Section 11.8: Power Series
|
|
- August Johnston
- 6 years ago
- Views:
Transcription
1 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 some cases, what it coverges to. More geerally however, istead of cosidered a series as a ifiite sum of umbers, we could cosider a ifiite sum of a expressio i a variable. If we could determie covergece properties of such series, this would allow us to make coclusios about ifiitely may series with just umbers simultaeously (by pluggig umbers i for the variable). We start with a defiitio. Defiitio 1.1. A power series is a series of the form c x =0 where x is a variable ad the c s are called the coefficiets of the series (ote that this series starts from 0, so there is a costat term to this polyomial). More geerally, a series of the form c (x a) =0 is called a power series cetered at a. If we are give a power series, it does ot immediately make sese to talk about covergece because x is a variable. However, if we choose a value of x, say a, the the power series becomes a regular series c a ad we ca asks questios about its covergece etc. A power series may coverge for some values of x ad diverge for others, so it ca be viewed as a fuctio whose domai is the set of all umbers for which it coverges. This leads us to the followig two questios: Questio 1.2. For what values does a power series coverge? Questio 1.3. How do we fid what values a power series coverges for? Sice a power series is essetially a series, we ca use most of the same tests we developed for regular series - the oly differece is that we must treat x like a variable ad determie the values it coverges for. We cosider a couple of easy examples to illustrate. 1
2 2 Example 1.4. (i) For what values of x does the series x coverge? For ay fixed value of x, this sum is just a geometric series, ad we kow all the values for which this coverges. Specifically, x coverges if ad oly if x < 1. (ii) For what values of x does the series x! coverge? Applyig the ratio test, we have x +1 (+1)! x! = x +1 (!) x ( + 1)! = x = 0 idepedet of the choice of x. Therefore, this series coverges for all values of x. (iii) For what values of x does the series!x coverge? Applyig the ratio test, we have ( + 1)!(x 2) (+1)!(x 2) = ( + 1)(x 2) which diverges if x 2. Thus this series coverges oly at x = 2. The last three examples suggest that there are three possibilities for the covergece of a power series - either all choices of x coverge, oly those o a certai iterval coverge or oly oe does. This is true i geeral as the followig result tells us. Result 1.5. For a give power series c (x a) there are oly three possibilities: (i) The series coverges oly whe x = a. (ii) The series coverges for all x. (iii) There is a positive umber R called the radius of covergece such that the series coverges if x a < R ad diverges if x a > R. If x a = R (so x = R a or x = a R), the the series may coverge or diverge.
3 If the radius of covergece is R ad the series is cetered aroud a, we say that the iterval (a R, a + R) is the iterval of covergece (where we iclude the edpoits if the series coverges at them). This leads us to the followig questio: Questio 1.6. How do we fid the iterval of covergece? I order to fid the iterval of covergece of a power series c (x a) we proceed as follows: (i) Observe that if we apply the ratio test, we are evaluatig the it c +1 (x a) +1 c x = (x a) c +1 c. Let L be this it. I order to coverge, the ratio test has to be less tha 1, so we eed to solve (x a) L < 1. Therefore, i order to coverge we must have (x a) 1. It follows that L the radius of covergece will be the reciprocal of the absolute values of the it (ii) This tells us that if the c +1. c L = c +1 c c (x a) coverges o the iterval (a 1/L, a+1/l). To fiish the problem off, we determie whether the series coverges at either of the edpoits, a 1/L ad a + 1/L. We illustrate with some examples: Example 1.7. Fid the radius ad iterval of covergece of the followig power series. (i) x 3 Applyig the ratio test we get x +1 (+1)3 +1 x 3 x = 3( + 1) = x 3 < 1, so x < 3. So the radius of covergece is 3. This meas that x 3 3
4 4 (ii) coverges i the iterval ( 3, 3). We ow eed to check the edpoits: x = 3 gives 3 3 = 1 which is the harmoic series, so diverges. x = 3 gives ( 3) = ( 1) 3 which is the alteratig harmoic series, so coverges. Therefore, the iterval of covergece is [ 3, 3). ( 2) (x + 3) Applyig the ratio test we get 2 +1 (x+3) (x+3) = 2 x + 3 = 2 x + 3 < 1, + 1 (iii) so x + 3 < 1/2. So the radius of covergece is 1/2. This meas that x 3 coverges i the iterval ( 3.5, 2.5). We ow eed to check the edpoits: x = 2.5 gives ( 2) ( 1 2 ) = ( 1) which coverges by the alteratig series test. x = 3.5 gives ( 2) ( 1 2 ) = 1 which diverges by the p series test. Thus the iterval of covergece is [ 2.5, 3.5). (x 4) Applyig the ratio test we get (+1) x 4 +1 (+1) 3 +1 x = ( + 1)( 3 + 1) x + 4 (( + 1) 3 + 1) = x + 4 < 1.
5 So the radius of covergece is 1. This meas that (x 4) coverges i the iterval (3, 5). We ow eed to check the edpoits: x = 3 gives ( 1) which coverges by the alteratig series test. x = 5 gives so coverges usig the p-series test ad the compariso test. Thus the iterval of covergece is [3, 5]. 5
Convergence: nth-term Test, Comparing Non-negative Series, Ratio Test
Covergece: th-term Test, Comparig No-egative Series, Ratio Test Power Series ad Covergece We have writte statemets like: l + x = x x2 + x3 2 3 + x + But we have ot talked i depth about what values of x
More information62. 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 +
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
More informationMa 530 Introduction to Power Series
Ma 530 Itroductio to Power Series Please ote that there is material o power series at Visual Calculus. Some of this material was used as part of the presetatio of the topics that follow. What is a Power
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 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 informationMTH 122 Calculus II Essex County College Division of Mathematics and Physics 1 Lecture Notes #20 Sakai Web Project Material
MTH 1 Calculus II Essex Couty College Divisio of Mathematics ad Physics 1 Lecture Notes #0 Sakai Web Project Material 1 Power Series 1 A power series is a series of the form a x = a 0 + a 1 x + a x + a
More informationTaylor Series (BC Only)
Studet Study Sessio Taylor Series (BC Oly) Taylor series provide a way to fid a polyomial look-alike to a o-polyomial fuctio. This is doe by a specific formula show below (which should be memorized): Taylor
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 informationQuiz. Use either the RATIO or ROOT TEST to determine whether the series is convergent or not.
Quiz. Use either the RATIO or ROOT TEST to determie whether the series is coverget or ot. e .6 POWER SERIES Defiitio. A power series i about is a series of the form c 0 c a c a... c a... a 0 c a where
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 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 informationAdditional Notes on Power Series
Additioal Notes o Power Series Mauela Girotti MATH 37-0 Advaced Calculus of oe variable Cotets Quick recall 2 Abel s Theorem 2 3 Differetiatio ad Itegratio of Power series 4 Quick recall We recall here
More informationSection 1.4. Power Series
Sectio.4. Power Series De itio. The fuctio de ed by f (x) (x a) () c 0 + c (x a) + c 2 (x a) 2 + c (x a) + ::: is called a power series cetered at x a with coe ciet sequece f g :The domai of this fuctio
More informationQuiz 5 Answers MATH 141
Quiz 5 Aswers MATH 4 8 AM Questio so the series coverges. + ( + )! ( + )! ( + )! = ( + )( + ) = 0 < We ca rewrite this series as the geometric series (x) which coverges whe x < or whe x
More informationMath 113 (Calculus 2) Section 12 Exam 4
Name: Row: Math Calculus ) Sectio Exam 4 8 0 November 00 Istructios:. Work o scratch paper will ot be graded.. For questio ad questios 0 through 5, show all your work i the space provided. Full credit
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 informationAn alternating series is a series where the signs alternate. Generally (but not always) there is a factor of the form ( 1) n + 1
Calculus II - Problem Solvig Drill 20: Alteratig Series, Ratio ad Root Tests Questio No. of 0 Istructios: () Read the problem ad aswer choices carefully (2) Work the problems o paper as eeded (3) Pick
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 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 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 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 informationSection 5.5. Infinite Series: The Ratio Test
Differece Equatios to Differetial Equatios Sectio 5.5 Ifiite Series: The Ratio Test I the last sectio we saw that we could demostrate the covergece of a series a, where a 0 for all, by showig that a approaches
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 informationSCORE. Exam 2. MA 114 Exam 2 Fall 2017
Exam Name: Sectio ad/or TA: Do ot remove this aswer page you will retur the whole exam. You will be allowed two hours to complete this test. No books or otes may be used. You may use a graphig calculator
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 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 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 informationCHAPTER 10 INFINITE SEQUENCES AND SERIES
CHAPTER 10 INFINITE SEQUENCES AND SERIES 10.1 Sequeces 10.2 Ifiite Series 10.3 The Itegral Tests 10.4 Compariso Tests 10.5 The Ratio ad Root Tests 10.6 Alteratig Series: Absolute ad Coditioal Covergece
More informationMath 113, Calculus II Winter 2007 Final Exam Solutions
Math, Calculus II Witer 7 Fial Exam Solutios (5 poits) Use the limit defiitio of the defiite itegral ad the sum formulas to compute x x + dx The check your aswer usig the Evaluatio Theorem Solutio: I this
More informationSeries Solutions (BC only)
Studet Study Sessio Solutios (BC oly) We have itetioally icluded more material tha ca be covered i most Studet Study Sessios to accout for groups that are able to aswer the questios at a faster rate. Use
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 informationf t dt. Write the third-degree Taylor polynomial for G
AP Calculus BC Homework - Chapter 8B Taylor, Maclauri, ad Power Series # Taylor & Maclauri Polyomials Critical Thikig Joural: (CTJ: 5 pts.) Discuss the followig questios i a paragraph: What does it mea
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 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 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 informationArkansas Tech University MATH 2924: Calculus II Dr. Marcel B. Finan
Arkasas Tech Uiversity MATH 94: Calculus II Dr Marcel B Fia 85 Power Series Let {a } =0 be a sequece of umbers The a power series about x = a is a series of the form a (x a) = a 0 + a (x a) + a (x a) +
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 informationThe Interval of Convergence for a Power Series Examples
The Iterval of Covergece for a Power Series Examples To review the process: How to Test a Power Series for Covergece. Fid the iterval where the series coverges absolutely. We have to use the Ratio or Root
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 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 informationSCORE. Exam 2. MA 114 Exam 2 Fall 2016
Exam 2 Name: Sectio ad/or TA: Do ot remove this aswer page you will retur the whole exam. You will be allowed two hours to complete this test. No books or otes may be used. You may use a graphig calculator
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 informationMath 341 Lecture #31 6.5: Power Series
Math 341 Lecture #31 6.5: Power Series We ow tur our attetio to a particular kid of series of fuctios, amely, power series, f(x = a x = a 0 + a 1 x + a 2 x 2 + where a R for all N. I terms of a series
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 informationMATH 6101 Fall Problems. Problems 11/9/2008. Series and a Famous Unsolved Problem (2-1)(2 + 1) ( 4) 12-Nov-2008 MATH
/9/008 MATH 60 Fall 008 Series ad a Famous Usolved Problem = = + + + + ( - )( + ) 3 3 5 5 7 7 9 -Nov-008 MATH 60 ( 4) = + 5 48 -Nov-008 MATH 60 3 /9/008 ( )! = + -Nov-008 MATH 60 4 3 4 5 + + + + + + +
More informationMath 113 Exam 4 Practice
Math Exam 4 Practice Exam 4 will cover.-.. 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
More informationMath 120 Answers for Homework 23
Math 0 Aswers for Homewor. (a) The Taylor series for cos(x) aroud a 0 is cos(x) x! + x4 4! x6 6! + x8 8! x0 0! + ( ) ()! x ( ) π ( ) ad so the series ()! ()! (π) is just the series for cos(x) evaluated
More informationMath 113 Exam 3 Practice
Math Exam Practice Exam 4 will cover.-., 0. ad 0.. Note that eve though. was tested i exam, questios from that sectios may also be o this exam. For practice problems o., refer to the last review. This
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 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 information( 1) n (4x + 1) n. n=0
Problem 1 (10.6, #). Fid the radius of covergece for the series: ( 1) (4x + 1). For what values of x does the series coverge absolutely, ad for what values of x does the series coverge coditioally? Solutio.
More informationPhysics 116A Solutions to Homework Set #1 Winter Boas, problem Use equation 1.8 to find a fraction describing
Physics 6A Solutios to Homework Set # Witer 0. Boas, problem. 8 Use equatio.8 to fid a fractio describig 0.694444444... Start with the formula S = a, ad otice that we ca remove ay umber of r fiite decimals
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 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 informationSCORE. Exam 2. MA 114 Exam 2 Fall 2016
MA 4 Exam Fall 06 Exam Name: Sectio ad/or TA: Do ot remove this aswer page you will retur the whole exam. You will be allowed two hours to complete this test. No books or otes may be used. You may use
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 information= lim. = lim. 3 dx = lim. [1 1 b 3 ]=1. 3. Determine if the following series converge or diverge. Justify your answers completely.
MTH Lecture 2: Solutios to Practice Problems for Exam December 6, 999 (Vice Melfi) ***NOTE: I ve proofread these solutios several times, but you should still be wary for typographical (or worse) errors..
More informationMATH 6101 Fall 2008 Series and a Famous Unsolved Problem
MATH 60 Fall 2008 Series ad a Famous Usolved Problem Problems = + + + + = (2- )(2+ ) 3 3 5 5 7 7 9 2-Nov-2008 MATH 60 2 Problems ( 4) = + 25 48 2-Nov-2008 MATH 60 3 Problems ( )! = + 2-Nov-2008 MATH 60
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 informationIn this section, we show how to use the integral test to decide whether a series
Itegral Test Itegral Test Example Itegral Test Example p-series 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 information1 Lecture 2: Sequence, Series and power series (8/14/2012)
Summer Jump-Start Program for Aalysis, 202 Sog-Yig Li Lecture 2: Sequece, Series ad power series (8/4/202). More o sequeces Example.. Let {x } ad {y } be two bouded sequeces. Show lim sup (x + y ) lim
More informationMTH 142 Exam 3 Spr 2011 Practice Problem Solutions 1
MTH 42 Exam 3 Spr 20 Practice Problem Solutios No calculators will be permitted at the exam. 3. A pig-pog ball is lauched straight up, rises to a height of 5 feet, the falls back to the lauch poit ad bouces
More information11.6 Absolute Convrg. (Ratio & Root Tests) & 11.7 Strategy for Testing Series
11.6 Absolute Covrg. (Ratio & Root Tests) & 11.7 Strategy for Testig Series http://screecast.com/t/ri3unwu84 Give ay series Σ a, we ca cosider the correspodig series 1 a a a a 1 2 3 whose terms are the
More informationTesting for Convergence
9.5 Testig for Covergece Remember: The Ratio Test: lim + If a is a series with positive terms ad the: The series coverges if L . The test is icoclusive if L =. a a = L This
More informationINFINITE SEQUENCES AND SERIES
INFINITE SEQUENCES AND SERIES INFINITE SEQUENCES AND SERIES I geeral, it is difficult to fid the exact sum of a series. We were able to accomplish this for geometric series ad the series /[(+)]. This is
More information1 Introduction to Sequences and Series, Part V
MTH 22 Calculus II Essex Couty College Divisio of Mathematics ad Physics Lecture Notes #8 Sakai Web Project Material Itroductio to Sequeces ad Series, Part V. The compariso test that we used prior, relies
More informationsin(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 informationONE-PAGE REVIEW. (x c) n is called the Taylor Series. MATH 1910 Recitation November 22, (Power Series) 11.7 (Taylor Series) and c
ONE-PAGE REVIEW 6 (Power Series 7 (Taylor Series MATH 90 Recitatio November 22, 206 ( A ifiite series of the form F(x is called the ceter (2 (2 The radius of covergece a (x c is called a power series (
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: Numerical Series
Chapter 6: Numerical Series 327 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
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 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 Cauchy-Riema equatios? Solutio. a) A fuctio f : G C is called
More informationSeries Review. a i converges if lim. i=1. a i. lim S n = lim i=1. 2 k(k + 2) converges. k=1. k=1
Defiitio: We say that the series S = Series Review i= a i is the sum of the first terms. i= a i coverges if lim S exists ad is fiite, where The above is the defiitio of covergece for series. order to see
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 informationMATH301 Real Analysis (2008 Fall) Tutorial Note #7. k=1 f k (x) converges pointwise to S(x) on E if and
MATH01 Real Aalysis (2008 Fall) Tutorial Note #7 Sequece ad Series of fuctio 1: Poitwise Covergece ad Uiform Covergece Part I: Poitwise Covergece Defiitio of poitwise covergece: A sequece of fuctios f
More information9.3 The INTEGRAL TEST; p-series
Lecture 9.3 & 9.4 Math 0B Nguye of 6 Istructor s Versio 9.3 The INTEGRAL TEST; p-series I this ad the followig sectio, you will study several covergece tests that apply to series with positive terms. Note
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 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 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 116 Practice for Exam 3
Math 6 Practice for Exam Geerated October 0, 207 Name: SOLUTIONS Istructor: Sectio Number:. This exam has 7 questios. Note that the problems are ot of equal difficulty, so you may wat to skip over ad retur
More informationPlease do NOT write in this box. Multiple Choice. Total
Istructor: Math 0560, Worksheet Alteratig Series Jauary, 3000 For realistic exam practice solve these problems without lookig at your book ad without usig a calculator. Multiple choice questios should
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 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 information1. C only. 3. none of them. 4. B only. 5. B and C. 6. all of them. 7. A and C. 8. A and B correct
M408D (54690/54695/54700), Midterm # Solutios Note: Solutios to the multile-choice questios for each sectio are listed below. Due to radomizatio betwee sectios, exlaatios to a versio of each of the multile-choice
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 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 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 informationCalculus with Analytic Geometry 2
Calculus with Aalytic Geometry Fial Eam Study Guide ad Sample Problems Solutios The date for the fial eam is December, 7, 4-6:3p.m. BU Note. The fial eam will cosist of eercises, ad some theoretical questios,
More informationLecture 2 Appendix B: Some sample problems from Boas, Chapter 1. Solution: We want to use the general expression for the form of a geometric series
Lecture Appedix B: ome sample problems from Boas, Chapter Here are some solutios to the sample problems assiged for Chapter, 6 ad 9 : 5 olutio: We wat to use the geeral expressio for the form of a geometric
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 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 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 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 informationCHAPTER 1 SEQUENCES AND INFINITE SERIES
CHAPTER SEQUENCES AND INFINITE SERIES SEQUENCES AND INFINITE SERIES (0 meetigs) Sequeces ad limit of a sequece Mootoic ad bouded sequece Ifiite series of costat terms Ifiite series of positive terms Alteratig
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 informationReview for Test 3 Math 1552, Integral Calculus Sections 8.8,
Review for Test 3 Math 55, Itegral Calculus Sectios 8.8, 0.-0.5. Termiology review: complete the followig statemets. (a) A geometric series has the geeral form k=0 rk.theseriescovergeswhe r is less tha
More informationInfinite Sequence and Series
Chapter 7 Ifiite Sequece ad Series 7. Sequeces A sequece ca be thought of as a list of umbers writte i a defiite order: a,a 2,a 3,a 4,...,a,... The umber a is called the first term, a 2 is the secod term,
More informationMATH 31B: MIDTERM 2 REVIEW
MATH 3B: MIDTERM REVIEW JOE HUGHES. Evaluate x (x ) (x 3).. Partial Fractios Solutio: The umerator has degree less tha the deomiator, so we ca use partial fractios. Write x (x ) (x 3) = A x + A (x ) +
More informationd) If the sequence of partial sums converges to a limit L, we say that the series converges and its
Ifiite Series. Defiitios & covergece Defiitio... Let {a } be a sequece of real umbers. a) A expressio of the form a + a +... + a +... is called a ifiite series. b) The umber a is called as the th term
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 information