatio) a 2 term. Thus: T () = c c c 3 2. We ll eed three equatios to solve for the three costats, yet we have oly the two iitial coditios T (0) = 1 ad T (1) = 2 to use. As i the last example, we ca geerate oe more value by usig the recurrece formula: T (2) = 5T (1) 6T (0) = 5(2) 6(1) + 12 = 16. This gives us the three equatios 1 = T (0) = c c c 3 (0) = c 1 + c 2 2 = T (1) = c c c 3 (1)2 1 = 3c 1 + 2c 2 + 2c 3 16 = T (2) = c c c 3 (2)2 2 = 9c 1 + 4c 2 + 8c 3 A little effort gives the solutio c 1 = 12, c 2 = 11, ad c 3 = 6. The exact solutio is the T () = Note that we could have safely cocluded that T () is i Θ(3 ) simply from its geeral solutio (without eve solvig for c 1, c 2, ad c 3 ). Check: I additio to T (2) = 16, the recurrece formula gives us T (3) = 5T (2) 6T (1) = 5(16) 6(2) + 24 = 92 Checkig these agaist our solutio, we fid T (2) = (2)2 2 = = 16 ad T (3) = (3)2 3 = = 92. Example 5. Predict the bigtheta behavior of a fuctio T () satisfyig the recurrece relatio T () = 7T ( 1) 10T ( 2) + (2 + 5)3. Solutio: The modified characteristic equatio is (x 2 7x + 10)(x 3) 1+1 = 0, or factorig, (x 2)(x 5)(x 3) 2 = 0. The geeral solutio is thus T () = c c c c 4 3. This will clearly be i Θ(5 ). Part 3. Chage of variable techique for decreasebycostatfactor recurrece relatios. May recursive algorithms (such as biary search or merge sort ) work by dividig the iput i half ad callig itself o the owhalfsized iput. Aalyzig the efficiecy of such algorithms leads to recurrece relatios that give T () i terms of T ( 2 ) istead of i terms of T ( 1). These decrease by a costatfactor recurrece relatios ca be coverted ito stadard liear recurrece relatios by applyig a simple chage of variable. Let s say we have a recurrece relatio of the form T () = at ( b ) + f() 4
5 where b is some positive iteger (usually b = 2). We will defie a ew fuctio S(k) by the rule S(k) = T (b k ). Our recurrece relatio the gives S(k) = T (b k ) = at ( ) + f(b k ) b k b = at (b k 1 ) + f(b k ) = as(k 1) + f(b k ) which is a firstorder liear recurrece relatio for S. This ca the be solved by the methods we ve already covered. Example 6. Solve the recurrece relatio { 2 = 1 T () = 3T ( 2 ) + log2 () > 1 Solutio: Usig the substitutio S(k) = T (2 k ) this recurrece formula becomes S(k) = T (2 k ) = 3T ( ) + (2 k )log 2 (2 k ) 2 k 2 = 3T (2 k 1 ) + (2 k )k = 3S(k 1) + k2 k The characteristic equatio of this relatio is (x 3)(x 2) 1+1 = 0, so its geeral solutio is S(k) = c 1 3 k + c 2 2 k + c 3 k2 k. What about iitial coditios for S? We ca use the recurrece formula for T to get T (1) = 2, T (2) = 3T (1) + 2 = 8, ad T (4) = 3T (2) + 8 = 32. So, S(0) = T (2 0 ) = T (1) = 2, S(1) = T (2 1 ) = T (2) = 8, ad S(2) = T (2 2 ) = T (4) = 32. We ca ow use these values to get equatios for c 1, c 2, ad c 3 : 2 = S(0) = c c = S(1) = c c c 3 (1)2 1 = 3c 1 + 2c 2 + 2c 3 32 = S(2) = c c c 3 (2)2 2 = 9c 1 + 4c 2 + 8c 3 Solvig this threebythree system of equatios (by whatever method you prefer) leads to the solutio c 1 = 8, c 2 = 6, ad c 3 = 2. This gives the solutio to the recurrece relatio for S as S(k) = 8 3 k 6 2 k 2k2 k But we wat a formula for T, ot S. Rememberig that S(k) = T (2 k ) ad assumig is a power of 2, we ca say T () = S(log 2 ) = 8 3 log log 2 2(log 2 )2 log 2 = 8 3 log 2 6 2log 2 valid for all values of that are powers of 2. Here we have used the familiar logarithm law that says 2 log 2 is equal to. What ca we do to simplify the term 3 log 2? This just requires a bit of logarithmic trickery: 3 log 2 = (2 log 2 3 ) log 2 = (2 log 2 ) log 2 3 = log 2 3 So, we ca rewrite our aswer as 5
6 T () = 8 log log 2 for = 1, 2, 4, 8,... Two otes about this solutio are i order: First, sice log 2 3 is clearly betwee 1 ad 2 (sice 3 is betwee 2 1 ad 2 2 ) we ca say that our solutio is i O( 2 ) (at least for the values = 1, 2, 4, 8,... ). Secod, we kow that our formula for T () is valid for powers of 2 that is, it correctly computes T (1), T (2), T (4), T (8), ad so o. What about the other values of? Thaks to somethig called the Smoothess Rule we ca at least say that our formula for T () has the correct bigtheta behavior. See your text for the exact statemet, but the basic idea is this: Smoothess Rule: Suppose T () is i Θ(f()) for values of that are powers of a costat b 2. The if f() is a ice fuctio (here ice icludes the polyomials, 2, 3, etc. but ot expoetial or factorial fuctios) we ca say that T () really is i Θ(f()). Homework Problems Solve each of the followig recurrece relatios. 1 = 0 (A) T () = 4 = 1 8T ( 1) 15T ( 2) > 1 5 = 0 (B) T () = 9 = 1 6T ( 1) 9T ( 2) > 1 (C) T () = { 1 = 0 2T ( 1) + 3 > 0 1 = 0 (D) T () = 2 = 1 4T ( 1) 3T ( 2) + 1 > 1 Hit: The ohomogeeous term is 1. { 2 = 1 (E) T () = T ( 3 ) + 7 > 1 { 2 = 1 (F) T () = 5T ( 2 ) + 3 > 1 6
Section 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 informationDivide and Conquer, Solving Recurrences, Integer Multiplication Scribe: Juliana Cook (2015), V. Williams Date: April 6, 2016
CS 6, Lecture 3 Divide ad Coquer, Solvig Recurreces, Iteger Multiplicatio Scribe: Juliaa Cook (05, V Williams Date: April 6, 06 Itroductio Today we will cotiue to talk about divide ad coquer, ad go ito
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 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 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 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 informationA Gentle Introduction to Algorithms: Part II
A Getle Itroductio to Algorithms: Part II Cotets of Part I:. Merge: (to merge two sorted lists ito a sigle sorted list.) 2. Bubble Sort 3. Merge Sort: 4. The BigO, BigΘ, BigΩ otatios: asymptotic bouds
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 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 informationPage 2 of 14 = T(2) + 2 = [ T(3)+1 ] + 2 Substitute T(3)+1 for T(2) = T(3) + 3 = [ T(4)+1 ] + 3 Substitute T(4)+1 for T(3) = T(4) + 4 After i
Page 1 of 14 Search C455 Chapter 4  Recursio Tree Documet last modified: 02/09/2012 18:42:34 Uses: Use recursio tree to determie a good asymptotic boud o the recurrece T() = Sum the costs withi each level
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 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 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 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 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 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 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 informationSearching Algorithm Efficiencies
Efficiecy of Liear Search Searchig Algorithm Efficiecies Havig implemeted the liear search algorithm, how would you measure its efficiecy? A useful measure (or metric) should be geeral, applicable to ay
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 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 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 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 information6.042/18.062J Mathematics for Computer Science. Recurrences
6.04/8.06J Mathematics for Computer Sciece Srii Devadas ad Eric Lehma March 7, 00 Lecture Notes Recurreces Recursio breakig a object dow ito smaller objects of the same type is a major theme i mathematics
More informationMATH /2003. Assignment 4. Due January 8, 2003 Late penalty: 5% for each school day.
MATH 260 2002/2003 Assigmet 4 Due Jauary 8, 2003 Late pealty: 5% for each school day. 1. 4.6 #10. A croissat shop has plai croissats, cherry croissats, chocolate croissats, almod croissats, apple croissats
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 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 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 informationCS 253: Algorithms. Chapter 4. DivideandConquer Recurrences Master Theorem. Credit: Dr. George Bebis
CS 5: Algorithms Chapter 4 DivideadCoquer Recurreces Master Theorem Credit: Dr. George Bebis Recurreces ad Ruig Time Recurreces arise whe a algorithm cotais recursive calls to itself Ruig time is represeted
More informationGregory Carey, 1998 Linear Transformations & Composites  1. Linear Transformations and Linear Composites
Gregory Carey, 1998 Liear Trasformatios & Composites  1 Liear Trasformatios ad Liear Composites I Liear Trasformatios of Variables Meas ad Stadard Deviatios of Liear Trasformatios A liear trasformatio
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 informationRecursion and Recurrences
Chapter 5 Recursio ad Recurreces 5.1 Growth Rates of Solutios to Recurreces Divide ad Coquer Algorithms Oe of the most basic ad powerful algorithmic techiques is divide ad coquer. Cosider, for example,
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 informationAlgebra Work Sheets. Contents
The work sheets are grouped accordig to math skill. Each skill is the arraged i a sequece of work sheets that build from simple to complex. Choose the work sheets that best fit the studet s eed ad will
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 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 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 informationData Structures. Outline
Data Structures Solvig Recurreces Tzachi (Isaac) Rose 1 Outlie Recurrece The Substitutio Method The Iteratio Method The Master Method Tzachi (Isaac) Rose 2 1 Recurrece A recurrece is a fuctio defied i
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 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 information6 Algorithm analysis
6 Algorithm aalysis Geerally, a algorithm has three cases Best case Average case Worse case. To demostrate, let us cosider the a really simple search algorithm which searches for k i the set A{a 1 a...
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 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 informationAlgorithms and Data Structures DV3. Arne Andersson
Algorithms ad Data Structures DV3 Are Adersso Today s lectures Textbook chapters 14, 5 Iformatiostekologi Itroductio Overview of Algorithmic Mathematics Recurreces Itroductio to Recurreces The Substitutio
More informationTHE REGRESSION MODEL IN MATRIX FORM. For simple linear regression, meaning one predictor, the model is. for i = 1, 2, 3,, n
We will cosider the liear regressio model i matrix form. For simple liear regressio, meaig oe predictor, the model is i = + x i + ε i for i =,,,, This model icludes the assumptio that the ε i s are a sample
More informationCHAPTER 3 THE TIME VALUE OF MONEY
CHAPTER 3 THE TIME VALUE OF MONEY OVERVIEW A dollar i the had today is worth more tha a dollar to be received i the future because, if you had it ow, you could ivest that dollar ad ear iterest. Of all
More informationPresent Value Factor To bring one dollar in the future back to present, one uses the Present Value Factor (PVF): Concept 9: Present Value
Cocept 9: Preset Value Is the value of a dollar received today the same as received a year from today? A dollar today is worth more tha a dollar tomorrow because of iflatio, opportuity cost, ad risk Brigig
More informationx(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 informationBasic Elements of Arithmetic Sequences and Series
MA40S PRECALCULUS UNIT G GEOMETRIC SEQUENCES CLASS NOTES (COMPLETED NO NEED TO COPY NOTES FROM OVERHEAD) Basic Elemets of Arithmetic Sequeces ad Series Objective: To establish basic elemets of arithmetic
More informationI. Chisquared Distributions
1 M 358K Supplemet to Chapter 23: CHISQUARED DISTRIBUTIONS, TDISTRIBUTIONS, AND DEGREES OF FREEDOM To uderstad tdistributios, we first eed to look at aother family of distributios, the chisquared distributios.
More informationLecture Notes CMSC 251
We have this messy summatio to solve though First observe that the value remais costat throughout the sum, ad so we ca pull it out frot Also ote that we ca write 3 i / i ad (3/) i T () = log 3 (log ) 1
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 information
Factoring x n 1: cyclotomic and Aurifeuillian polynomials Paul Garrett
(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 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 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 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 informationLearning outcomes. Algorithms and Data Structures. Time Complexity Analysis. Time Complexity Analysis How fast is the algorithm? Prof. Dr.
Algorithms ad Data Structures Algorithm efficiecy Learig outcomes Able to carry out simple asymptotic aalysisof algorithms Prof. Dr. Qi Xi 2 Time Complexity Aalysis How fast is the algorithm? Code the
More information5.4 Amortization. Question 1: How do you find the present value of an annuity? Question 2: How is a loan amortized?
5.4 Amortizatio Questio 1: How do you fid the preset value of a auity? Questio 2: How is a loa amortized? Questio 3: How do you make a amortizatio table? Oe of the most commo fiacial istrumets a perso
More informationhp calculators HP 30S Base Conversions Numbers in Different Bases Practice Working with Numbers in Different Bases
Numbers i Differet Bases Practice Workig with Numbers i Differet Bases Numbers i differet bases Our umber system (called HiduArabic) is a decimal system (it s also sometimes referred to as deary system)
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 informationAsymptotic Notation, Recurrences: Substitution, Iteration, Master Method. Lecture 2
Asymptotic Notatio, Recurreces: Substitutio, Iteratio, Master Method Lecture 2 Solvig recurreces The aalysis of merge sort from Lecture 1 required us to solve a recurrece. Recurreces are like solvig itegrals,
More informationChapter 6: Variance, the law of large numbers and the MonteCarlo method
Chapter 6: Variace, the law of large umbers ad the MoteCarlo method Expected value, variace, ad Chebyshev iequality. If X is a radom variable recall that the expected value of X, E[X] is the average value
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 informationThe geometric series and the ratio test
The geometric series ad the ratio test Today we are goig to develop aother test for covergece based o the iterplay betwee the it compariso test we developed last time ad the geometric series. A ote about
More informationLiteral Equations and Formulas
. Literal Equatios ad Formulas. OBJECTIVE 1. Solve a literal equatio for a specified variable May problems i algebra require the use of formulas for their solutio. Formulas are simply equatios that express
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 informationHypothesis testing. Null and alternative hypotheses
Hypothesis testig Aother importat use of samplig distributios is to test hypotheses about populatio parameters, e.g. mea, proportio, regressio coefficiets, etc. For example, it is possible to stipulate
More informationSecond Order Linear Partial Differential Equations. Part III
Secod Order iear Partial Differetial Equatios Part III Oedimesioal Heat oductio Equatio revisited; temperature distributio of a bar with isulated eds; ohomogeeous boudary coditios; temperature distributio
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 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 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 informationSolutions to Exercises Chapter 4: Recurrence relations and generating functions
Solutios to Exercises Chapter 4: Recurrece relatios ad geeratig fuctios 1 (a) There are seatig positios arraged i a lie. Prove that the umber of ways of choosig a subset of these positios, with o two chose
More informationLinear Algebra II. 4 Determinants. Notes 4 1st November Definition of determinant
MTH6140 Liear Algebra II Notes 4 1st November 2010 4 Determiats The determiat is a fuctio defied o square matrices; its value is a scalar. It has some very importat properties: perhaps most importat is
More information7.1 Finding Rational Solutions of Polynomial Equations
4 Locker LESSON 7. Fidig Ratioal Solutios of Polyomial Equatios Name Class Date 7. Fidig Ratioal Solutios of Polyomial Equatios Essetial Questio: How do you fid the ratioal roots of a polyomial equatio?
More informationNumerical Solution of Equations
School of Mechaical Aerospace ad Civil Egieerig Numerical Solutio of Equatios T J Craft George Begg Buildig, C4 TPFE MSc CFD Readig: J Ferziger, M Peric, Computatioal Methods for Fluid Dyamics HK Versteeg,
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 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 informationINFINITE SERIES KEITH CONRAD
INFINITE SERIES KEITH CONRAD. Itroductio The two basic cocepts of calculus, differetiatio ad itegratio, are defied i terms of limits (Newto quotiets ad Riema sums). I additio to these is a third fudametal
More information+ 1= x + 1. These 4 elements form a field.
Itroductio to fiite fields II Fiite field of p elemets F Because we are iterested i doig computer thigs it would be useful for us to costruct fields havig elemets. Let s costruct a field of elemets; we
More informationA probabilistic proof of a binomial identity
A probabilistic proof of a biomial idetity Joatho Peterso Abstract We give a elemetary probabilistic proof of a biomial idetity. The proof is obtaied by computig the probability of a certai evet i two
More information1 StateSpace Canonical Forms
StateSpace Caoical Forms For ay give system, there are essetially a ifiite umber of possible state space models that will give the idetical iput/output dyamics Thus, it is desirable to have certai stadardized
More informationYour organization has a Class B IP address of 166.144.0.0 Before you implement subnetting, the Network ID and Host ID are divided as follows:
Subettig Subettig is used to subdivide a sigle class of etwork i to multiple smaller etworks. Example: Your orgaizatio has a Class B IP address of 166.144.0.0 Before you implemet subettig, the Network
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 informationMath Background: Review & Beyond. Design & Analysis of Algorithms COMP 482 / ELEC 420. Solving for Closed Forms. Obtaining Recurrences
Math Backgroud: Review & Beyod. Asymptotic otatio Desig & Aalysis of Algorithms COMP 48 / ELEC 40 Joh Greier. Math used i asymptotics 3. Recurreces 4. Probabilistic aalysis To do: [CLRS] 4 # T() = O()
More informationSolving equations. Pretest. Warmup
Solvig equatios 8 Pretest Warmup We ca thik of a algebraic equatio as beig like a set of scales. The two sides of the equatio are equal, so the scales are balaced. If we add somethig to oe side of the
More information7. Sample Covariance and Correlation
1 of 8 7/16/2009 6:06 AM Virtual Laboratories > 6. Radom Samples > 1 2 3 4 5 6 7 7. Sample Covariace ad Correlatio The Bivariate Model Suppose agai that we have a basic radom experimet, ad that X ad Y
More informationBuilding Blocks Problem Related to Harmonic Series
TMME, vol3, o, p.76 Buildig Blocks Problem Related to Harmoic Series Yutaka Nishiyama Osaka Uiversity of Ecoomics, Japa Abstract: I this discussio I give a eplaatio of the divergece ad covergece of ifiite
More informationAlgebra Vocabulary List (Definitions for Middle School Teachers)
Algebra Vocabulary List (Defiitios for Middle School Teachers) A Absolute Value Fuctio The absolute value of a real umber x, x is xifx 0 x = xifx < 0 http://www.math.tamu.edu/~stecher/171/f02/absolutevaluefuctio.pdf
More informationChapter 5 Unit 1. IET 350 Engineering Economics. Learning Objectives Chapter 5. Learning Objectives Unit 1. Annual Amount and Gradient Functions
Chapter 5 Uit Aual Amout ad Gradiet Fuctios IET 350 Egieerig Ecoomics Learig Objectives Chapter 5 Upo completio of this chapter you should uderstad: Calculatig future values from aual amouts. Calculatig
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 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 informationRADICALS AND SOLVING QUADRATIC EQUATIONS
RADICALS AND SOLVING QUADRATIC EQUATIONS Evaluate Roots Overview of Objectives, studets should be able to:. Evaluate roots a. Siplify expressios of the for a b. Siplify expressios of the for a. Evaluate
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 informationCURIOUS MATHEMATICS FOR FUN AND JOY
WHOPPING COOL MATH! CURIOUS MATHEMATICS FOR FUN AND JOY APRIL 1 PROMOTIONAL CORNER: Have you a evet, a workshop, a website, some materials you would like to share with the world? Let me kow! If the work
More informationDEFINITION OF INVERSE MATRIX
Lecture. Iverse matrix. To be read to the music of Back To You by Brya dams DEFINITION OF INVERSE TRIX Defiitio. Let is a square matrix. Some matrix B if it exists) is said to be iverse to if B B I where
More informationIncremental calculation of weighted mean and variance
Icremetal calculatio of weighted mea ad variace Toy Fich faf@cam.ac.uk dot@dotat.at Uiversity of Cambridge Computig Service February 009 Abstract I these otes I eplai how to derive formulae for umerically
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 informationhp calculators HP 12C Platinum Statistics  correlation coefficient The correlation coefficient HP12C Platinum correlation coefficient
HP 1C Platium Statistics  correlatio coefficiet The correlatio coefficiet HP1C Platium correlatio coefficiet Practice fidig correlatio coefficiets ad forecastig HP 1C Platium Statistics  correlatio coefficiet
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 information1 Computing the Standard Deviation of Sample Means
Computig the Stadard Deviatio of Sample Meas Quality cotrol charts are based o sample meas ot o idividual values withi a sample. A sample is a group of items, which are cosidered all together for our aalysis.
More information