CS103A Handout 23 Winter 2002 February 22, 2002 Solving Recurrence Relations


 Bennett Newton
 1 years ago
 Views:
Transcription
1 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 hoc techique. However, oe very importat class of recurrece relatios ca be explicitly solved i a systematic way. These are the recurrece relatios that express the terms of a sequece as a liear combiatio of previous terms. Defiitio: A liear homogeeous recurrece relatio of degree k with costat coefficiets is a recurrece relatio of the form: a = c 1 a 1 + c a +L+ c k a k where c 1,c,c 3,K,c k are real umbers, ad c k 0. The recurrece relatio i the defiitio is liear sice the righthad side is a sum of multiple of the previous terms of the sequece. The recurrece relatio is homogeeous sice o terms of the recurrece relatio fail to ivolve a previous term of the sequece i some way. The coefficiets of the terms of the sequece are all costats, rather tha fuctios that deped o. The degree is k because a is expressed i terms of the previous k terms of the sequece. A cosequece of strog iductio is that a sequece satisfyig the recurrece relatio i the defiitio is uiquely determied by this recurrece relatio ad the k iitial coditio: a = c 1 a 1 + c a +L+ c k a k a 0 = C 0 a 1 = C 1 KK a k 1 = C k 1 Solvig the Little Mosters The basic approach for solvig liear homogeeous recurrece relatios is to look for solutios of the form a = r, where r is a costat. Note that a = r is a solutio to the recurrece if ad oly if: r = c 1 r 1 + c r + c 3 r 3 +L+ c k r k Whe both sides of the equatio are divided by r k ad the righthad side is subtracted from the left, we obtai the equivalet equatio: r k c 1 r k 1 c r k c 3 r k 3 L c k 1 r c k = 0
2 Cosequetly, the sequece with a = r is a solutio if ad oly if r is a solutio to this last equatio, which is called the characteristic equatio of the recurrece relatio. The solutios of this equatio are called characteristic roots of the recurrece relatio. As we ll see, these characteristic roots ca be used to give a explicit formula for all the solutios of the recurrece relatio. First, we should develop results that deal with liear homogeeous recurrece relatios with costat coefficiets of degree two. The results for secod order relatios ca be exteded to solve higherorder equatios. The mathematics ivolved i provig everythig is really messy, so eve though I ll refer to it i lecture, I do t wat to place it i here, because you might icorrectly thik you re resposible for kowig the proof, ad you re ot. Let s tur our udivided attetio to liear homogeeous recurrece relatios of degree two. First, cosider the case where there are two distict characteristic roots. A Fact Without Proof: Let c 1 ad c be real umbers. Suppose that r rc 1 c = 0 has two distict roots r 1 ad r. The the sequece { a } is a solutio of the recurrece relatio a = c 1 a 1 + c a if ad oly if a = b 1 r 1 + b r for = 0,1,,3,4,5,K where b 1 ad b are costats determied by the iitial coditios of the recurrece relatio. Problem: Fid a explicit solutio to the tilig recurrece relatio we developed last time. You may recall that it wet asomethig like this: T = T 1 + T T 0 = 1 T 1 = 1 Well, we surmise that the solutio will be a liear combiatio or terms havig the form T = r. Followig the rule from above, we just plug i T = r ito out recurrece relatio ad see what costraits are placed o r. Let s liste i: T = T 1 + T guess that T = r r = r 1 + r r = r + 1 r r 1= 0 r 1 = 1+ 5 ;r = 1 5 Well, so the guess that T = r is a good oe provided that r be equal to oe of the two roots above. I fact, ay liear combiatio of 1+ 5 ad iitial coditios for a momet that is, T = b 1 will also satisfy the recurrece if you igore the b for ay costat coefficiets as factors. It s oly whe you supply the iitial coditios that b 1 ad b are required to adopt a specific value. I fact, the iitial coditios here dictate that:
3 3 b b 1 + b = 1 + b 1 5 = 1 Ad after solvig for b 1 ad b do we fially arrive at the uique solutio to our recurrece problem: T = How very satisfyig. Problem: Fid a closed form solutio to the bit strig problem modeled i our last hadout. Welllllll, the recurrece relatio, save the iitial coditios, is exactly the same as it that for the tilig problem; that traslates to a solutio of the same basic form. But the fact that the iitial coditios are differet hits that the costats multiplyig the idividual terms: 1+ 5 B = b b ote same form, but possibly differet costats The differet iitial coditios place differet costrait o the values of b 1 ad b do we. Now the followig system of equatios must be solved: b b 1 + b = 1 + b 1 5 = We ed up with somethig like this as our solutio i this case: B = Problem: Solve the recurrece relatio give below J = J J 0 = 3J 1 = 5 You may be askig, Jerry, where i the world is the J 1 term? Relax it s i there it s just that its multiplyig coefficiet is zero, so I did t bother writig it i. It s still a homogeeous equatio of degree two, so I solve it like ay other recurrece of this type.
4 4 J = J J 0 = 3J 1 = 5 J = J r = 1 r 1 = 0 r 1 = 1;r = 1 J = b 1 ( 1) +b ( 1) =b 1 +b ( 1) Fuy little twist droppig the ( 1), sice it s trasparet to us ayway. b 1 + b = 5 b 1 b = 5 3 b 1 = 3,b = 5 3 J = ( 1) Solvig Coupled Recurrece Relatios The first recurreces hadout icluded examples where solutios ivolved coupled recurrece relatios. The circular Tower of Haoi Problem, the 3 tilig problem, the pillar problem, ad the Martia DNA problem were all complex eough to require (or at least beefit from) the ivetio of a secod coutig problem ad a secod recurrece variable. Remember the pillar recurrece? If ot, here it is oce more: 1 S = S 1 + S + 4T 1 = 0 = 1 0 T = 1 S 1 + T 1 = 0 = 1 Because S ad T are defied i terms of oe aother, it s possible to use the secod recurrece of the two to elimiate all occurreces or T from the first. It takes a algebra ad a little igeuity, but whe we rid of the secod variable, we are ofte left with a sigle recurrece relatio which ca be solved like other recurreces we ve see before. The above system of recurreces reduces to a sigle liear, homogeeous, costatcoefficiet equatio oce we get rid of T. Do t believe me? Read o, Thomas. Problem: Solve the above system of recurreces for S by elimiatig T ad solvig the liear equatio that results. I wat to completely elimiate all traces of T ad arrive at (what will tur out to be) a (cubic) recurrece relatio for just S, ad the solve it. First thigs first: We wat to get rid of the T 1 term from the recurrece relatio for S, ad to do that, I make the eat little observatio that the secod equality below follows from the first by replacig by 1: S = S 1 + S + 4T 1 S 1 = S + S 3 + 4T Subtractig the secod equatio from the first, we get:
5 5 ( ) ( S + S 3 + 4T ) S S 1 = S 1 + S + 4T 1 = S 1 S S 3 + 4T 1 4T S = 3S 1 S S 3 + 4T 1 4T ( ) = 3S 1 S S T 1 T Believe it or ot, this is progress, because I have somethig very iterestig to say about T 1 T it s always equal to S. Just look at the crafty maipulatios I work up from the T recurrece relatio: T = S 1 + T 1 T 1 = S + T T 1 T = S + T T = S How crafty! Now I ca eradicate ay metio of T from the S recurrece relatio, ad I do so likea this: S = 3S 1 S S 3 + 4( T 1 T ) = 3S 1 S S 3 + 4S = 3S 1 + 3S S 3 Therefore, a ucoupled recurrece relatio for S ca be expressed as follows: 1 S = S 1 + S 0 + 4T 1 = 9 3S 1 + 3S S 3 = 0 = 1 = 3 Admittedly, that was a lot of work, but this is pretty excitig, because we re o the verge of takig what is clearly a homogeeous, costatcoefficiet equatio ad comig up with a closed form solutio. As always, guess a solutio of the form S = a ad substitute to see what values of a make everythig work out regardless of the iitial coditios. More algebra: S S = a = 3S 1 + 3S S 3 S =a a = 3a 1 + 3a a 3 a 3 = 3a + 3a 1 0 = a 3 3a 3a + 1 ( ) ( )( a 3) 0 = ( a + 1) a 4a = ( a + 1) a + 3 a = 1, ± 3 That meas that the geeral solutio to our recurrece (igorig the iitial coditios) is ( ) + y( 3) + z 1 S = x + 3 require that: ( ), where x, y, ad z ca be ay real umbers. Boudary coditios
6 6 S 0 = x + y + z = 1 S 1 = ( + 3 )x + ( 3)y z = S = ( )x + ( 1 4 3)y + z = 9 This is a system of three liear equatios for three ukows, ad it admits exactly oe solutio. Solvig for x, y, ad z is a matter of algebra, ad after all that algebra is over, we coverge o a closed formula of S = ( ), which is ( ) ( 6 3 ) ( ) +1 rouded to the earest iteger. Neat! Problem: Revisit the Martia DNA problem, ad show that the umber of valid Martia DNA strads of legth is give as a + b = F 3 + (yes, the Fiboacci umber!) Let s brig back the recurrece that defied a ad b. 0 = 0 a = = 1 a 1 + b 1 1 = 0 b = 3 = 1 a 1 + 3b 1 Elimiate oe of the recurrece terms i order to get a closed solutio (though this time we ll ultimately eed to solve for both variables, because you re iterested i their sum.) a = a 1 + b 1 b = a 1 + 3b 1 Notice that the first oe tells us somethig about a a 1, so let s subtract a shifted versio of the secod equatio from the origial to get at somethig that s all b ad o a. b = a 1 + 3b 1 b 1 = a + 3b b b 1 = a 1 a ( ) + 3b 1 3b ( ) + 3b 1 3b b = b 1 + a 1 a The first equatio tells us that a 1 a = b. Substitutio yields b = b 1 + 4b + 3b 1 3b = 4b 1 + b The characteristic equatio here is r 4r 1 = 0 ; b = c 1 ( + 5 ) + c ( 5). Because b 0 = 1 ad b 1 = 3, we determie c 1, c by solvig the followig two equatios:
7 7 c 1 + c = 1 ( + 5)c 1 + ( 5)c = 3 c 1 = 5 + 5, c = 5 5. b = ( + 5) ( 5). You might thik you have to do the same thig for a, ad i theory you re right i that you eed to solve it, but we more or less have. Because b = a 1 + 3b 1 we actually kow that b = a 1 + 3b 1 a 1 = b 3b 1 a 1 = 1 b 3 b 1 a = 1 b +1 3 b a + b = 1 b +1 3 b + b = 1 b +1 1 b Recall that we re really just iterested i a + b, ad sice it ca be defied just i terms of the b, we get: a + b = 1 b +1 3 b + b = 1 b +1 1 b = 1 = 1 = 1 = = ( + 5) (( + 5) 1) ( + 5) ( 1 + 5) ( + 5) ( + 5) ( + 5) ( 5) (( 5) 1) 5 ( 5) ( 1 5) 5 ( 5) ( ) ( + 5) ( ) ( 5) That s typically the closedform you d leave it i, but we also wat to show that a + b = F 3 +. Here goes:
8 8 F 3 + = 1 5 = 1 5 = 1 5 = = = = ( + 5) ( + 5) ( + 5) ( + 5) ( + 5) Therefore, a + b = F 3 +, ad we rejoice. ( + 5) ( + 5) ( + 5) Problem: What about the Circular Tower of Haoi recurrece? Why ca t we solve that oe as easily? Well, you probably already oticed that both the pillar ad the DNA recurreces did t have ay ihomogeeous terms aywhere. That s ot the case with the Circular Tower of Haoi recurrece, so while we are certaily ivited to elimiate oe of the variables from the set of recurreces, there s ot much hope for solvig it like we did for Martia DNA. Q = 0; R if = 0 if > 0 R = 0; Q + Q if = 0 if > 0 Clearly, it s a piece of cake to defie Q i terms of previous Qs, but these 1s just are t goig to go away. so that Q = R ( ) + 1 = Q 1 + Q + 1 = Q 1 + Q Q = 1 Q + Q if = 0 if = 1 if > 1 We re sorta bummig because of that ihomogeeous 3. There is a trick to solvig this oe, but it s ot somethig I m formally goig to require you to kow, sice there s little pedagogical value i memorizig tricks. For those of you with othig to do o a Friday
9 9 ight, try substitutig Q = A 1 1 ito the above system (makig sure to adjust those base cases as well.) Solve for A, the take whatever you got there ad subtract 1 from it to get Q. Geeral Approach to Solvig all Liear FirstOrder Recurreces There is a geeral techique that ca reduce virtually ay recurrece of the form a T = b T 1 + c to a sum. The idea is to multiply both sides by a summatio factor, s, to arrive at s a T = s b T 1 + s c. This factor s is chose to make s b = s 1 a 1. The if we write S = s a T we have a sumrecurrece for S as: S = S 1 + s c Hece S = s 0 a 0 T 0 + s k c k = s 1b 1 T 0 + s k c k 1 k 1 T = s s a 1 b 1 T 0 + s k c k. 1 k 1 k ad the solutio to the origial recurrece becomes So, you ask: How ca we be clever eough to fid the perfect s? Well, the relatio s = s 1 a 1 be ufolded by repeated substitutio for the s i to tell us that the fractio: b ca s = a 1 a a 3 Ka 1 b b 1 b Kb (or ay coveiet costat multiple of this value) will be a suitable summatio factor. Problem: Solve V = 5 =0 V 1 + 3! 1 factor. by fidig the appropriate summatio Derivatio of the solutio just follows protocol. You're specifically told to use summatio factors here, so we should do that. Notice here that a = 1 ad b =, so that the summatio factor should be chose as: s = 1 1! = 1! V 0 = 5 for sure, but the recurrece formula V = V 1 + 3! becomes 1! V = 1 ( 1)! V Let T = 1! V for the time beig, just so we ca solve a easier recurrece relatio. We tackle T = T 1 + 3, ad would't you kow it: T = 3 + T 0 = ! V 0 = T = 1! V. We eed V =!T, so therefore we have that V =! ( 3 + 5). Woo!
10 0 =0 Problem: Solve C = summatio factor. C k 0 k 1 by fidig the appropriate Let s write out the recurreces for C ad C 1 a little more explicitly. C = C k 0 k 1 C 1 = + 1 C k 0 k Subtractig the secod oe from the first oe is a good idea, but oly after the multiply through by a factor that ll make each of the summatios equal to each other: C = C k 0 k 1 1 C 1 = = 1 + C k 0 k C k 0 k Subtractig the secod oe from the first, we arrive at: C 1 C 1 = C k 1+ 0 k 1 = + C 1 C = + 1 C 1 + C = ( + 1)C 1 + C k 0 k If we choose a summatio factor of s = a 1 a a 3 Ka 1 = ( 1) ( ) ( 3)K1 = b b 1 b Kb ( + 1)( 1)K3 through, we arrive at: + 1 ( ) C = ( + 1) ( ) C = C 1 + ( )C ( ) C + 1 = C ( + 1) + 1 ( ) ad multiply
11 If we let ( + 1)D = C, the we fially arrive at a equatio that looks reasoable: D = D Repeated substitutio yields the followig: D = D = D = D = D M 1 = D 0 + k k ( ) = H +1 1 = H = H +1 Recall that C = ( + 1)D, so that C = + 1 ( ) H = ( +1)H
Soving 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 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 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 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 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 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 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 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 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 /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 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 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 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 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 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 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 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 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 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 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 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 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 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 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 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 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 informationChapter 7. In the questions below, describe each sequence recursively. Include initial conditions and assume that the sequences begin with a 1.
Use the followig to aswer questios 6: Chapter 7 I the questios below, describe each sequece recursively Iclude iitial coditios ad assume that the sequeces begi with a a = 5 As: a = 5a,a = 5 The Fiboacci
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 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 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 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 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 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 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 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 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 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 informationDerivation of the Poisson distribution
Gle Cowa RHUL Physics 1 December, 29 Derivatio of the Poisso distributio I this ote we derive the fuctioal form of the Poisso distributio ad ivestigate some of its properties. Cosider a time t i which
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 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 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 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 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 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 informationElementary Theory of Russian Roulette
Elemetary Theory of Russia Roulette iterestig patters of fractios Satoshi Hashiba Daisuke Miematsu Ryohei Miyadera Itroductio. Today we are goig to study mathematical theory of Russia roulette. If some
More informationDiscrete Mathematics and Probability Theory Spring 2014 Anant Sahai Note 13
EECS 70 Discrete Mathematics ad Probability Theory Sprig 2014 Aat Sahai Note 13 Itroductio At this poit, we have see eough examples that it is worth just takig stock of our model of probability ad may
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 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 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 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 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 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 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 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 informationNPTEL STRUCTURAL RELIABILITY
NPTEL Course O STRUCTURAL RELIABILITY Module # 0 Lecture 1 Course Format: Web Istructor: Dr. Aruasis Chakraborty Departmet of Civil Egieerig Idia Istitute of Techology Guwahati 1. Lecture 01: Basic Statistics
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 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 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 informationProperties of MLE: consistency, asymptotic normality. Fisher information.
Lecture 3 Properties of MLE: cosistecy, asymptotic ormality. Fisher iformatio. I this sectio we will try to uderstad why MLEs are good. Let us recall two facts from probability that we be used ofte throughout
More information5: Introduction to Estimation
5: Itroductio to Estimatio Cotets Acroyms ad symbols... 1 Statistical iferece... Estimatig µ with cofidece... 3 Samplig distributio of the mea... 3 Cofidece Iterval for μ whe σ is kow before had... 4 Sample
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 informationEssential Question How can you use properties of exponents to simplify products and quotients of radicals?
. Properties of Ratioal Expoets ad Radicals Essetial Questio How ca you use properties of expoets to simplify products ad quotiets of radicals? Reviewig Properties of Expoets Work with a parter. Let a
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 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 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 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 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 informationA Simplified Binet Formula for kgeneralized Fibonacci Numbers
A Simplified Biet Formula for kgeeralized Fiboacci Numbers Gregory P. B. Dresde Departmet of Mathematics Washigto ad Lee Uiversity Lexigto, VA 440 dresdeg@wlu.edu Abstract I this paper, we preset a particularly
More informationDivergence of p 1/p. Adrian Dudek. adrian.dudek[at]anu.edu.au
Divergece of / Adria Dudek adria.dudek[at]au.edu.au Whe I was i high school, my maths teacher cheekily told me that it s ossible to add u ifiitely may umbers ad get a fiite umber. She the illustrated this
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 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 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 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 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 informationYour grandmother and her financial counselor
Sectio 10. Arithmetic Sequeces 963 Objectives Sectio 10. Fid the commo differece for a arithmetic sequece. Write s of a arithmetic sequece. Use the formula for the geeral of a arithmetic sequece. Use the
More informationMath 475, Problem Set #6: Solutions
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),
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 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 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 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 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 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 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 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 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 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 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 informationSection 1.6: Proof by Mathematical Induction
Sectio.6 Proof by Iductio Sectio.6: Proof by Mathematical Iductio Purpose of Sectio: To itroduce the Priciple of Mathematical Iductio, both weak ad the strog versios, ad show how certai types of theorems
More informationSolving Logarithms and Exponential Equations
Solvig Logarithms ad Epoetial Equatios Logarithmic Equatios There are two major ideas required whe solvig Logarithmic Equatios. The first is the Defiitio of a Logarithm. You may recall from a earlier topic:
More informationEngineering 323 Beautiful Homework Set 3 1 of 7 Kuszmar Problem 2.51
Egieerig 33 eautiful Homewor et 3 of 7 Kuszmar roblem.5.5 large departmet store sells sport shirts i three sizes small, medium, ad large, three patters plaid, prit, ad stripe, ad two sleeve legths log
More informationTHE UNLIKELY UNION OF PARTITIONS AND DIVISORS
THE UNLIKELY UNION OF PARTITIONS AND DIVISORS Abdulkadir Hasse, Thomas J. Osler, Mathematics Departmet ad Tirupathi R. Chadrupatla, Mechaical Egieerig Rowa Uiversity Glassboro, NJ 828 I the multiplicative
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 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 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 informationProject Deliverables. CS 361, Lecture 28. Outline. Project Deliverables. Administrative. Project Comments
Project Deliverables CS 361, Lecture 28 Jared Saia Uiversity of New Mexico Each Group should tur i oe group project cosistig of: About 612 pages of text (ca be loger with appedix) 612 figures (please
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 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 informationLecture 7: Borel Sets and Lebesgue Measure
EE50: Probability Foudatios for Electrical Egieers JulyNovember 205 Lecture 7: Borel Sets ad Lebesgue Measure Lecturer: Dr. Krisha Jagaatha Scribes: Ravi Kolla, Aseem Sharma, Vishakh Hegde I this lecture,
More information