MATH 147 Homework 4. ( = lim. n n)( n + 1 n) n n n. 1 = lim
|
|
- Felix Fisher
- 5 years ago
- Views:
Transcription
1 MATH 147 Homework 4 1. Defie the sequece {a } by a =. a) Prove that a +1 a = 0. b) Prove that {a } is ot a Cauchy sequece. Solutio: a) We have: ad so we re doe. a +1 a = + 1 = ) + 1 ) = = 0 b) To prove that {a } is ot a Cauchy sequece, we eed to prove the followig statemet: There is some ɛ > 0 such that for all atural umbers N, there are two atural umbers ad m with > N ad m > N such that a a m ɛ. This is tricky. If you do t immediately see why the setece above is what we eed to prove, go back to the defiitio of Cauchy sequece ad work it out carefully. It s worth it. We eed to show that there is some ɛ with a certai property. So, let ɛ = 1 we ll show that this particular choice of ɛ will work. I fact, every value of ɛ > 0 is OK, but we ll choose this oe.) We do t get to pick N, though that s foisted upo us. But with our give N, we eed to fid ad m bigger tha N such that a a m ɛ = 1. That is to say, we eed to fid ad m bigger tha N such that m 1. Not too hard, really, but let s do it carefully ad rigorously. Let s pick = N + 1) 2 ad m = N + 2) 2. Sice N is a atural umber, we kow N > 0, so i particular = N + 1) 2 = N + 1 > N ad m = N + 2) 2 = N + 2 > N. Better yet, we have a a m 1 1, so we have succeeded! The sequece {a } is ot Cauchy.
2 2. a) For ay real umber a > 1, prove that the fuctio f) = a is a strictly icreasig fuctio of that is, if > y the a > a y. b) For ay real umbers a > 1 ad b > 0, show that there is a uique real umber such that a = b. Call this real umber. c) For ay real umber a > 1, show that the fuctio f) = log a is cotiuous o the iterval 0, ). Solutio: a) If ad y are itegers, the this is easy: if m >, the sice a > 1, we get a m > a. If we broade our horizos ad let = m/ ad y = t/u be ratioal umbers, the m/ > t/u meas mu > t, so a mu > a t. It s temptig to raise each side to the power 1/u, but we eed to make sure that that wo t chage the directio of the iequality. But this is easy: for ay positive iteger N ad real umbers z > 1 ad w > 1, if z N > w N, the z > w because z w implies z N w N. So we ve verified the statemet if ad y are ratioal. I geeral, for ay real umbers ad y with > y, choose ratioal umbers p ad q so that > p > q > y. The a a p by elemetary properties of its choose a sequece { } of ratioal umbers covergig to with > p for all, ad ote that a a p ). Similarly, a q a y. Sice we already kow that a p > a q, we re doe. b) Sice a is a strictly icreasig fuctio, there ca be at most oe real umber with a = b. Sice the sequece a is ubouded ad icreasig, ad the sequece a is decreasig ad coverges to 0, the fuctio a achieves values both larger ad smaller tha b. By the Itermediate Value Theorem, it therefore achieves the value b. c) First, ote that log a + log a y = log a y, because if = a X ad y = a Y, the y = a X a Y = a X+Y by elemetary properties of sequeces. Similarly, log a log a y = log a /y. Choose ay ɛ > 0, ad ay b 0, ), ad ay sequece {b } covergig to b. We wat to fid N such that if > N, the < ɛ. Choose N large eough so that b b < mi{ b a ɛ b, a ɛ b b } for all > N. The a bit of maipulatio shows that b /b is betwee a ɛ ad a ɛ, so: = /b) < ɛ ad we re doe.
3 3. Let a R be a real umber with a > 1. From the defiitios we used, prove the log base chage formula for ay base b > 1: log b = log a [Note: If you use the fact that a ) y = a y for irratioal or y, you must prove it.] Solutio: We will use the fact metioed i the Note, so we ll prove that first, to get it out of the way. Let ad y be real umbers, ad let a > 0 be a real umber. Choose ay sequeces of ratioal umbers { } ad {y } with = ad y = y. The we compute: ) y a ) y = a ) ym = m a ) = m a ) ym ) = m aym = m aym ) = a y as desired. Notice that m a ym ) = m a ym ) because the sequece { y m } for varyig is a sequece of ratioal umbers covergig to y m. Now we ca verify the log base chage rule. To check it, otice that log b is the uique real umber such that b log b =. So to prove the rule, we just eed to prove the followig: b log a = By defiitio of logarithms, we kow that b = a
4 Substitutig this ito our alleged log base chage formula yields: b log a = a ) log a = a log a = a log a = just like we wated. 4. Let a < b be real umbers, ad let f : [a, b] R be a cotiuous fuctio. Prove that the image f[a, b]) = {r R r = f)for some [a, b]} is also a closed iterval. Solutio: By the Etreme Value Theorem, there are elemets M ad m of [a, b] such that fm) = sup{f) [a, b]} ad fm) = if{f) [a, b]}. Therefore, we have f[a, b]) [fm), fm)]. Coversely, if C [fm), fm)] is ay elemet, the sice fm) < C < fm), ad because f is cotiuous, the Itermediate Value Theorem says that there is some c [a, b] such that fc) = [a, b]. Therefore f[a, b]) is the closed iterval [fm), fm)]. 5. Compute a +b for ay real umbers a ad b with a > 1. Solutio: The it is zero, regardless of what a ad b are. If you did t kow this already, you ca guess it by pluggig very big values of ito this formula.) To prove this, otice that the ratio of cosecutive terms is as follows: We compute: + 1 a +1 + b + 1 a + b = + 1 a + b a +1 + b = ) = 1
5 ad a + b a +1 + b = 1 a a +1 + ba a +1 + b = 1 a 1 + ba b a +1 + b ) = 1 a So both those its eist, ad we ca deduce: + 1 = + 1 = 1 a a + b a +1 + b a + b a +1 + b The relevace of this calculatio is that we ca ow compare our sequece to a geometric sequece, whose it we ca easily calculate. To wit: there is some atural umber N such that if > N, we have: + 1 a + b a +1 + b < α where 1 < α < 1. Use ɛ = α 1, ad remember that everythig is positive.) a a Therefore, sice the quatity o the left is the ratio betwee cosecutive terms of the sequece > N, we have: a +b, we deduce by a simple iductio that for all 0 < a + b < ) N α N a N + b Sice the sequece o the right coverges to zero, the Squeeze Theorem tells us that the sequece i the middle also teds to zero. But that s what we wated to prove. 6. Let f : R R be the fuctio f) = log a. Compute f) for ay a > 1. Solutio: The it is zero, regardless of what a is.
6 To prove this, choose ay ɛ > 0. We eed to fid a real umber N such that if > N, the log a < ɛ. The first, easiest thig to ote is that we do t eed the absolute value bars: for a > 1 ad > 1, both ad log a are positive. Net, ote that log a log a + 1) sice log a ad are both icreasig fuctios of. The otatio meas the largest iteger that is ot bigger tha. I other words, it s, rouded dow to the earest iteger.) If we ow set k = log a + 1), the we get: log a k a k 1 If we write l = k, the we further deduce: log a l + 1 a l 1 We d like to use the previous problem to compute the it of the iteger thig o the right side to be zero, ad the deduce that the left side must also go to zero. We compute: l l + 1 a l 1 = l l a l 1 + l 1 a l 1 = 0 But this is ot quite eough! Maybe l does t go to ifiity as does. Of course, this does ot happe: l = log a + 1), so if is ubouded above, the is also ubouded above, ad so l is ubouded above. We therefore kow that for all real umbers N, there is some B such that if > B, the l > N. If we choose N large eough so that l+1 < ɛ, we a l 1 therefore deduce that if > B, the log a < ɛ, as desired.
Infinite Sequences and Series
Chapter 6 Ifiite Sequeces ad Series 6.1 Ifiite Sequeces 6.1.1 Elemetary Cocepts Simply speakig, a sequece is a ordered list of umbers writte: {a 1, a 2, a 3,...a, a +1,...} where the elemets a i represet
More informationMath 61CM - Solutions to homework 3
Math 6CM - Solutios to homework 3 Cédric De Groote October 2 th, 208 Problem : Let F be a field, m 0 a fixed oegative iteger ad let V = {a 0 + a x + + a m x m a 0,, a m F} be the vector space cosistig
More informationTopics. Homework Problems. MATH 301 Introduction to Analysis Chapter Four Sequences. 1. Definition of convergence of sequences.
MATH 301 Itroductio to Aalysis Chapter Four Sequeces Topics 1. Defiitio of covergece of sequeces. 2. Fidig ad provig the limit of sequeces. 3. Bouded covergece theorem: Theorem 4.1.8. 4. Theorems 4.1.13
More informationAssignment 5: Solutions
McGill Uiversity Departmet of Mathematics ad Statistics MATH 54 Aalysis, Fall 05 Assigmet 5: Solutios. Let y be a ubouded sequece of positive umbers satisfyig y + > y for all N. Let x be aother sequece
More informationMA131 - Analysis 1. Workbook 3 Sequences II
MA3 - Aalysis Workbook 3 Sequeces II Autum 2004 Cotets 2.8 Coverget Sequeces........................ 2.9 Algebra of Limits......................... 2 2.0 Further Useful Results........................
More informationSeunghee Ye Ma 8: Week 5 Oct 28
Week 5 Summary I Sectio, we go over the Mea Value Theorem ad its applicatios. I Sectio 2, we will recap what we have covered so far this term. Topics Page Mea Value Theorem. Applicatios of the Mea Value
More informationSequence A sequence is a function whose domain of definition is the set of natural numbers.
Chapter Sequeces Course Title: Real Aalysis Course Code: MTH3 Course istructor: Dr Atiq ur Rehma Class: MSc-I Course URL: wwwmathcityorg/atiq/fa8-mth3 Sequeces form a importat compoet of Mathematical Aalysis
More informationExponential Functions and Taylor Series
MATH 4530: Aalysis Oe Expoetial Fuctios ad Taylor Series James K. Peterso Departmet of Biological Scieces ad Departmet of Mathematical Scieces Clemso Uiversity March 29, 2017 MATH 4530: Aalysis Oe Outlie
More informationRead carefully the instructions on the answer book and make sure that the particulars required are entered on each answer book.
THE UNIVERSITY OF WARWICK FIRST YEAR EXAMINATION: Jauary 2009 Aalysis I Time Allowed:.5 hours Read carefully the istructios o the aswer book ad make sure that the particulars required are etered o each
More informationHOMEWORK #10 SOLUTIONS
Math 33 - Aalysis I Sprig 29 HOMEWORK # SOLUTIONS () Prove that the fuctio f(x) = x 3 is (Riema) itegrable o [, ] ad show that x 3 dx = 4. (Without usig formulae for itegratio that you leart i previous
More information6.3 Testing Series With Positive Terms
6.3. TESTING SERIES WITH POSITIVE TERMS 307 6.3 Testig Series With Positive Terms 6.3. Review of what is kow up to ow I theory, testig a series a i for covergece amouts to fidig the i= sequece of partial
More information(A sequence also can be thought of as the list of function values attained for a function f :ℵ X, where f (n) = x n for n 1.) x 1 x N +k x N +4 x 3
MATH 337 Sequeces Dr. Neal, WKU Let X be a metric space with distace fuctio d. We shall defie the geeral cocept of sequece ad limit i a metric space, the apply the results i particular to some special
More informationMath 113 Exam 3 Practice
Math Exam Practice Exam will cover.-.9. This sheet has three sectios. The first sectio will remid you about techiques ad formulas that you should kow. The secod gives a umber of practice questios for you
More informationFind a formula for the exponential function whose graph is given , 1 2,16 1, 6
Math 4 Activity (Due by EOC Apr. ) Graph the followig epoetial fuctios by modifyig the graph of f. Fid the rage of each fuctio.. g. g. g 4. g. g 6. g Fid a formula for the epoetial fuctio whose graph is
More informationMA541 : Real Analysis. Tutorial and Practice Problems - 1 Hints and Solutions
MA54 : Real Aalysis Tutorial ad Practice Problems - Hits ad Solutios. Suppose that S is a oempty subset of real umbers that is bouded (i.e. bouded above as well as below). Prove that if S sup S. What ca
More informationRecitation 4: Lagrange Multipliers and Integration
Math 1c TA: Padraic Bartlett Recitatio 4: Lagrage Multipliers ad Itegratio Week 4 Caltech 211 1 Radom Questio Hey! So, this radom questio is pretty tightly tied to today s lecture ad the cocept of cotet
More information1+x 1 + α+x. x = 2(α x2 ) 1+x
Math 2030 Homework 6 Solutios # [Problem 5] For coveiece we let α lim sup a ad β lim sup b. Without loss of geerality let us assume that α β. If α the by assumptio β < so i this case α + β. By Theorem
More informationREAL ANALYSIS II: PROBLEM SET 1 - SOLUTIONS
REAL ANALYSIS II: PROBLEM SET 1 - SOLUTIONS 18th Feb, 016 Defiitio (Lipschitz fuctio). A fuctio f : R R is said to be Lipschitz if there exists a positive real umber c such that for ay x, y i the domai
More informationSolutions to Homework 7
Solutios to Homework 7 Due Wedesday, August 4, 004. Chapter 4.1) 3, 4, 9, 0, 7, 30. Chapter 4.) 4, 9, 10, 11, 1. Chapter 4.1. Solutio to problem 3. The sum has the form a 1 a + a 3 with a k = 1/k. Sice
More information10.1 Sequences. n term. We will deal a. a n or a n n. ( 1) n ( 1) n 1 2 ( 1) a =, 0 0,,,,, ln n. n an 2. n term.
0. Sequeces A sequece is a list of umbers writte i a defiite order: a, a,, a, a is called the first term, a is the secod term, ad i geeral eclusively with ifiite sequeces ad so each term Notatio: the sequece
More informationSolutions to Tutorial 3 (Week 4)
The Uiversity of Sydey School of Mathematics ad Statistics Solutios to Tutorial Week 4 MATH2962: Real ad Complex Aalysis Advaced Semester 1, 2017 Web Page: http://www.maths.usyd.edu.au/u/ug/im/math2962/
More informationMAS111 Convergence and Continuity
MAS Covergece ad Cotiuity Key Objectives At the ed of the course, studets should kow the followig topics ad be able to apply the basic priciples ad theorems therei to solvig various problems cocerig covergece
More informationMATH 112: HOMEWORK 6 SOLUTIONS. Problem 1: Rudin, Chapter 3, Problem s k < s k < 2 + s k+1
MATH 2: HOMEWORK 6 SOLUTIONS CA PRO JIRADILOK Problem. If s = 2, ad Problem : Rudi, Chapter 3, Problem 3. s + = 2 + s ( =, 2, 3,... ), prove that {s } coverges, ad that s < 2 for =, 2, 3,.... Proof. The
More informationn n 2 n n + 1 +
Istructor: Marius Ioescu 1. Let a =. (5pts) (a) Prove that for every ε > 0 there is N 1 such that a +1 a < ε if N. Solutio: Let ε > 0. The a +1 a < ε is equivalet with + 1 < ε. Simplifyig, this iequality
More informationSequences I. Chapter Introduction
Chapter 2 Sequeces I 2. Itroductio A sequece is a list of umbers i a defiite order so that we kow which umber is i the first place, which umber is i the secod place ad, for ay atural umber, we kow which
More informationExponential Functions and Taylor Series
Expoetial Fuctios ad Taylor Series James K. Peterso Departmet of Biological Scieces ad Departmet of Mathematical Scieces Clemso Uiversity March 29, 207 Outlie Revistig the Expoetial Fuctio Taylor Series
More informationIf a subset E of R contains no open interval, is it of zero measure? For instance, is the set of irrationals in [0, 1] is of measure zero?
2 Lebesgue Measure I Chapter 1 we defied the cocept of a set of measure zero, ad we have observed that every coutable set is of measure zero. Here are some atural questios: If a subset E of R cotais a
More informationSolutions to Math 347 Practice Problems for the final
Solutios to Math 347 Practice Problems for the fial 1) True or False: a) There exist itegers x,y such that 50x + 76y = 6. True: the gcd of 50 ad 76 is, ad 6 is a multiple of. b) The ifiimum of a set is
More informationUniversity of Manitoba, Mathletics 2009
Uiversity of Maitoba, Mathletics 009 Sessio 5: Iequalities Facts ad defiitios AM-GM iequality: For a, a,, a 0, a + a + + a (a a a ) /, with equality iff all a i s are equal Cauchy s iequality: For reals
More informationTopic 1 2: Sequences and Series. A sequence is an ordered list of numbers, e.g. 1, 2, 4, 8, 16, or
Topic : Sequeces ad Series A sequece is a ordered list of umbers, e.g.,,, 8, 6, or,,,.... A series is a sum of the terms of a sequece, e.g. + + + 8 + 6 + or... Sigma Notatio b The otatio f ( k) is shorthad
More informationMath 299 Supplement: Real Analysis Nov 2013
Math 299 Supplemet: Real Aalysis Nov 203 Algebra Axioms. I Real Aalysis, we work withi the axiomatic system of real umbers: the set R alog with the additio ad multiplicatio operatios +,, ad the iequality
More informationDefinition An infinite sequence of numbers is an ordered set of real numbers.
Ifiite sequeces (Sect. 0. Today s Lecture: Review: Ifiite sequeces. The Cotiuous Fuctio Theorem for sequeces. Usig L Hôpital s rule o sequeces. Table of useful its. Bouded ad mootoic sequeces. Previous
More information5 Sequences and Series
Bria E. Veitch 5 Sequeces ad Series 5. Sequeces A sequece is a list of umbers i a defiite order. a is the first term a 2 is the secod term a is the -th term The sequece {a, a 2, a 3,..., a,..., } is a
More informationMA131 - Analysis 1. Workbook 2 Sequences I
MA3 - Aalysis Workbook 2 Sequeces I Autum 203 Cotets 2 Sequeces I 2. Itroductio.............................. 2.2 Icreasig ad Decreasig Sequeces................ 2 2.3 Bouded Sequeces..........................
More information2.1. The Algebraic and Order Properties of R Definition. A binary operation on a set F is a function B : F F! F.
CHAPTER 2 The Real Numbers 2.. The Algebraic ad Order Properties of R Defiitio. A biary operatio o a set F is a fuctio B : F F! F. For the biary operatios of + ad, we replace B(a, b) by a + b ad a b, respectively.
More informationMath 104: Homework 2 solutions
Math 04: Homework solutios. A (0, ): Sice this is a ope iterval, the miimum is udefied, ad sice the set is ot bouded above, the maximum is also udefied. if A 0 ad sup A. B { m + : m, N}: This set does
More informationBertrand s Postulate
Bertrad s Postulate Lola Thompso Ross Program July 3, 2009 Lola Thompso (Ross Program Bertrad s Postulate July 3, 2009 1 / 33 Bertrad s Postulate I ve said it oce ad I ll say it agai: There s always a
More informationLecture Notes for Analysis Class
Lecture Notes for Aalysis Class Topological Spaces A topology for a set X is a collectio T of subsets of X such that: (a) X ad the empty set are i T (b) Uios of elemets of T are i T (c) Fiite itersectios
More informationThe Growth of Functions. Theoretical Supplement
The Growth of Fuctios Theoretical Supplemet The Triagle Iequality The triagle iequality is a algebraic tool that is ofte useful i maipulatig absolute values of fuctios. The triagle iequality says that
More informationA detailed proof of the irrationality of π
Matthew Straugh Math 4 Midterm A detailed proof of the irratioality of The proof is due to Iva Nive (1947) ad essetial to the proof are Lemmas ad 3 due to Charles Hermite (18 s) First let us itroduce some
More informationPart A, for both Section 200 and Section 501
Istructios Please write your solutios o your ow paper. These problems should be treated as essay questios. A problem that says give a example or determie requires a supportig explaatio. I all problems,
More informationDifferent kinds of Mathematical Induction
Differet ids of Mathematical Iductio () Mathematical Iductio Give A N, [ A (a A a A)] A N () (First) Priciple of Mathematical Iductio Let P() be a propositio (ope setece), if we put A { : N p() is true}
More informationINEQUALITIES BJORN POONEN
INEQUALITIES BJORN POONEN 1 The AM-GM iequality The most basic arithmetic mea-geometric mea (AM-GM) iequality states simply that if x ad y are oegative real umbers, the (x + y)/2 xy, with equality if ad
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, o-egative fuctio o the closed iterval [a, b] Fid
More informationAdvanced Analysis. Min Yan Department of Mathematics Hong Kong University of Science and Technology
Advaced Aalysis Mi Ya Departmet of Mathematics Hog Kog Uiversity of Sciece ad Techology September 3, 009 Cotets Limit ad Cotiuity 7 Limit of Sequece 8 Defiitio 8 Property 3 3 Ifiity ad Ifiitesimal 8 4
More information2 Banach spaces and Hilbert spaces
2 Baach spaces ad Hilbert spaces Tryig to do aalysis i the ratioal umbers is difficult for example cosider the set {x Q : x 2 2}. This set is o-empty ad bouded above but does ot have a least upper boud
More informationMA131 - Analysis 1. Workbook 9 Series III
MA3 - Aalysis Workbook 9 Series III Autum 004 Cotets 4.4 Series with Positive ad Negative Terms.............. 4.5 Alteratig Series.......................... 4.6 Geeral Series.............................
More information4x 2. (n+1) x 3 n+1. = lim. 4x 2 n+1 n3 n. n 4x 2 = lim = 3
Exam Problems (x. Give the series (, fid the values of x for which this power series coverges. Also =0 state clearly what the radius of covergece is. We start by settig up the Ratio Test: x ( x x ( x x
More informationFall 2013 MTH431/531 Real analysis Section Notes
Fall 013 MTH431/531 Real aalysis Sectio 8.1-8. Notes Yi Su 013.11.1 1. Defiitio of uiform covergece. We look at a sequece of fuctios f (x) ad study the coverget property. Notice we have two parameters
More informationZeros of Polynomials
Math 160 www.timetodare.com 4.5 4.6 Zeros of Polyomials I these sectios we will study polyomials algebraically. Most of our work will be cocered with fidig the solutios of polyomial equatios of ay degree
More informationf n (x) f m (x) < ɛ/3 for all x A. By continuity of f n and f m we can find δ > 0 such that d(x, x 0 ) < δ implies that
Lecture 15 We have see that a sequece of cotiuous fuctios which is uiformly coverget produces a limit fuctio which is also cotiuous. We shall stregthe this result ow. Theorem 1 Let f : X R or (C) be a
More informationCardinality Homework Solutions
Cardiality Homework Solutios April 16, 014 Problem 1. I the followig problems, fid a bijectio from A to B (you eed ot prove that the fuctio you list is a bijectio): (a) A = ( 3, 3), B = (7, 1). (b) A =
More informationLecture 6: Integration and the Mean Value Theorem. slope =
Math 8 Istructor: Padraic Bartlett Lecture 6: Itegratio ad the Mea Value Theorem Week 6 Caltech 202 The Mea Value Theorem The Mea Value Theorem abbreviated MVT is the followig result: Theorem. Suppose
More informationChapter 6 Infinite Series
Chapter 6 Ifiite Series I the previous chapter we cosidered itegrals which were improper i the sese that the iterval of itegratio was ubouded. I this chapter we are goig to discuss a topic which is somewhat
More informationBasic Sets. Functions. MTH299 - Examples. Example 1. Let S = {1, {2, 3}, 4}. Indicate whether each statement is true or false. (a) S = 4. (e) 2 S.
Basic Sets Example 1. Let S = {1, {2, 3}, 4}. Idicate whether each statemet is true or false. (a) S = 4 (b) {1} S (c) {2, 3} S (d) {1, 4} S (e) 2 S. (f) S = {1, 4, {2, 3}} (g) S Example 2. Compute the
More informationMath 2784 (or 2794W) University of Connecticut
ORDERS OF GROWTH PAT SMITH Math 2784 (or 2794W) Uiversity of Coecticut Date: Mar. 2, 22. ORDERS OF GROWTH. Itroductio Gaiig a ituitive feel for the relative growth of fuctios is importat if you really
More informationReal Analysis Fall 2004 Take Home Test 1 SOLUTIONS. < ε. Hence lim
Real Aalysis Fall 004 Take Home Test SOLUTIONS. Use the defiitio of a limit to show that (a) lim si = 0 (b) Proof. Let ε > 0 be give. Defie N >, where N is a positive iteger. The for ε > N, si 0 < si
More informationMath 140A Elementary Analysis Homework Questions 3-1
Math 0A Elemetary Aalysis Homework Questios -.9 Limits Theorems for Sequeces Suppose that lim x =, lim y = 7 ad that all y are o-zero. Detarime the followig limits: (a) lim(x + y ) (b) lim y x y Let s
More informationSequences and Series of Functions
Chapter 6 Sequeces ad Series of Fuctios 6.1. Covergece of a Sequece of Fuctios Poitwise Covergece. Defiitio 6.1. Let, for each N, fuctio f : A R be defied. If, for each x A, the sequece (f (x)) coverges
More informationCarleton College, Winter 2017 Math 121, Practice Final Prof. Jones. Note: the exam will have a section of true-false questions, like the one below.
Carleto College, Witer 207 Math 2, Practice Fial Prof. Joes Note: the exam will have a sectio of true-false questios, like the oe below.. True or False. Briefly explai your aswer. A icorrectly justified
More information} is said to be a Cauchy sequence provided the following condition is true.
Math 4200, Fial Exam Review I. Itroductio to Proofs 1. Prove the Pythagorea theorem. 2. Show that 43 is a irratioal umber. II. Itroductio to Logic 1. Costruct a truth table for the statemet ( p ad ~ r
More information10.6 ALTERNATING SERIES
0.6 Alteratig Series Cotemporary Calculus 0.6 ALTERNATING SERIES I the last two sectios we cosidered tests for the covergece of series whose terms were all positive. I this sectio we examie series whose
More informationChapter 6 Overview: Sequences and Numerical Series. For the purposes of AP, this topic is broken into four basic subtopics:
Chapter 6 Overview: Sequeces ad Numerical Series I most texts, the topic of sequeces ad series appears, at first, to be a side topic. There are almost o derivatives or itegrals (which is what most studets
More informationIt is often useful to approximate complicated functions using simpler ones. We consider the task of approximating a function by a polynomial.
Taylor Polyomials ad Taylor Series It is ofte useful to approximate complicated fuctios usig simpler oes We cosider the task of approximatig a fuctio by a polyomial If f is at least -times differetiable
More informationNATIONAL UNIVERSITY OF SINGAPORE FACULTY OF SCIENCE SEMESTER 1 EXAMINATION ADVANCED CALCULUS II. November 2003 Time allowed :
NATIONAL UNIVERSITY OF SINGAPORE FACULTY OF SCIENCE SEMESTER EXAMINATION 003-004 MA08 ADVANCED CALCULUS II November 003 Time allowed : hours INSTRUCTIONS TO CANDIDATES This examiatio paper cosists of TWO
More informationPart I: Covers Sequence through Series Comparison Tests
Part I: Covers Sequece through Series Compariso Tests. Give a example of each of the followig: (a) A geometric sequece: (b) A alteratig sequece: (c) A sequece that is bouded, but ot coverget: (d) A sequece
More informationMath 132, Fall 2009 Exam 2: Solutions
Math 3, Fall 009 Exam : Solutios () a) ( poits) Determie for which positive real umbers p, is the followig improper itegral coverget, ad for which it is diverget. Evaluate the itegral for each value of
More informationChapter 7: Numerical Series
Chapter 7: Numerical Series Chapter 7 Overview: Sequeces ad Numerical Series I most texts, the topic of sequeces ad series appears, at first, to be a side topic. There are almost o derivatives or itegrals
More informationMath 163 REVIEW EXAM 3: SOLUTIONS
Math 63 REVIEW EXAM 3: SOLUTIONS These otes do ot iclude solutios to the Cocept Check o p8. They also do t cotai complete solutios to the True-False problems o those pages. Please go over these problems
More informationf(x) dx as we do. 2x dx x also diverges. Solution: We compute 2x dx lim
Math 3, Sectio 2. (25 poits) Why we defie f(x) dx as we do. (a) Show that the improper itegral diverges. Hece the improper itegral x 2 + x 2 + b also diverges. Solutio: We compute x 2 + = lim b x 2 + =
More informationNotes #3 Sequences Limit Theorems Monotone and Subsequences Bolzano-WeierstraßTheorem Limsup & Liminf of Sequences Cauchy Sequences and Completeness
Notes #3 Sequeces Limit Theorems Mootoe ad Subsequeces Bolzao-WeierstraßTheorem Limsup & Limif of Sequeces Cauchy Sequeces ad Completeess This sectio of otes focuses o some of the basics of sequeces of
More informationA sequence of numbers is a function whose domain is the positive integers. We can see that the sequence
Sequeces A sequece of umbers is a fuctio whose domai is the positive itegers. We ca see that the sequece,, 2, 2, 3, 3,... is a fuctio from the positive itegers whe we write the first sequece elemet as
More informationMATH 413 FINAL EXAM. f(x) f(y) M x y. x + 1 n
MATH 43 FINAL EXAM Math 43 fial exam, 3 May 28. The exam starts at 9: am ad you have 5 miutes. No textbooks or calculators may be used durig the exam. This exam is prited o both sides of the paper. Good
More informationSection 11.8: Power Series
Sectio 11.8: Power Series 1. Power Series I this sectio, we cosider geeralizig the cocept of a series. Recall that a series is a ifiite sum of umbers a. We ca talk about whether or ot it coverges ad i
More informationMath 10A final exam, December 16, 2016
Please put away all books, calculators, cell phoes ad other devices. You may cosult a sigle two-sided sheet of otes. Please write carefully ad clearly, USING WORDS (ot just symbols). Remember that the
More informationNEW FAST CONVERGENT SEQUENCES OF EULER-MASCHERONI TYPE
UPB Sci Bull, Series A, Vol 79, Iss, 207 ISSN 22-7027 NEW FAST CONVERGENT SEQUENCES OF EULER-MASCHERONI TYPE Gabriel Bercu We itroduce two ew sequeces of Euler-Mascheroi type which have fast covergece
More informationPolynomial Functions and Their Graphs
Polyomial Fuctios ad Their Graphs I this sectio we begi the study of fuctios defied by polyomial expressios. Polyomial ad ratioal fuctios are the most commo fuctios used to model data, ad are used extesively
More information1 Generating functions for balls in boxes
Math 566 Fall 05 Some otes o geeratig fuctios Give a sequece a 0, a, a,..., a,..., a geeratig fuctio some way of represetig the sequece as a fuctio. There are may ways to do this, with the most commo ways
More informationMIDTERM 3 CALCULUS 2. Monday, December 3, :15 PM to 6:45 PM. Name PRACTICE EXAM SOLUTIONS
MIDTERM 3 CALCULUS MATH 300 FALL 08 Moday, December 3, 08 5:5 PM to 6:45 PM Name PRACTICE EXAM S Please aswer all of the questios, ad show your work. You must explai your aswers to get credit. You will
More informationLesson 10: Limits and Continuity
www.scimsacademy.com Lesso 10: Limits ad Cotiuity SCIMS Academy 1 Limit of a fuctio The cocept of limit of a fuctio is cetral to all other cocepts i calculus (like cotiuity, derivative, defiite itegrals
More informationContinuous Functions
Cotiuous Fuctios Q What does it mea for a fuctio to be cotiuous at a poit? Aswer- I mathematics, we have a defiitio that cosists of three cocepts that are liked i a special way Cosider the followig defiitio
More informationInduction: Solutions
Writig Proofs Misha Lavrov Iductio: Solutios Wester PA ARML Practice March 6, 206. Prove that a 2 2 chessboard with ay oe square removed ca always be covered by shaped tiles. Solutio : We iduct o. For
More informationCHAPTER I: Vector Spaces
CHAPTER I: Vector Spaces Sectio 1: Itroductio ad Examples This first chapter is largely a review of topics you probably saw i your liear algebra course. So why cover it? (1) Not everyoe remembers everythig
More informationPROBLEM SET 5 SOLUTIONS 126 = , 37 = , 15 = , 7 = 7 1.
Math 7 Sprig 06 PROBLEM SET 5 SOLUTIONS Notatios. Give a real umber x, we will defie sequeces (a k ), (x k ), (p k ), (q k ) as i lecture.. (a) (5 pts) Fid the simple cotiued fractio represetatios of 6
More information1 Approximating Integrals using Taylor Polynomials
Seughee Ye Ma 8: Week 7 Nov Week 7 Summary This week, we will lear how we ca approximate itegrals usig Taylor series ad umerical methods. Topics Page Approximatig Itegrals usig Taylor Polyomials. Defiitios................................................
More informationSOLUTIONS TO EXAM 3. Solution: Note that this defines two convergent geometric series with respective radii r 1 = 2/5 < 1 and r 2 = 1/5 < 1.
SOLUTIONS TO EXAM 3 Problem Fid the sum of the followig series 2 + ( ) 5 5 2 5 3 25 2 2 This series diverges Solutio: Note that this defies two coverget geometric series with respective radii r 2/5 < ad
More informationMetric Space Properties
Metric Space Properties Math 40 Fial Project Preseted by: Michael Brow, Alex Cordova, ad Alyssa Sachez We have already poited out ad will recogize throughout this book the importace of compact sets. All
More informationM17 MAT25-21 HOMEWORK 5 SOLUTIONS
M17 MAT5-1 HOMEWORK 5 SOLUTIONS 1. To Had I Cauchy Codesatio Test. Exercise 1: Applicatio of the Cauchy Codesatio Test Use the Cauchy Codesatio Test to prove that 1 diverges. Solutio 1. Give the series
More informationFinal Solutions. 1. (25pts) Define the following terms. Be as precise as you can.
Mathematics H104 A. Ogus Fall, 004 Fial Solutios 1. (5ts) Defie the followig terms. Be as recise as you ca. (a) (3ts) A ucoutable set. A ucoutable set is a set which ca ot be ut ito bijectio with a fiite
More informationLecture 17Section 10.1 Least Upper Bound Axiom
Lecture 7Sectio 0. Least Upper Boud Axiom Sectio 0.2 Sequeces of Real Numbers Jiwe He Real Numbers. Review Basic Properties of R: R beig Ordered Classificatio N = {0,, 2,...} = {atural umbers} Z = {...,
More information1 Introduction. 1.1 Notation and Terminology
1 Itroductio You have already leared some cocepts of calculus such as limit of a sequece, limit, cotiuity, derivative, ad itegral of a fuctio etc. Real Aalysis studies them more rigorously usig a laguage
More informationPLEASE MARK YOUR ANSWERS WITH AN X, not a circle! 1. (a) (b) (c) (d) (e) 3. (a) (b) (c) (d) (e) 5. (a) (b) (c) (d) (e) 7. (a) (b) (c) (d) (e)
Math 0560, Exam 3 November 6, 07 The Hoor Code is i effect for this examiatio. All work is to be your ow. No calculators. The exam lasts for hour ad 5 mi. Be sure that your ame is o every page i case pages
More informationChapter 10: Power Series
Chapter : Power Series 57 Chapter Overview: Power Series The reaso series are part of a Calculus course is that there are fuctios which caot be itegrated. All power series, though, ca be itegrated because
More informationPRACTICE FINAL/STUDY GUIDE SOLUTIONS
Last edited December 9, 03 at 4:33pm) Feel free to sed me ay feedback, icludig commets, typos, ad mathematical errors Problem Give the precise meaig of the followig statemets i) a f) L ii) a + f) L iii)
More informationMath 128A: Homework 1 Solutions
Math 8A: Homework Solutios Due: Jue. Determie the limits of the followig sequeces as. a) a = +. lim a + = lim =. b) a = + ). c) a = si4 +6) +. lim a = lim = lim + ) [ + ) ] = [ e ] = e 6. Observe that
More informationn=1 a n is the sequence (s n ) n 1 n=1 a n converges to s. We write a n = s, n=1 n=1 a n
Series. Defiitios ad first properties A series is a ifiite sum a + a + a +..., deoted i short by a. The sequece of partial sums of the series a is the sequece s ) defied by s = a k = a +... + a,. k= Defiitio
More informationPUTNAM TRAINING INEQUALITIES
PUTNAM TRAINING INEQUALITIES (Last updated: December, 207) Remark This is a list of exercises o iequalities Miguel A Lerma Exercises If a, b, c > 0, prove that (a 2 b + b 2 c + c 2 a)(ab 2 + bc 2 + ca
More informationSection 7 Fundamentals of Sequences and Series
ectio Fudametals of equeces ad eries. Defiitio ad examples of sequeces A sequece ca be thought of as a ifiite list of umbers. 0, -, -0, -, -0...,,,,,,. (iii),,,,... Defiitio: A sequece is a fuctio which
More informationPower Series: A power series about the center, x = 0, is a function of x of the form
You are familiar with polyomial fuctios, polyomial that has ifiitely may terms. 2 p ( ) a0 a a 2 a. A power series is just a Power Series: A power series about the ceter, = 0, is a fuctio of of the form
More informationSequences. Notation. Convergence of a Sequence
Sequeces A sequece is essetially just a list. Defiitio (Sequece of Real Numbers). A sequece of real umbers is a fuctio Z (, ) R for some real umber. Do t let the descriptio of the domai cofuse you; it
More information