Sequences II. Chapter Convergent Sequences


 Martina Whitehead
 2 years ago
 Views:
Transcription
1 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., ǫ = 0.0 ad ǫ = 0.00 fid a N such that a < ǫ wheever > N. Defiitio Let a R. A sequece a ) teds to a if, for each ǫ > 0, there exists a atural umber N such that a a < ǫ for all > N. See figure 3. for a illustratio of this defiitio. We use the otatio a ) a, a a, as ad lim a = a ad say that a ) coverges to a, or the limit of the sequece a ) as teds to ifiity is a. Example Prove a ) = ) +. Let ǫ > 0. We have to fid a atural umber N so that a = + < ǫ whe > N. We have + = + = + <. Good Nough Ay N that works is good eough  it does t have to be the smallest possible N. Recycle Have a closer look at figure 3., what has bee chaged from figure2.6ofchapter2? Itturs out that this defiitio is very similar to the defiitio of a ull sequece. Elephats Revisited A ull sequece is a special case of a coverget sequece. So memorise this defiitio ad get the other oe for free. Hece it suffices to fid N so that < ǫ wheever > N. But < ǫ if ad oly if ǫ < so it is eough to choose N to be a atural umber with N > ǫ. The, if > N we have a = + = + = + < < N < ǫ. Lemma a ) a if ad oly if a a) 0. 27
2 28 CHAPTER 3. SEQUENCES II e e a 0 N Figure 3.: Coverget sequeces; first choose ε, the fid N. Proof. We kow that a a) 0 meas that for each ǫ > 0, there exists a atural umber N such that a a < ǫ whe > N. But this is exactly the defiitio of a ) a. We have spoke of the limit of a sequece but ca a sequece have more tha oe limit? The aswer had better be No or our defiitio is suspect. Theorem Uiqueess of Limits A sequece caot coverge to more tha oe limit. Exercise Prove the theorem by assumig a ) a, a ) b with a < b ad obtaiig a cotradictio. [Hit: try drawig a graph of the sequeces with a ad b marked o] Theorem Every coverget sequece is bouded. Exercise 2 Prove the theorem above. Coectio It wo t have escaped your otice that the Sum Rule for ull sequeces is just a special case of the Sum Rule for sequeces. The same goes for the Product Rule. Why do t we have a Quotiet Rule for ull sequeces? 3.2 Algebra of Limits Theorem Sum Rule, Product Rule ad Quotiet Rule Let a,b R. Suppose a ) a ad b ) b. The ca +db ) ca+db a b Sum Rule for Sequeces a b ) ab Product Rule for Sequeces ) a b, ifb 0 Quotiet Rule for Sequeces
3 3.2. ALGEBRA OF LIMITS 29 There is aother useful way we ca express all these rules: If a ) ad b ) are coverget the lim lim ca +db ) = c lim a +d lim b lim a b ) = lim a lim b a b ) = lim a ), if lim lim b ) b ) 0 Sum Rule Product Rule Quotiet Rule Example I full detail 2 +)6 ) lim = lim ) ) usig the Quotiet Rule [ + 2 ) 6 = lim lim 2+ 5 = 3 ) )] usig the Product ad Sum Rules )) +lim 2 6 lim )) = +0)6 0) 2+0 = 3 2+5lim 3 ) Uless you are asked to show where you use each of the rules you ca keep your solutios simpler. Either of the followig is fie: ) ) 2 +)6 ) + 6 lim 2 3 = lim = +0)6 0) = or 2 +)6 ) = ) ) )6 0) = Bigger ad Better By iductio, the Sum ad Product Rules ca be exteded to cope with ay fiite umber of coverget sequeces. For example, for three sequeces: lim abc) = lim a lim b lim c Exercise 3 Show that a a)b b)+ab b)+ba a) = a b ab Exercise 4 Use the idetity i Exercise 3 ad the rules for ull sequeces to prove the Product Rule for sequeces. Exercise 5 Write a proof of the Quotiet Rule. You might like to structure your proof as follows.. Note that bb ) b 2 ad show that bb > b2 2 for sufficietly large. Do t Worry You caot use the Quotiet Rule if some of the b s are zero. Fortuately, the fact that b 0 esures that there ca oly be a fiite umber of these. Ca you see why? So you ca apply the Rule to the shifted sequece a N+/b N+) for some wisely chose N.
4 30 CHAPTER 3. SEQUENCES II 2. The show that evetually 0 b. 3. Now tackle a b = a b. b b ) 2 b b b 2 adtherefore b Cuig Required Do you kow a cuig way to rewrite ? Exercise 6 Fid the limit of each of the sequeces defied below ) 2) Coectio The Sadwich Rule for ull sequeces represets the case whe l = Further Useful Results Theorem Sadwich Theorem for Sequeces Suppose a ) l ad b ) l. If a c b the c ) l. This improved Sadwich theorem ca be tackled by rewritig the hypothesis as 0 c a b a ad applyig the earlier Sadwich theorem. Exercise 7 Prove the Sadwich Theorem for sequeces. Max ad Mi I your proof you may well use the fact that each fiite set has a maximum ad a miimum. Is this true of ifiite sets? There are goig to be may occasios whe we are iterested i the behaviour of a sequece after a certai poit, regardless of what wet o before that. This ca be doe by choppig off the first N terms of a sequece a ) to get a shifted sequece b ) give by b = a N+. We ofte write this as a N+ ), so that a N+ ) = a N+, a N+2, a N+3, a N+4,... which starts at the term a N+. We use it i the defiitio below. Defiitio A sequece a ) satisfies a certai property evetually if there is a atural umber N such that the sequece a N+ ) satisfies that property. For istace, a sequece a ) is evetually bouded if there exists N such that the sequece a N+ ) is bouded. Lemma If a sequece is evetually bouded the it is bouded. Exercise 8 Prove this lemma. The ext result, called the Shift Rule, tells you that a sequece coverges if ad oly if it coverges evetually. So you ca chop off or add o ay fiite
5 3.3. FURTHER USEFUL RESULTS 3 umber of terms at the begiig of a sequece without affectig the coverget behaviour of its ifiite tail. Theorem Shift Rule Let N be a atural umber. Let a ) be a sequece. The a a if ad oly if the shifted sequece a N+ a. Proof. Fix ǫ > 0. If a ) a we kow there exists N such that a a < ǫ wheever > N. Whe > N, weseethatn+ > N, therefore a N+ a < ǫ. Hece a N+ ) a. Coversely, suppose that a N+ ) a. The there exists N 2 such that a N+ a < ǫ wheever > N 2. Whe > N + N 2 the N > N 2 so a a = a N+ N) a < ǫ. Hece a ) a. Corollary Sadwich Theorem with Shift Rule Suppose a ) l ad b ) l. If evetually a c b the c ) l. Example We kow / 0 therefore /+5) 0. Exercise 9 Show that the Shift Rule also works for sequeces which ted to ifiity: a ) if ad oly if a N+ ). If all the terms of a coverget sequece sit withi a certai iterval, does its limit lie i that iterval, or ca it escape? For istace, if the terms of a coverget sequece are all positive, is its limit positive too? Lemma Suppose a ) a. If a 0 for all the a 0. Exercise 0 Prove this result. [Hit: Assume that a < 0 ad let ǫ = a > 0. The use the defiitio of covergece to arrive at a cotradictio.] Exercise Prove or disprove the followig statemet: Suppose a ) a. If a > 0 for all the a > 0. Theorem Iequality Rule Suppose a ) a ad b ) b. If evetually) a b the a b. Exercise 2 Prove this result usig the previous Lemma. [Hit: Cosider b a ).]
6 32 CHAPTER 3. SEQUENCES II Limits o Limits Limits caot escape from closed itervals. They ca escape from ope itervals  but oly as far as the ed poits. Cautio Note that the subsequece a i ) is idexed by i ot. I all cases i i. Why is this?) Remember these facts whe subsequeces appear! Prove the obvious It may seem obvious that every subsequece of a coverget sequece coverges, but you should still check that you kow how to prove it! Corollary Closed Iterval Rule Suppose a ) a. If evetually) A a B the A a B. If A < a < B it is ot the case that A < a < B. For example 0 < + < but Subsequeces A subsequece of a ) is a sequece cosistig of some or all) of its terms i their origial order. For istace, we ca pick out the terms with eve idex to get the subsequece a 2,a 4,a 6,a 8,a 0,... or we ca choose all those whose idex is a perfect square a,a 4,a 9,a 6,a 25,... I the first case we chose the terms i positios 2,4,6,8,...ad i the secod those i positios,4,9,6,25,... I geeral, if we take ay strictly icreasig sequece of atural umbers i ) =, 2, 3, 4,... we ca defie a subsequece of a ) by a i ) = a,a 2,a 3,a 4,... Defiitio A subsequece of a ) is a sequece of the form a i ), where i ) is a strictly icreasig sequece of atural umbers. Effectively, the sequece i ) picks out which terms of a ) get to belog to the subsequece. Thik back to the defiitio of covergece of a sequece. Why is it immediate from the defiitio that if a sequece a ) coverges to a the all subsequece a i ) coverge to a? This is a fact which we will be usig costatly i the rest of the course. Notice that the shifted sequece a N+ ) is a subsequece of a ). Exercise 3 Let a ) = 2 ). Write dow the first four terms of the three subsequeces a +4 ), a 3 ) ad a 2 ). Here is aother result which we will eed i later chapters. Exercise 4 Suppose we have a sequece a ) ad are tryig to prove that it coverges. Assume that we have show that the subsequeces a 2 ) ad a 2+ ) both coverge to the same limit a. Prove that a ) a coverges. Exercise 5 Aswer Yes or No to the followig questios, but be sure that you kow why ad that you are t just guessig.
7 3.5. * APPLICATION  SPEED OF CONVERGENCE * 33. A sequece a ) is kow to be icreasig, but ot strictly icreasig. a) Might there be a strictly icreasig subsequece of a )? b) Must there be a strictly icreasig subsequece of a )? 2. If a sequece is bouded, must every subsequece be bouded? 3. If the subsequece a 2, a 3,..., a +,... is bouded, does it follow that the sequece a ) is bouded? 4. If the subsequece a 3, a 4,..., a +2,... is bouded does it follow that the sequece a ) is bouded? 5. If the subsequece a N+, a N+2,..., a N+,... is bouded does it follow that the sequece a ) is bouded? Lemma Every subsequece of a bouded sequece is bouded. Proof. Let a ) be a bouded sequece. The there exist L ad U such that L a U for all. It follows that if a i ) is a subsequece of a ) the L a i U for all i. Hece a i ) is bouded. You might be surprised to lear that every sequece, o matter how boucy ad illbehaved, cotais a icreasig or decreasig subsequece. Theorem Every sequece has a mootoic subsequece. 3.5 * Applicatio  Speed of Covergece * Ofte sequeces are defied recursively, that is, later terms are defied i terms of earlier oes. Cosider a sequece a ) where a 0 = ad a + = a +2, so the sequece begis a 0 =,a = 3,a 2 = 3+2. Exercise 6 Use iductio to show that a 2 for all. Sie Time Agai The fact that a sequece has a guarateed mootoic subsequece does t mea that the subsequece is easy to fid. Try idetifyig a icreasig or decreasig subsequece of si ad you ll see what I mea. Now assume that a ) coverges to a limit, say, a. The: a = lim a = lim a+ ) 2 2 ) = ) 2 lim a + 2 = a 2 2 So to fid a we have to solve the quadratic equatio a 2 a 2 = 0. We ca rewrite this as a+)a 2) = 0, so either a = or a = 2. But which oe is it? The Iequality Theorem comes to our rescue here. Sice a for all it follows that a, therefore a = 2. We will ow ivestigate the speed that a approaches 2.
8 34 CHAPTER 3. SEQUENCES II Exercise 7 Show that 2 a + = 2 a 2+. Use this idetity ad iductio 2+a to show that 2 a ) 2+ for all. How may iteratios are eeded so 3) that a is withi 0 00 is its limit 2? A excellet method for calculatig square roots is the NewtoRaphso method which you may have met at Alevel. Whe applied to the problem of calculatig 2 this leads to the sequece give by: a 0 = 2 ad a + = a + a 2. Exercise 8 Use a calculator to calculate a,a 2,a 3,a 4. Compare them with 2. Exercise 9 Use iductio to show that a 2 for all. Assumig that a ) coverges, show that the limit must be 2. We will ow show that the sequece coverges to 2 like a bat out of hell. Exercise 20 Show that a + 2) = a 2) 2 2a. Usig this idetity show by iductio that a How may iteratios do you eed before you ca guaretee to calculate 2 to withi a error of 0 00 approximately 00 decimal places)? SequecesasiExercise7aresaidtocovergeexpoetially adthoseasi Exercise 20 are said to coverge quadratically sice the error is squared at each iteratio. The stadard methods for calculatig π were expoetial just as is the Archimedes method) util the mid 970s whe a quadratically coverget approximatio was discovered. Check Your Progress By the ed of this chapter you should be able to: Defie what it meas for a sequece to coverge to a limit. Prove that every coverget sequece is bouded. State, prove ad use the followig results about coverget sequeces: If a ) a ad b ) b the: Sum Rule: ca +db ) ca+db Product Rule: a b ) ab Quotiet Rule: a /b ) a/b if b 0 Sadwich Theorem: if a = b ad a c b the c ) a Closed Iterval Rule: if A a B the A a B Explai the term subsequece ad give a rage of examples.
Infinite 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 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 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 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 informationSection 9.2 Series and Convergence
Sectio 9. Series ad Covergece Goals of Chapter 9 Approximate Pi Prove ifiite series are aother importat applicatio of limits, derivatives, approximatio, slope, ad cocavity of fuctios. Fid challegig atiderivatives
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 information4.3. The Integral and Comparison Tests
4.3. THE INTEGRAL AND COMPARISON TESTS 9 4.3. The Itegral ad Compariso Tests 4.3.. The Itegral Test. Suppose f is a cotiuous, positive, decreasig fuctio o [, ), ad let a = f(). The the covergece or divergece
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 information2.7 Sequences, Sequences of Sets
2.7. SEQUENCES, SEQUENCES OF SETS 67 2.7 Sequeces, Sequeces of Sets 2.7.1 Sequeces Defiitio 190 (sequece Let S be some set. 1. A sequece i S is a fuctio f : K S where K = { N : 0 for some 0 N}. 2. For
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 informationMA2108S Tutorial 5 Solution
MA08S Tutorial 5 Solutio Prepared by: LuJigyi LuoYusheg March 0 Sectio 3. Questio 7. Let x := / l( + ) for N. (a). Use the difiitio of limit to show that lim(x ) = 0. Proof. Give ay ɛ > 0, sice ɛ > 0,
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 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 informationif A S, then X \ A S, and if (A n ) n is a sequence of sets in S, then n A n S,
Lecture 5: Borel Sets Topologically, the Borel sets i a topological space are the σalgebra geerated by the ope sets. Oe ca build up the Borel sets from the ope sets by iteratig the operatios of complemetatio
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 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 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 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 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 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 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 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 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 information1. a n = 2. a n = 3. a n = 4. a n = 5. a n = 6. a n =
Versio PREVIEW Homework Berg (5860 This pritout should have 9 questios. Multiplechoice questios may cotiue o the ext colum or page fid all choices before aswerig. CalCb0b 00 0.0 poits Rewrite the fiite
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 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 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 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 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 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 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 informationMeasurable Functions
Measurable Fuctios Dug Le 1 1 Defiitio It is ecessary to determie the class of fuctios that will be cosidered for the Lebesgue itegratio. We wat to guaratee that the sets which arise whe workig with these
More informationTAYLOR SERIES, POWER SERIES
TAYLOR SERIES, POWER SERIES The followig represets a (icomplete) collectio of thigs that we covered o the subject of Taylor series ad power series. Warig. Be prepared to prove ay of these thigs durig the
More information5. SEQUENCES AND SERIES
5. SEQUENCES AND SERIES 5.. Limits of Sequeces Let N = {0,,,... } be the set of atural umbers ad let R be the set of real umbers. A ifiite real sequece u 0, u, u, is a fuctio from N to R, where we write
More informationThe Limit of a Sequence
3 The Limit of a Sequece 3. Defiitio of limit. I Chapter we discussed the limit of sequeces that were mootoe; this restrictio allowed some shortcuts ad gave a quick itroductio to the cocept. But may importat
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 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 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 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 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 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 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 information0,1 is an accumulation
Sectio 5.4 1 Accumulatio Poits Sectio 5.4 BolzaoWeierstrass ad HeieBorel Theorems Purpose of Sectio: To itroduce the cocept of a accumulatio poit of a set, ad state ad prove two major theorems of real
More informationf(x + T ) = f(x), for all x. The period of the function f(t) is the interval between two successive repetitions.
Fourier Series. Itroductio Whe the Frech mathematicia Joseph Fourier (76883) was tryig to study the flow of heat i a metal plate, he had the idea of expressig the heat source as a ifiite series of sie
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 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 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 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 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 information3. Greatest Common Divisor  Least Common Multiple
3 Greatest Commo Divisor  Least Commo Multiple Defiitio 31: The greatest commo divisor of two atural umbers a ad b is the largest atural umber c which divides both a ad b We deote the greatest commo gcd
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 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 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 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 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 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 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 informationStatistics Lecture 14. Introduction to Inference. Administrative Notes. Hypothesis Tests. Last Class: Confidence Intervals
Statistics 111  Lecture 14 Itroductio to Iferece Hypothesis Tests Admiistrative Notes Sprig Break! No lectures o Tuesday, March 8 th ad Thursday March 10 th Exteded Sprig Break! There is o Stat 111 recitatio
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 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 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 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 4: Cheeger s Inequality
Spectral Graph Theory ad Applicatios WS 0/0 Lecture 4: Cheeger s Iequality Lecturer: Thomas Sauerwald & He Su Statemet of Cheeger s Iequality I this lecture we assume for simplicity that G is a dregular
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 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 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 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 informationNATIONAL SENIOR CERTIFICATE GRADE 12
NATIONAL SENIOR CERTIFICATE GRADE MATHEMATICS P EXEMPLAR 04 MARKS: 50 TIME: 3 hours This questio paper cosists of 8 pages ad iformatio sheet. Please tur over Mathematics/P DBE/04 NSC Grade Eemplar INSTRUCTIONS
More informationApproximating the Sum of a Convergent Series
Approximatig the Sum of a Coverget Series Larry Riddle Ages Scott College Decatur, GA 30030 lriddle@agesscott.edu The BC Calculus Course Descriptio metios how techology ca be used to explore covergece
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 informationMath C067 Sampling Distributions
Math C067 Samplig Distributios Sample Mea ad Sample Proportio Richard Beigel Some time betwee April 16, 2007 ad April 16, 2007 Examples of Samplig A pollster may try to estimate the proportio of voters
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 informationCHAPTER 3 DIGITAL CODING OF SIGNALS
CHAPTER 3 DIGITAL CODING OF SIGNALS Computers are ofte used to automate the recordig of measuremets. The trasducers ad sigal coditioig circuits produce a voltage sigal that is proportioal to a quatity
More informationKey Ideas Section 81: Overview hypothesis testing Hypothesis Hypothesis Test Section 82: Basics of Hypothesis Testing Null Hypothesis
Chapter 8 Key Ideas Hypothesis (Null ad Alterative), Hypothesis Test, Test Statistic, Pvalue Type I Error, Type II Error, Sigificace Level, Power Sectio 81: Overview Cofidece Itervals (Chapter 7) are
More informationOverview of some probability distributions.
Lecture Overview of some probability distributios. I this lecture we will review several commo distributios that will be used ofte throughtout the class. Each distributio is usually described by its probability
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 informationApproximating Area under a curve with rectangles. To find the area under a curve we approximate the area using rectangles and then use limits to find
1.8 Approximatig Area uder a curve with rectagles 1.6 To fid the area uder a curve we approximate the area usig rectagles ad the use limits to fid 1.4 the area. Example 1 Suppose we wat to estimate 1.
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 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 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 informationa 4 = 4 2 4 = 12. 2. Which of the following sequences converge to zero? n 2 (a) n 2 (b) 2 n x 2 x 2 + 1 = lim x n 2 + 1 = lim x
0 INFINITE SERIES 0. Sequeces Preiary Questios. What is a 4 for the sequece a? solutio Substitutig 4 i the expressio for a gives a 4 4 4.. Which of the followig sequeces coverge to zero? a b + solutio
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 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 informationCase Study. Normal and t Distributions. Density Plot. Normal Distributions
Case Study Normal ad t Distributios Bret Halo ad Bret Larget Departmet of Statistics Uiversity of Wiscosi Madiso October 11 13, 2011 Case Study Body temperature varies withi idividuals over time (it ca
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 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 informationEconomics 140A Confidence Intervals and Hypothesis Testing
Ecoomics 140A Cofidece Itervals ad Hypothesis Testig Obtaiig a estimate of a parameter is ot the al purpose of statistical iferece because it is highly ulikely that the populatio value of a parameter is
More informationContents. 7 Sequences and Series. 7.1 Sequences and Convergence. Calculus II (part 3): Sequences and Series (by Evan Dummit, 2015, v. 2.
Calculus II (part 3): Sequeces ad Series (by Eva Dummit, 05, v..00) Cotets 7 Sequeces ad Series 7. Sequeces ad Covergece......................................... 7. Iite Series.................................................
More informationThe Harmonic Series Diverges Again and Again
The Harmoic Series Diverges Agai ad Agai Steve J. Kifowit Prairie State College Terra A. Stamps Prairie State College The harmoic series, = = 3 4 5, is oe of the most celebrated ifiite series of mathematics.
More informationChapter One BASIC MATHEMATICAL TOOLS
Chapter Oe BAIC MATHEMATICAL TOOL As the reader will see, the study of the time value of moey ivolves substatial use of variables ad umbers that are raised to a power. The power to which a variable is
More informationMath 105: Review for Final Exam, Part II  SOLUTIONS
Math 5: Review for Fial Exam, Part II  SOLUTIONS. Cosider the fuctio fx) =x 3 l x o the iterval [/e, e ]. a) Fid the x ad ycoordiates of ay ad all local extrema ad classify each as a local maximum or
More informationUniversity of California, Los Angeles Department of Statistics. Distributions related to the normal distribution
Uiversity of Califoria, Los Ageles Departmet of Statistics Statistics 100B Istructor: Nicolas Christou Three importat distributios: Distributios related to the ormal distributio Chisquare (χ ) distributio.
More informationConfidence Intervals for the Mean of Nonnormal Data Class 23, 18.05, Spring 2014 Jeremy Orloff and Jonathan Bloom
Cofidece Itervals for the Mea of Noormal Data Class 23, 8.05, Sprig 204 Jeremy Orloff ad Joatha Bloom Learig Goals. Be able to derive the formula for coservative ormal cofidece itervals for the proportio
More informationTaylor Series and Polynomials
Taylor Series ad Polyomials Motivatios The purpose of Taylor series is to approimate a fuctio with a polyomial; ot oly we wat to be able to approimate, but we also wat to kow how good the approimatio is.
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 informationThe following example will help us understand The Sampling Distribution of the Mean. C1 C2 C3 C4 C5 50 miles 84 miles 38 miles 120 miles 48 miles
The followig eample will help us uderstad The Samplig Distributio of the Mea Review: The populatio is the etire collectio of all idividuals or objects of iterest The sample is the portio of the populatio
More informationwhere: T = number of years of cash flow in investment's life n = the year in which the cash flow X n i = IRR = the internal rate of return
EVALUATING ALTERNATIVE CAPITAL INVESTMENT PROGRAMS By Ke D. Duft, Extesio Ecoomist I the March 98 issue of this publicatio we reviewed the procedure by which a capital ivestmet project was assessed. The
More informationGCSE STATISTICS. 4) How to calculate the range: The difference between the biggest number and the smallest number.
GCSE STATISTICS You should kow: 1) How to draw a frequecy diagram: e.g. NUMBER TALLY FREQUENCY 1 3 5 ) How to draw a bar chart, a pictogram, ad a pie chart. 3) How to use averages: a) Mea  add up all
More information