Math 113 Exam 3 Practice

Transcription

1 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 to work o. The third sectio give the aswers of the questios i sectio. Review Sequeces Sequeces are a importat part of Chapter, because so much of what we do ivolves them. You eed to be able to take a sequece ad determie its behavior. Is is icreasig or decreasig? Does it coverge i the limit? What does it coverge to? Some theorems may be of help here:. If a sequece coverges, it is bouded.. If a sequece is bouded ad is (evetually icreasig or decreasig, the it coverges.. If a sequece {a } matches a fuctio f (i.e. f( = a ad the the limit of the sequece is also L. lim f(x = L, Rule is useful because we ca use everythig we kow about limits of fuctios to fid limits of sequeces. Sice L Hopital s rule is oe of them, you should expect to use it. There are other rules about sums of sequeces ad products of sequeces, etc. You are advised to review them i the text. Importat Sequeces Some limits occur ofte eough that it is advisable to kow about them i advace. For example, Dr. McKay expects all of his studets to kow the followig:. If c is a real, positive umber, the c /.. If c is a real, positive umber, the c 0.. c! /. ( 5. + c e c. All of the above limits except the third ca be prove usig L Hopital s rule. The third is a bit tricky but ca be c doe by oticig that the ( + st term of the series is times the th term. + If you ecouter these limits i a problem, you are welcome to use what you kow about them ad move o. Recursive Sequeces Most sequeces that we deal with have a rule we ca apply to fid the th term of the sequece. Recursive sequeces, however, oly have a rule that allows us to fid the th term if we kow all of the other terms that come before it. The most famous recursive sequece is the Fibboaci sequece give by a =, a =, a = a + a. Geerally, it is difficult to tell what a recursive sequece does. The theorems listed above ca show that a recursive sequece coverges. For example if the sequece ca be show to be bouded ad is icreasig, the it must have a limit. If a recursive sequece coverges, the fidig the limit of the sequece is easy: Let L represet the limit. Sice the sequece coverges to L, every sequece elemet i the recursio formula coverges to L also. For example, suppose we wish to fid the limit of a =, a =, a = a.

2 If we are reasoably sure the limit exists, we ca replace a by L to get L = L 0 You ca the use this to solve for L. Make sure the series coverges however, or you may arrive at the wrog coclusio. For example, suppose we have the recursive sequece a =, a = a. Usig this techique gives L = L, which yields L =. However, it is ot hard to see that this sequece diverges. Series I this sectio we leared about coverget ad diverget series. A series coverges if the sequece of partial sums coverge. There are some particular types of series that we leared about: Geometric Series ar a. We leared that the geometric series coverges to if r < ad diverges otherwise. We saw several applicatios where we could write a problem i terms of a geometric series. r Harmoic Series. We saw by examiatio of the partial sums s that this diverges. The itegral test also = shows the divergece of this series. ( + Alteratig Harmoic Series We kow by a later test that the alteratig Harmoic series coverges. We kow by the Maclauri series of l( + x that it coverges to = l(. Telescopig Series This is a series where the partial sum collapses to the sum of a few terms. We ca the take the limit of the partial sum to see what the series coverges to. Note that i. the oly series whose sums we could calculate were geometric ad telescopig. Tests for Covergece We leared about the followig tests for covergece: Divergece Test If a 0 the a diverges. This is a excellet test to start with because the limit is ofte easy to calculate. Keep i mid, however, if the limit is 0, the the Divergece test tells you othig. You must try some other test. p series If you recogize a series as a p series, p the you ca use the fact that a p series coverges whe (ad oly whe p >. Geometric series ar = a r but oly whe r <. Compariso Test To use the compariso test, we eed to have a large group of test series available. We also eed to kow if these test series coverge or ot. The most commo test series for the compariso test are the p series ad the geometric series. If the series acts like a p series, or acts like a geometric series, the you may wish to use the compariso test. Remember, if 0 a b ad b coverges, the a coverges. a diverges, the b diverges. Limit Compariso Test This test works well for the type of problems that also work with the compariso test, but is somewhat easier. You still eed the test series, but you do t eed to work to make the terms of the series greater tha or less tha some kow series. You oly eed to check the limit a lim. b If it is fiite ad positive, the both series coverge or both diverge. Sice you already kow about oe of them, you the kow about the other.

3 Itegral Test If we are tryig to determie whether a coverges, ad there is a fuctio f(x with f( = a, the the sum coverges iff a f(x dx coverges. (We assume that both the series {a } ad f(x are positive. So the itegral test is hady if the associated fuctio ca be itegrated without too much difficulty. Alteratig Series Test To use the alteratig series test, you eed to verify three thigs: The series is alteratig. (This ca usually be doe by ispectio. The terms of the series coverge to 0. (Hopefully you did this whe you applied the Divergece test. Fially, the terms of the absolute values are decreasig. The secod statemet does ot ecessarily imply the third. If this is true, the the alteratig series test tells us the series coverges. Ratio Test If lim a + a = L, the the series is absolutely coverget if L < ad diverget if L >. If L =, the test fails. This test works really well whe a factorial is preset i a. Root Test If lim a = L, the the series is absolutely coverget if L < ad diverget if L >. If L =, the test fails. This test works really well whe there are powers of i a. Remember, the Itegral test ad the compariso tests oly work whe the series has o-egative terms. If you have a series where the terms are both positive ad egative, the you must be able to say whether the series coverges absolutely, coverges coditioally, or diverges. It is oe of these. These are mutually exclusive coditios. Estimatig the tail I a ifiite series, the tail is a term usually used to idicate the last part of the series. For example, if we wish to approximate the sum of the followig coverget series, the we ca write it as k ( +, ( + + =k+ ( + The part that is still a ifiite sum is called the tail. The sum of the tail is called the error of our approximatio. If we ca test covergece of a series by the itegral test, the there is a easy way to fid a estimate of the tail: Assume f(x is defied o [b, for some b, ad f( = a. The k+ f(x dx =k+ a k f(x dx. For example suppose that we sum the first 5 terms of the above series: How close is this? We fid that ad 4 ( + ( + = = Thus, the error is betwee these two umbers. 4 (x + dx = 484 = , (x + dx = 4 =

4 If we ca use the compariso test to fid covergece, the we ca sometimes still use the above formula, but oly for upper bouds. For example, if I am tryig to estimate the fact that meas that =k+ =k+ +, + < + < k x dx. If our sum is alteratig, with the absolute values decreasig to 0, the the tail is bouded by the absolute value of its first term. That is, if a > 0 for all, ( a a k+. =k+ (Note that it does ot matter if there is ( or ( + i the sum. All that matters is that the sum is alteratig, ad satisfies the alteratig series test. Power Series Recall that a power series is a series of the form a (x c. The value c is called the ceter of the power series, ad the values a are called the coefficiets. A power series is a way to represet a fuctio. However, the power series may have a differet domai tha the fuctio does. To fid the domai of the power series, (called the iterval of covergece, we do the followig:. Apply the ratio or root test to the power series. If the limit is 0, the power series coverges everywhere ad the radius of covergece is. If the limit is, the power series coverges oly at the ceter, ad the radius of covergece is 0. Otherwise, set the limit to be less tha, ad rework the iequality so it says x c < R. R is the radius of covergece.. The power series is ow guarateed to coverge absolutely o (c R, c + R, ad diverge o (, c R (c + R,. We ow test the power series at the edpoits. Plug the edpoits c R ad c + R ito the power series ad use oe of the other 5 tests (ot Ratio, ot Root to determie whether they coverge. State the iterval of covergece usig paretheses to idicate the power series does ot coverge at a edpoit, ad a bracket to idicate it does. Fidig sums of series Fidig a power series that represets a specific fuctio is the ext topic. The first oe we leared was the geometric series: x = x iff x (,. We the foud the sum of several series by differetiatig, itegratig, multiplyig by x, etc. I additio to the geometric series, we foud the followig power series. You should kow them ad their iterval of covergece. ta x = l( + x = cos x = ( x +, [, ] + ( x +, (, ] + ( x, (, (!

5 Questios Try to study the review otes ad memorize ay relevat equatios before tryig to work these equatios. If you caot solve a problem without the book or otes, you will ot be able to solve that problem o the exam. Determie whether each sequece i to 4 is coverget. State what it coverges to, if applicable. Is the sequece icreasig or decreasig? Is the sequece bouded?. a = a = cos(π/. a = si + 4. a = Fid the value that the sequece give by a =, a = a + coverges to. Determie whether the series is coverget or diverget. If it is coverget, fid its sum = = k k k 5 For problems 9 to 7, determie whether the series coverges or diverges. State the test you used. e = ( l = k= k= k= = = ta k + k k (k + k l k l( 7. = cos + 8. Show that 5 is a upper boud o the error of = 4 if the sum is approximated by the first + 7 two terms. 9. Approximate the sum of 0... = + by summig the first 0 terms. Fid a boud o the error of your approximatio. For problems 0 through 5, determie whether the series is absolutely coverget, coditioally coverget, or diverget. = ( l (! = = cos(π = ( ( + ( (l = ( 5 + = 6. Show that 4 is a upper boud o the error of = 4 if the sum is approximated by the first + 7 two terms. 7. Suppose the power series a (x+ has a radius = of covergece R = 5. List all possible itervals of covergece. 8. Fid the radius ad iterval of covergece of (x = 9. Fid the radius ad iterval of covergece of ( 4 (x + = 0. Fid the radius ad iterval of covergece of x =. Fid the radius ad iterval of covergece of x =

6 . Fid the radius ad iterval of covergece of! x 00 =. Fid a power series represetatio i powers of x for the fuctio f(x = + x with iterval of covergece. 4. Fid a power series represetatio i powers of (x for the fuctio f(x = +x ad give the iterval of covergece. 5. Fid a power series represetatio i powers of (x for l( + x. 6. What is the power series represetatio of x ( x? Aswers. coverges to 0, decreasig, bouded. diverges, ot icreasig or decreasig, bouded.. coverges to 0, ot icreasig or decreasig, bouded. 4. diverges, icreasig, bouded below diverges by the Divergece Test 7. Coverges to / (Telescopig sum 8. Coverges to 5/4 (geometric series 9. Use the itegral test x e x dx = 5 e Therefore it coverges by the itegral test 0. Use the itegral test ( l x dx = Therefore it coverges by the itegral test.. Use the itegral test x ta x π dx = + x Therefore it coverges by the itegral test. x (x + dx = 9 Therefore it coverges by the itegral test.. We have to be careful here sice the fuctio is ot defied at k =. By a chage of variables, k = + we see that k l k = ( + l( + ad k= = we ca the use the itegral test. The book otes that we ca also simply chage the limits of itegratio, though it does ot state this as a theorem. dx = ( (x + l(x + Therefore it diverges by the itegral test < + 6 = ( ad ( coverges (geometric r =. Thus + 6 coverges by Com- pariso Test. 5. lim Hece = Compariso Test. 6. l >. Thus Test. = cos + = l( = ad diverges. diverges by Limit diverges by Compariso cos 7. + < + < ad coverges (p-series. Thus coverges by Compariso Test. 8. Sice 4 +7 < it is sufficiet to show that 4 5 is a boud o the sum. The R 4 x dx = 4 4. = 9. 0 = + = Note that 0 x + x dx =.05, x + x dx = Thus, the sum lies i the iterval (.84409, Coverges by the Alteratig Series test. By the Itegral Test, it does ot coverge absolutely. So it coverges coditioally.. Diverges by the Test for Divergece.. Coverges by the Alteratig Series test. By the Limit Compariso test (with b =, it does ot coverge absolutely. So it coverges coditioally.. Coverges absolutely by the Limit Compariso test (with b =.

7 4. Coverges by the Alteratig Series test (Use L Hôpital s rule. By the Itegral Test, it does ot coverge absolutely. So it coverges coditioally. 5. Coverges absolutely by the Limit Compariso test (with b =. / 6. R = 4 +7 = 4 = x 4 dx = 4 7. (-6,4, (-6,4], [-6,4, [-6,4] 8. a+ a 0 for all x: R =, I = (, 9. a+ a 4 x < : R = 4, I = ( 7 4, 9 4 ] 0. a+ a x < : R =, I = (,. a+ a x < : R =, I = [, ]. a+ a for all x: R = 0, I = {0} (. +x = ( x = = ( x + for x (, 4. +x = +(x = ( = (x ( x ( (x for x (, 5. Itegrate the previous solutio to get l ( + x = C + ( (+ (x + : (C = l x 6. ( x = x d dx = x = ( x x = ( = x d dx x x + =