1. 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

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1 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 roblems are give at the ed of the documet MC# (0 ts.) 7. oe of them 8. all of them. B oly MC#2 (0 ts.) 6. A ad B 8. A ad B. A ad B MC# (5 ts.) 2. it = 0 4. it = 0. it = 0 MC#4 (5 ts.) 6. a = 2 2. a = a = 2 2 Questio # (25 oits) Dee the series X =2 l ( ) l : + a) Is the series absolutely coverget? Justify your aswer ad your use of ay test. Solutio: This roblem is similar to Questio # i Quiz 2. We must d whether the series X =2 l + coverges or ot. To do so, ote that the rst term i the deomiator grows faster tha the secod term ad is therefore the domiat cotributio as!. We ca comare this series with X usig the it comariso test. Let b = b =! c! =2 l l l l + + l ad c = l l =! l q + l =. We have that sice l! = 0 (this P P is justied by L'Hosital's rule). Therefore, l coverges if ad oly if coverges. Now ote that P =2 Z l l dx = 2 x l x + diverges by the itegral test sice the imroer itegral Z l 2 du (substitutio with u = l x) u l P diverges. So l diverges ad the origial series is ot absolutely coverget.

2 [There are may other ways to solve this roblem by comarig agaist P a series which we kow is diverget. For examle, sice l + l 2 l ad diverges by 2 l the itegral test, we d that the origial series is ot absolutely coverget.] b) If ot, is the series coditioally coverget or diverget? Justify your use of ay test. Solutio: We use the alteratig series test. Let b = l l : + First ote that b! 0 as!. Next, sice ad l are icreasig fuctios of we must have that b + b for all. The alteratig series test the imlies that the series is coverget. Sice it is ot absolutely coverget, it is coditioally coverget. Questio #2 (20 oits) Cosider the ower series X (x 2) : = a) Fid R, the radius of covergece of the series. Solutio: We will use the ratio test. Let The series coverges whe ad diverges whe! a + x 2 a = = a! = = (x 2) :! + x 2 (x 2)+ ( + ) + (x 2) < jx 2j >. Therefore, the radius of covergece is R =. b) Fid I, the iterval of covergece of the series. At the least, ame ay tests you use. i h h i ; 5, ; 5, ; 5 Solutio: The iterval of covergece is ecessarily either or ; 5. For x = the series is X ( ) ( ) = = X It is the alteratig harmoic series, which we kow coverges (this ca be show by usig the alteratig series test). For x = 5, the series is X = X = ; = = :, 2

3 h which diverges as it is the harmoic series (i.e., a -series with = ). To coclude, the iterval of covergece is I =. ; 5 Questio # (25 oits) Cosider the fuctio f (x) = l + 2 x2 : a) Fid its secod-degree Taylor olyomial T 2 (x) cetered at a = 2. Solutio: The olyomial T 2 (x) cetered at 2 ecessarily has the form Sice T 2 (x) = f (2) + f 0 (2)! (x 2) + f 00 (2) (x 2) 2 : 2! f 0 (x) = x + ; f 00 (x) = 2 x2 2 x2 + 2 x2 2 we get that f (2) = l, f 0 (2) = 2/, f 00 (2) = /9. So T 2 (x) = l() + 2 (x 2) 8 (x 2)2 : b) Fid the Maclauri series of f (i.e., series cetered at a = 0) ad determie its radius of covergece. [Hit: Try usig the kow ower series reresetatio to d the aswer quickly.] Solutio: First ote that X x = f 0 (x) = =0 x ; jxj < x + 2 x2 = x ( 2 x2 : Substitutig 2 x2 for x i the series exasio of x gives that whe obtai f (x) = ( ) x X = x 2 x2 2 x2 =0 2 x2 <, i.e., whe jxj < 2. We itegrate the ower series term by term to Z! ( ) X Z ( ) X ( ) 2 x2 dx = 2 x2+ dx = C + 2 (2 + 2) x2+2 : =0 =0 x X =0 To d the costat of itegratio C, ote that C = f(0) = l = 0. To coclude, X ( ) f(x) = 2 (2 + 2) x2+2 ; R = 2: =0

4 Versio 00 tester sriivasa (54690) This rit-out should have 4 questios. Multile-choice questios may cotiue o the ext colum or age fid all choices before aswerig. CalC2g0a oits Which, if ay, of the followig statemets are true? A. If a = 0, the a coverges. B. The Ratio Test ca be used to determie whether / coverges. C. If a is diverget, the a is diverget.. C oly correct 2. A oly. all of them 4. A ad B oly 5. B oly 6. B ad C oly 7. oe of them 8. A ad C oly Exlaatio: A. False: whe a = /, the a = 0, but a = = = diverges by the Itegral Test. B. False: whe a = /, the a + a = ( + ) as,, so the Ratio Test is icoclusive. C. True: if a were coverget, the a would be absolutely coverget, hece coverget. CalC2fc oits Determie which, if ay, of the followig series coverge. 6 (A) ( + 6) (B) (C) = = =. C oly 2. A oly ( ) ( ) !. oe of them 4. B oly 5. B ad C 6. all of them 7. A ad C ( ) 8. A ad B correct Exlaatio: To check for covergece we shall use either the Ratio test or the Root test which meas comutig oe or other of a + a, for each of the give series. a /

5 Versio 00 tester sriivasa (54690) 2 (A) The root test is the better oe to use because a / = as, so series (A) coverges. (B) The root test agai is the better oe to use because a / = 6 ( ) < as, so series (B) coverges. (C) The ratio test is the better oe to use because a ( + ( + )! ) = a!( + ) +. But while Thus ( + )!! = +, ( ) ( + ) + =. + + ( ) a + = a + e > as, so series (C) diverges. Cosequetly, of the give ifiite series, coverge. Fid the value of oly A ad B CalC7g2d oits. it = 0 correct 2. it = x x.. oe of the other aswers 4. it = 5. it = 6. it = Exlaatio: Set f(x) = x, g(x) = x = e xl. The f, g are everywhere differetiable fuctios such that f(x) = g(x) = Thus L Hosital s Rule alies, i which case Now f(x) g(x) = f (x) g (x). f (x) = x 2, g (x) = (l ) x. But the f (x) g (x) = x 2 (l) x, which, u to a costat, is the same it we started with excet that x i the umerator has become x 2. Cosequetly, if we aly L Hosital s rule sufficietly ofte, we fially ed u with Thus f(x) g(x) = x x = 0. CalC2b50s oits costat x. If the th artial sum S of a ifiite series = a

6 Versio 00 tester sriivasa (54690) is give by fid a for >. S = 2,. a = 2 2 correct 2. a = 2 2 ( ) 2. a = 2 4. a = 2 2 ( ) 2 5. a = 6. a = 2 ( 2 2 Exlaatio: By defiitio, the th artial sum of ) = a is give by I articular, S = a + a a. a = { S S, >, S, =. Thus a = S S = 2 2 = 2 ( ) 2 2 whe >. Cosequetly, for >. a = 2 2