62. Power series Definition 16. (Power series) Given a sequence {c n }, the series. c n x n = c 0 + c 1 x + c 2 x 2 + c 3 x 3 +

 Georgia Johns
 9 months ago
 Views:
Transcription
1 62. Power series Defiitio 16. (Power series) Give a sequece {c }, the series c x = c 0 + c 1 x + c 2 x 2 + c 3 x 3 + is called a power series i the variable x. The umbers c are called the coefficiets of the series I geeral, the series will coverge or diverge, depedig o the choice of x. The power series always coverges for x = 0 to the umber c 0. Example 98. For what values of x does the power series series x / coverge? Solutio: By the root test, x = x x as
2 62. POWER SERIES 125 So, the series coverges for all 1 < x < 1 ad diverges as x > 1 or x < 1. The root test is icoclusive for x = ±1. These values have to ivestigated by differet meas. For x = 1, the power series becomes the harmoic series 1/ which is diverget. For x = 1, the power series becomes the alteratig harmoic series ( 1) / which is coverget. Thus, the power series coverges if x [ 1, 1) ad diverges otherwise. Give a umber a, cosider a power series i the variable y = x a c y = c (x a) It is also called a power series cetered at a or a power series about a. Let S be the set of all values of x for which a power series i x coverges ad let S a be the set of all values of x for which the correspodig power series i (x a) coverges. What is the relatio betwee S ad S a? Sice the series are obtaied from oe aother by merely shiftig the value of the variable by a umber a, x x a, the set S a is therefore obtaied by addig the umber a to every elemet of S: (60) x S a x a S = S a = {x x a S} For example, the series (x 2) / coverges if x 2 [ 1, 1) or x [1, 3) ad diverges otherwise by Example 98. Thus, the problem of fidig the set S a is equivalet to the problem of fidig the set S Power series as a fuctio. Suppose that a power series i x coverges o a set S. The it defies a fuctio o S: f(x) = c x, x S The set S is called the domai of such a fuctio. Fuctios defied by power series are most commo i applicatios. May of them have special otatios (like elemetary fuctios si, cos, exp, etc). Their properties are well studied. I what follows it will be show that familiar elemetary fuctios si(x), cos(x), exp(x), etc ca also be represeted as power series. Example 99. Fid the domai of the Bessel fuctio of order 0 that is defied by the power series J 0 (x) = ( 1) 2 2 (!) 2 x2
3 SEQUENCES AND SERIES where, by commo covetio, 0! = 1. Solutio: Sice a = c x 2 cotais the factorial, the ratio test is more coveiet: a +1 a = x 2 c +1 c = x (!) 2 2 2(+1) (( + 1)!) 2 = x ( + 1) 2 0 as. So, the series coverges for all x. Values of a fuctio defied by a power series ca be estimated by partial sums which are polyomials i the variable x: f(x) f (x) = c k x k = c 0 + c 1 x + c 2 x c x k=0 Thus, partial sums defie a sequece of polyomials that coverges to the fuctio o S, f (x) f(x) for all x S. The accuracy of the approximatio is determied by the remaider R (x) = f(x) f (x). The accuracy assessmet is discussed i Sectio Sice the remider R (x) is a fuctio o S, the error of the approximatio is ot geerally uiform, i.e. it depeds o x Radius of covergece. The set S o which a power series is coverget is a importat characteristic ad its properties have to be studied. Lemma 2. (Properties of a power series) (i). If a power series c x coverges whe x = b 0, the it coverges wheever x b. (ii). If a power series c x diverges whe x = d 0, the it diverges wheever x d. Proof: If c b coverges, the by the ecessary coditio for covergece, c b 0 as. This meas, i particular, that for ε = 1 there exists a iteger N such that c b < ε = 1 for all > N. Thus, for > N, c x = c b x = c b x < x b b b which shows that the series c x coverges by compariso with the geometric series q where q = x/b ad x/b < 1 or x < b. Suppose that c d diverges. If x is ay umber such that x > d, the c x caot coverge because, by part (i) of the lemma, the covergece of c x implies the covergece of c d. Therefore c x diverges. This lemma allows us to establish the followig descriptio of the set S.
4 62. POWER SERIES 127 Theorem 45. (Covergece properties of a power series) For a power series c x there are oly three possibilities: 1. The series coverges oly whe x = 0 2. The series coverges for all x 3. There is a positive umber R such that the series coverges if x < R ad diverges if x > R. Proof: Suppose that either case 1 or case 2 is true. There there are umbers b 0 ad d 0 such that the power series coverges for x = b ad diverges for x = d. By Lemma 2 the set of covergece S lies i the iterval: x d for all x S. This shows that d is a upper boud for the set S. By the completeess axiom, S has a least upper boud R = sup S. If x > R, the x S ad c x diverges. If x < R, the x is ot a upper boud for S ad there exist a umber b S such that b > x. Sice b S, c x coverges by Lemma 2. Theorem 45 shows that a power series coverges i a sigle ope iterval ( R, R) ad diverges outside this iterval. The set S may or may ot iclude the poits x = ±R. This questio requires a special ivestigatio just like i Example 98. So, the umber R is characteristic for covergece properties of a power series. Defiitio 17. (Radius of covergece) The radius of covergece of a power series c x is a positive umber R > 0 such that the series coverges i the ope iterval ( R, R) ad diverges outside it. A power series is said to have a zero radius of covergece, R = 0, if it coverges oly whe x = 0. A power series is said to have a ifiite radius of covergece, R =, if it coverges for all values of x. The ratio or root test ca be used to determie the radius of covergece. Corollary 3. (Radius of covergece of a power series) Give a power series c x, if if c +1 lim c lim = α = R = 1 α c = α = R = 1 α where R = 0 if α = ad R = if α = 0 Proof: Put a = c x i the ratio test (Theorem 60.4). The a +1 / a = x c +1 / c x α. The series coverges if x α < 1 which shows that R = 1/α. Similarly, usig the root test (Theorem 40), a = x c x α < 1, which shows that R = 1/α.
5 SEQUENCES AND SERIES Remark. If the sequeces i Corollary 3 do ot coverge, the Theorem 41 should be used where a = c x. Oce the radius of covergece has bee foud ad 0 < R <, the cases x = ±R have to be ivestigated by some other meas (as the root or ratio test is icoclusive i this case) to determie the iterval of covergece S of a power series. Example 100. Fid the radius of covergece ad the iterval of covergece of the power series c x where c = ( q) / + 1 ad q > 0 Solutio: c +1 c = q q = q = q 1 + 1/ 1 + 2/ q = α Therefore R = 1/α = 1/q. If x = 1/q, the c x = ( 1) / + 1 = ( 1) b. The sequece b coverges mootoically to zero so that ( 1) b coverges by the alteratig series test. If x = 1/q, the c x = 1/ + 1 > 1/ 2, 1. The p series 1/ 1/2 diverges (p = 1/2 < 1) so that 1/ + 1 diverges by the compariso test. Thus, the iterval of covergece is S = [ 1/q, 1/q). Example 101. Fid the radius of covergece ad the iterval of covergece of the power series 2 (x + 1) /q where q > 0. Solutio: Put y = x +1. If S is the iterval of covergece of c y where c = 2 /q the the iterval of covergece i questio is obtaied by addig 1 to all umbers i S accordig to the rule (60). c = 1 q 2 = 1 q ( ) 2 1 q = α So R = 1/α = q. If y = q, c y = 2 ad the series 2 diverges (a = 2 does ot coverge to 0). If y = q, the c x = ( 1) 2 ad the series diverges because a = ( 1) 2 does ot coverge to 0. The series coverges oly if y = x + 1 < q ad, hece, the iterval of covergece is x ( q 1, q 1) (the iterval ( q, q) shifted by 1).
sin(n) + 2 cos(2n) n 3/2 3 sin(n) 2cos(2n) n 3/2 a n =
60. Ratio ad root tests 60.1. Absolutely coverget series. Defiitio 13. (Absolute covergece) A series a is called absolutely coverget if the series of absolute values a is coverget. The absolute covergece
More informationMAT1026 Calculus II Basic Convergence Tests for Series
MAT026 Calculus II Basic Covergece Tests for Series Egi MERMUT 202.03.08 Dokuz Eylül Uiversity Faculty of Sciece Departmet of Mathematics İzmir/TURKEY Cotets Mootoe Covergece Theorem 2 2 Series of Real
More informationSUMMARY OF SEQUENCES AND SERIES
SUMMARY OF SEQUENCES AND SERIES Importat Defiitios, Results ad Theorems for Sequeces ad Series Defiitio. A sequece {a } has a limit L ad we write lim a = L if for every ɛ > 0, there is a correspodig iteger
More informationINFINITE SEQUENCES AND SERIES
11 INFINITE SEQUENCES AND SERIES INFINITE SEQUENCES AND SERIES 11.4 The Compariso Tests I this sectio, we will lear: How to fid the value of a series by comparig it with a kow series. COMPARISON TESTS
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 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 informationInfinite 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 information( a) ( ) 1 ( ) 2 ( ) ( ) 3 3 ( ) =!
.8,.9: Taylor ad Maclauri Series.8. Although we were able to fid power series represetatios for a limited group of fuctios i the previous sectio, it is ot immediately obvious whether ay give fuctio has
More informationBC: Q401.CH9A Convergent and Divergent Series (LESSON 1)
BC: Q40.CH9A Coverget ad Diverget Series (LESSON ) INTRODUCTION Sequece Notatio: a, a 3, a,, a, Defiitio: A sequece is a fuctio f whose domai is the set of positive itegers. Defiitio: A ifiite series (or
More informationAre the following series absolutely convergent? n=1. n 3. n=1 n. ( 1) n. n=1 n=1
Absolute covergece Defiitio A series P a is called absolutely coverget if the series of absolute values P a is coverget. If the terms of the series a are positive, absolute covergece is the same as covergece.
More informationAP Calculus Chapter 9: Infinite Series
AP Calculus Chapter 9: Ifiite Series 9. Sequeces a, a 2, a 3, a 4, a 5,... Sequece: A fuctio whose domai is the set of positive itegers = 2 3 4 a = a a 2 a 3 a 4 terms of the sequece Begi with the patter
More information5.6 Absolute Convergence and The Ratio and Root Tests
5.6 Absolute Covergece ad The Ratio ad Root Tests Bria E. Veitch 5.6 Absolute Covergece ad The Ratio ad Root Tests Recall from our previous sectio that diverged but ( ) coverged. Both of these sequeces
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 informationAns: a n = 3 + ( 1) n Determine whether the sequence converges or diverges. If it converges, find the limit.
. Fid a formula for the term a of the give sequece: {, 3, 9, 7, 8 },... As: a = 3 b. { 4, 9, 36, 45 },... As: a = ( ) ( + ) c. {5,, 5,, 5,, 5,,... } As: a = 3 + ( ) +. Determie whether the sequece coverges
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 informationMa 530 Infinite Series I
Ma 50 Ifiite Series I Please ote that i additio to the material below this lecture icorporated material from the Visual Calculus web site. The material o sequeces is at Visual Sequeces. (To use this li
More informationThe Ratio Test. THEOREM 9.17 Ratio Test Let a n be a series with nonzero terms. 1. a. n converges absolutely if lim. n 1
460_0906.qxd //04 :8 PM Page 69 SECTION 9.6 The Ratio ad Root Tests 69 Sectio 9.6 EXPLORATION Writig a Series Oe of the followig coditios guaratees that a series will diverge, two coditios guaratee that
More informationLECTURE SERIES WITH NONNEGATIVE TERMS (II). SERIES WITH ARBITRARY TERMS
LECTURE 4 SERIES WITH NONNEGATIVE TERMS II). SERIES WITH ARBITRARY TERMS Series with oegative terms II) Theorem 4.1 Kummer s Test) Let x be a series with positive terms. 1 If c ) N i 0, + ), r > 0 ad 0
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 informationCarleton College, Winter 2017 Math 121, Practice Final Prof. Jones. Note: the exam will have a section of truefalse questions, like the one below.
Carleto College, Witer 207 Math 2, Practice Fial Prof. Joes Note: the exam will have a sectio of truefalse questios, like the oe below.. True or False. Briefly explai your aswer. A icorrectly justified
More informationNotice that this test does not say anything about divergence of an alternating series.
MATH 572H Sprig 20 Worksheet 7 Topics: absolute ad coditioal covergece; power series. Defiitio. A series b is called absolutely coverget if the series b is coverget. If the series b coverges, while b diverges,
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 informationINFINITE SERIES PROBLEMSSOLUTIONS. 3 n and 1. converges by the Comparison Test. and. ( 8 ) 2 n. 4 n + 2. n n = 4 lim 1
MAC 3 ) Note that 6 3 + INFINITE SERIES PROBLEMSSOLUTIONS 6 3 +
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 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 information(a) (b) All real numbers. (c) All real numbers. (d) None. to show the. (a) 3. (b) [ 7, 1) (c) ( 7, 1) (d) At x = 7. (a) (b)
Chapter 0 Review 597. E; a ( + )( + ) + + S S + S + + + + + + S lim + l. D; a diverges by the Itegral l k Test sice d lim [(l ) ], so k l ( ) does ot coverge absolutely. But it coverges by the Alteratig
More informationMath 210A Homework 1
Math 0A Homework Edward Burkard Exercise. a) State the defiitio of a aalytic fuctio. b) What are the relatioships betwee aalytic fuctios ad the CauchyRiema equatios? Solutio. a) A fuctio f : G C is called
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 informationMath 116 Practice for Exam 3
Math 6 Practice for Eam 3 Geerated April 4, 26 Name: SOLUTIONS Istructor: Sectio Number:. This eam has questios. Note that the problems are ot of equal difficulty, so you may wat to skip over ad retur
More informatione to approximate (using 4
Review: Taylor Polyomials ad Power Series Fid the iterval of covergece for the series Fid a series for f ( ) d ad fid its iterval of covergece Let f( ) Let f arcta a) Fid the rd degree Maclauri polyomial
More informationJANE PROFESSOR WW Prob Lib1 Summer 2000
JANE PROFESSOR WW Prob Lib Summer 000 Sample WeBWorK problems. WeBWorK assigmet Series6CompTests due /6/06 at :00 AM..( pt) Test each of the followig series for covergece by either the Compariso Test or
More informationSolutions to Tutorial 5 (Week 6)
The Uiversity of Sydey School of Mathematics ad Statistics Solutios to Tutorial 5 (Wee 6 MATH2962: Real ad Complex Aalysis (Advaced Semester, 207 Web Page: http://www.maths.usyd.edu.au/u/ug/im/math2962/
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 informationDefinition 4.2. (a) A sequence {x n } in a Banach space X is a basis for X if. unique scalars a n (x) such that x = n. a n (x) x n. (4.
4. BASES I BAACH SPACES 39 4. BASES I BAACH SPACES Sice a Baach space X is a vector space, it must possess a Hamel, or vector space, basis, i.e., a subset {x γ } γ Γ whose fiite liear spa is all of X ad
More informationSeries III. Chapter Alternating Series
Chapter 9 Series III With the exceptio of the Null Sequece Test, all the tests for series covergece ad divergece that we have cosidered so far have dealt oly with series of oegative terms. Series with
More informationMath 140A Elementary Analysis Homework Questions 31
Math 0A Elemetary Aalysis Homework Questios .9 Limits Theorems for Sequeces Suppose that lim x =, lim y = 7 ad that all y are ozero. Detarime the followig limits: (a) lim(x + y ) (b) lim y x y Let s
More informationREVIEW 1, MATH n=1 is convergent. (b) Determine whether a n is convergent.
REVIEW, MATH 00. Let a = +. a) Determie whether the sequece a ) is coverget. b) Determie whether a is coverget.. Determie whether the series is coverget or diverget. If it is coverget, fid its sum. a)
More informationIn this section, we show how to use the integral test to decide whether a series
Itegral Test Itegral Test Example Itegral Test Example pseries Compariso Test Example Example 2 Example 3 Example 4 Example 5 Exa Itegral Test I this sectio, we show how to use the itegral test to decide
More informationSequences, Series, and All That
Chapter Te Sequeces, Series, ad All That. Itroductio Suppose we wat to compute a approximatio of the umber e by usig the Taylor polyomial p for f ( x) = e x at a =. This polyomial is easily see to be 3
More informationa 3, a 4, ... are the terms of the sequence. The number a n is the nth term of the sequence, and the entire sequence is denoted by a n
60_090.qxd //0 : PM Page 59 59 CHAPTER 9 Ifiite Series Sectio 9. EXPLORATION Fidig Patters Describe a patter for each of the followig sequeces. The use your descriptio to write a formula for the th term
More information11.6 Absolute Convergence and the Ratio and Root Tests
.6 Absolute Covergece ad the Ratio ad Root Tests The most commo way to test for covergece is to igore ay positive or egative sigs i a series, ad simply test the correspodig series of positive terms. Does
More informationPractice Test Problems for Test IV, with Solutions
Practice Test Problems for Test IV, with Solutios Dr. Holmes May, 2008 The exam will cover sectios 8.2 (revisited) to 8.8. The Taylor remaider formula from 8.9 will ot be o this test. The fact that sums,
More informationMath 106 Fall 2014 Exam 3.2 December 10, 2014
Math 06 Fall 04 Exam 3 December 0, 04 Determie if the series is coverget or diverget by makig a compariso (DCT or LCT) with a suitable b Fill i the blaks with your aswer For Coverget or Diverget write
More informationf x x c x c x c... x c...
CALCULUS BC WORKSHEET ON POWER SERIES. Derive the Taylor series formula by fillig i the blaks below. 4 5 Let f a a c a c a c a4 c a5 c a c What happes to this series if we let = c? f c so a Now differetiate
More information2.4.2 A Theorem About Absolutely Convergent Series
0 Versio of August 27, 200 CHAPTER 2. INFINITE SERIES Add these two series: + 3 2 + 5 + 7 4 + 9 + 6 +... = 3 l 2. (2.20) 2 Sice the reciprocal of each iteger occurs exactly oce i the last series, we would
More information9.3 Power Series: Taylor & Maclaurin Series
9.3 Power Series: Taylor & Maclauri Series If is a variable, the a ifiite series of the form 0 is called a power series (cetered at 0 ). a a a a a 0 1 0 is a power series cetered at a c a a c a c a c 0
More informationMath 106 Fall 2014 Exam 3.1 December 10, 2014
Math 06 Fall 0 Exam 3 December 0, 0 Determie if the series is coverget or diverget by makig a compariso DCT or LCT) with a suitable b Fill i the blaks with your aswer For Coverget or Diverget write Coverget
More informationMTH 246 TEST 3 April 4, 2014
MTH 26 TEST April, 20 (PLEASE PRINT YOUR NAME!!) Name:. (6 poits each) Evaluate lim! a for the give sequece fa g. (a) a = 2 2 5 2 5 (b) a = 2 7 2. (6 poits) Fid the sum of the telescopig series p p 2.
More informationAnalytic Continuation
Aalytic Cotiuatio The stadard example of this is give by Example Let h (z) = 1 + z + z 2 + z 3 +... kow to coverge oly for z < 1. I fact h (z) = 1/ (1 z) for such z. Yet H (z) = 1/ (1 z) is defied for
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 TrueFalse problems o those pages. Please go over these problems
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 informationQuiz No. 1. ln n n. 1. Define: an infinite sequence A function whose domain is N 2. Define: a convergent sequence A sequence that has a limit
Quiz No.. Defie: a ifiite sequece A fuctio whose domai is N 2. Defie: a coverget sequece A sequece that has a limit 3. Is this sequece coverget? Why or why ot? l Yes, it is coverget sice L=0 by LHR. INFINITE
More informationTRUE/FALSE QUESTIONS FOR SEQUENCES
MAT1026 CALCULUS II 21.02.2012 Dokuz Eylül Üiversitesi Fe Fakültesi Matematik Bölümü Istructor: Egi Mermut web: http://kisi.deu.edu.tr/egi.mermut/ TRUE/FALSE QUESTIONS FOR SEQUENCES Write TRUE or FALSE
More informationComparison Study of Series Approximation. and Convergence between Chebyshev. and Legendre Series
Applied Mathematical Scieces, Vol. 7, 03, o. 6, 3337 HIKARI Ltd, www.mhikari.com http://d.doi.org/0.988/ams.03.3430 Compariso Study of Series Approimatio ad Covergece betwee Chebyshev ad Legedre Series
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 information2 n = n=1 a n is convergent and we let. i=1
Lecture 3 : Series So far our defiitio of a sum of umbers applies oly to addig a fiite set of umbers. We ca exted this to a defiitio of a sum of a ifiite set of umbers i much the same way as we exteded
More informationSection 11.6 Absolute and Conditional Convergence, Root and Ratio Tests
Sectio.6 Absolute ad Coditioal Covergece, Root ad Ratio Tests I this chapter we have see several examples of covergece tests that oly apply to series whose terms are oegative. I this sectio, we will lear
More information6. Uniform distribution mod 1
6. Uiform distributio mod 1 6.1 Uiform distributio ad Weyl s criterio Let x be a seuece of real umbers. We may decompose x as the sum of its iteger part [x ] = sup{m Z m x } (i.e. the largest iteger which
More informationSubject: Differential Equations & Mathematical ModelingIII
Power Series Solutios of Differetial Equatios about Sigular poits Subject: Differetial Equatios & Mathematical ModeligIII Lesso: Power series solutios of differetial equatios about Sigular poits Lesso
More informationSection A assesses the Units Numerical Analysis 1 and 2 Section B assesses the Unit Mathematics for Applied Mathematics
X0/70 NATIONAL QUALIFICATIONS 005 MONDAY, MAY.00 PM 4.00 PM APPLIED MATHEMATICS ADVANCED HIGHER Numerical Aalysis Read carefully. Calculators may be used i this paper.. Cadidates should aswer all questios.
More informationMATH 6101 Fall Problems. Problems 11/9/2008. Series and a Famous Unsolved Problem (21)(2 + 1) ( 4) 12Nov2008 MATH
/9/008 MATH 60 Fall 008 Series ad a Famous Usolved Problem = = + + + + (  )( + ) 3 3 5 5 7 7 9 Nov008 MATH 60 ( 4) = + 5 48 Nov008 MATH 60 3 /9/008 ( )! = + Nov008 MATH 60 4 3 4 5 + + + + + + +
More informationSolutions to quizzes Math Spring 2007
to quizzes Math 4 Sprig 7 Name: Sectio:. Quiz a) x + x dx b) l x dx a) x + dx x x / + x / dx (/3)x 3/ + x / + c. b) Set u l x, dv dx. The du /x ad v x. By Itegratio by Parts, x(/x)dx x l x x + c. l x
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 informationThe picture in figure 1.1 helps us to see that the area represents the distance traveled. Figure 1: Area represents distance travelled
1 Lecture : Area Area ad distace traveled Approximatig area by rectagles Summatio The area uder a parabola 1.1 Area ad distace Suppose we have the followig iformatio about the velocity of a particle, how
More informationLecture 3 The Lebesgue Integral
Lecture 3: The Lebesgue Itegral 1 of 14 Course: Theory of Probability I Term: Fall 2013 Istructor: Gorda Zitkovic Lecture 3 The Lebesgue Itegral The costructio of the itegral Uless expressly specified
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 informationLimit superior and limit inferior c Prof. Philip Pennance 1 Draft: April 17, 2017
Limit erior ad limit iferior c Prof. Philip Peace Draft: April 7, 207. Defiitio. The limit erior of a sequece a is the exteded real umber defied by lim a = lim a k k Similarly, the limit iferior of a
More informationTHE SOLUTION OF NONLINEAR EQUATIONS f( x ) = 0.
THE SOLUTION OF NONLINEAR EQUATIONS f( ) = 0. Noliear Equatio Solvers Bracketig. Graphical. Aalytical Ope Methods Bisectio False Positio (RegulaFalsi) Fied poit iteratio Newto Raphso Secat The root of
More informationIn number theory we will generally be working with integers, though occasionally fractions and irrationals will come into play.
Number Theory Math 5840 otes. Sectio 1: Axioms. I umber theory we will geerally be workig with itegers, though occasioally fractios ad irratioals will come ito play. Notatio: Z deotes the set of all itegers
More informationComplex Analysis Spring 2001 Homework I Solution
Complex Aalysis Sprig 2001 Homework I Solutio 1. Coway, Chapter 1, sectio 3, problem 3. Describe the set of poits satisfyig the equatio z a z + a = 2c, where c > 0 ad a R. To begi, we see from the triagle
More informationChapter 3. Strong convergence. 3.1 Definition of almost sure convergence
Chapter 3 Strog covergece As poited out i the Chapter 2, there are multiple ways to defie the otio of covergece of a sequece of radom variables. That chapter defied covergece i probability, covergece i
More information2.4  Sequences and Series
2.4  Sequeces ad Series Sequeces A sequece is a ordered list of elemets. Defiitio 1 A sequece is a fuctio from a subset of the set of itegers (usually either the set 80, 1, 2, 3,... < or the set 81, 2,
More informationInfinite Series Sequences: terms nth term Listing Terms of a Sequence 2 n recursively defined n+1 Pattern Recognition for Sequences Ex:
Ifiite Series Sequeces: A sequece i defied s fuctio whose domi is the set of positive itegers. Usully it s esier to deote sequece i subscript form rther th fuctio ottio.,, 3, re the terms of the sequece
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 informationCouncil for Innovative Research
ABSTRACT ON ABEL CONVERGENT SERIES OF FUNCTIONS ERDAL GÜL AND MEHMET ALBAYRAK Yildiz Techical Uiversity, Departmet of Mathematics, 34210 Eseler, Istabul egul34@gmail.com mehmetalbayrak12@gmail.com I this
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 informationIt is always the case that unions, intersections, complements, and set differences are preserved by the inverse image of a function.
MATH 532 Measurable Fuctios Dr. Neal, WKU Throughout, let ( X, F, µ) be a measure space ad let (!, F, P ) deote the special case of a probability space. We shall ow begi to study realvalued fuctios defied
More informationCalculus BC and BCD Drill on Sequences and Series!!! By Susan E. Cantey Walnut Hills H.S. 2006
Calculus BC ad BCD Drill o Sequeces ad Series!!! By Susa E. Catey Walut Hills H.S. 2006 Sequeces ad Series I m goig to ask you questios about sequeces ad series ad drill you o some thigs that eed to be
More informationMATH 2300 review problems for Exam 2
MATH 2300 review problems for Exam 2. A metal plate of costat desity ρ (i gm/cm 2 ) has a shape bouded by the curve y = x, the xaxis, ad the lie x =. (a) Fid the mass of the plate. Iclude uits. Mass =
More informations = and t = with C ij = A i B j F. (i) Note that cs = M and so ca i µ(a i ) I E (cs) = = c a i µ(a i ) = ci E (s). (ii) Note that s + t = M and so
3 From the otes we see that the parts of Theorem 4. that cocer us are: Let s ad t be two simple oegative Fmeasurable fuctios o X, F, µ ad E, F F. The i I E cs ci E s for all c R, ii I E s + t I E s +
More informationSolutions to Final Exam Review Problems
. Let f(x) 4+x. Solutios to Fial Exam Review Problems Math 5C, Witer 2007 (a) Fid the Maclauri series for f(x), ad compute its radius of covergece. Solutio. f(x) 4( ( x/4)) ( x/4) ( ) 4 4 + x. Sice the
More informationFUNDAMENTALS OF REAL ANALYSIS by
FUNDAMENTALS OF REAL ANALYSIS by Doğa Çömez Backgroud: All of Math 450/1 material. Namely: basic set theory, relatios ad PMI, structure of N, Z, Q ad R, basic properties of (cotiuous ad differetiable)
More informationExample 2. Find the upper bound for the remainder for the approximation from Example 1.
Lesso 8 Error Approimatios 0 Alteratig Series Remaider: For a coverget alteratig series whe approimatig the sum of a series by usig oly the first terms, the error will be less tha or equal to the absolute
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 informationAP Calculus. Notes and Homework for Chapter 9
AP Calculus Notes ad Homework for Chapter 9 (Mr. Surowski) I do t feel that the textbook does a particularly good job at itroducig the material o ifiite series, power series (Maclauri ad Taylor series),
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 informationProperties of Fuzzy Length on Fuzzy Set
Ope Access Library Joural 206, Volume 3, e3068 ISSN Olie: 2333972 ISSN Prit: 23339705 Properties of Fuzzy Legth o Fuzzy Set Jehad R Kider, Jaafar Imra Mousa Departmet of Mathematics ad Computer Applicatios,
More informationIntegrable Functions. { f n } is called a determining sequence for f. If f is integrable with respect to, then f d does exist as a finite real number
MATH 532 Itegrable Fuctios Dr. Neal, WKU We ow shall defie what it meas for a measurable fuctio to be itegrable, show that all itegral properties of simple fuctios still hold, ad the give some coditios
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 information6 Integers Modulo n. integer k can be written as k = qn + r, with q,r, 0 r b. So any integer.
6 Itegers Modulo I Example 2.3(e), we have defied the cogruece of two itegers a,b with respect to a modulus. Let us recall that a b (mod ) meas a b. We have proved that cogruece is a equivalece relatio
More informationTopic 9  Taylor and MacLaurin Series
Topic 9  Taylor ad MacLauri Series A. Taylors Theorem. The use o power series is very commo i uctioal aalysis i act may useul ad commoly used uctios ca be writte as a power series ad this remarkable result
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 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 informationOnce we have a sequence of numbers, the next thing to do is to sum them up. Given a sequence (a n ) n=1
. Ifiite Series Oce we have a sequece of umbers, the ext thig to do is to sum them up. Give a sequece a be a sequece: ca we give a sesible meaig to the followig expressio? a = a a a a While summig ifiitely
More informationDavenportSchinzel Sequences and their Geometric Applications
Advaced Computatioal Geometry Sprig 2004 DaveportSchizel Sequeces ad their Geometric Applicatios Prof. Joseph Mitchell Scribe: Mohit Gupta 1 Overview I this lecture, we itroduce the cocept of DaveportSchizel
More informationand the sum of its first n terms be denoted by. Convergence: An infinite series is said to be convergent if, a definite unique number., finite.
INFINITE SERIES Seqece: If a set of real mbers a seqece deoted by * + * Or * + * occr accordig to some defiite rle, the it is called + if is fiite + if is ifiite Series: is called a series ad is deoted
More informationMost text will write ordinary derivatives using either Leibniz notation 2 3. y + 5y= e and y y. xx tt t
Itroductio to Differetial Equatios Defiitios ad Termiolog Differetial Equatio: A equatio cotaiig the derivatives of oe or more depedet variables, with respect to oe or more idepedet variables, is said
More informationSingular Continuous Measures by Michael Pejic 5/14/10
Sigular Cotiuous Measures by Michael Peic 5/4/0 Prelimiaries Give a set X, a σalgebra o X is a collectio of subsets of X that cotais X ad ad is closed uder complemetatio ad coutable uios hece, coutable
More informationSome sufficient conditions of a given. series with rational terms converging to an irrational number or a transcdental number
Some sufficiet coditios of a give arxiv:0807.376v2 [math.nt] 8 Jul 2008 series with ratioal terms covergig to a irratioal umber or a trascdetal umber Yu Gao,Jiig Gao Shaghai Putuo college, Shaghai Jiaotog
More information