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Course: M 408D 408D, Spring 2012
School: University of Texas
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Word Count: 1997

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gilbert HW02 (55035) This print-out should have 21 questions. Multiple-choice questions may continue on the next column or page find all choices before answering. 1 assuming that the pattern of the first few terms continues. 2 n 2. 001 3 n 10.0 points Find a formula for the general term an of the sequence o = a {an} n=1 = 2, 6, 10, 14, . . . 3. 3n = a 2 n 3 . a n = 2 , assuming that...

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gilbert HW02 (55035) This print-out should have 21 questions. Multiple-choice questions may continue on the next column or page find all choices before answering. 1 assuming that the pattern of the first few terms continues. 2 n 2. 001 3 n 10.0 points Find a formula for the general term an of the sequence o = a {an} n=1 = 2, 6, 10, 14, . . . 3. 3n = a 2 n 3 . a n = 2 , assuming that the pattern of the first few terms continues. 3 4 . a n = 3 2 1. an = n + 3 6. 2. an = 4n 2 correct 3. an = 5n 3 7. 4. an = 3n 1 5. an = n + 4 Explanation: By inspection, consecutive terms an1 an in the sequence 3 = n1 n1 correct 4 n 3n = a 4 n Explanation: By inspection, consecutive terms an1 and an in the sequence and o {an} n=1 = a n1 {an} n=1 = 2 1, 3, have the property that 2, 6, 10, 14, . . . have the property that a = ra n1 n an an1 = d = 4 . 4 8 , 27 , . . . o 9 = 2 3 a n1 Thus Thus 2 an = ran1 = r an2 = . . . an = an1 + d = an2 + 2d = . . . = a1 + (n 1)d = 2 + 4(n 1) . Consequently, a n = 4n 2 =r n 1 a1 2n1 a1 . = 3 Consequently, . n1 an = 32 keywords: 002 since a1 = 1. 10.0 points Find a formula for the general term a n of the sequence {an}n =1 = 2 1 , 3 , 9 4 8 , 27 , . . . o, keywords: sequence, common ratio 003 10.0 points . HW02 gilbert (55035) 2 5 Compute the value of lim n when lim lim n n 1. limit = 12 3. converges with limit = 57 4. converges with limit = 32 = 2. b n 5 correct 5. sequence does not converge Explanation: After division by n we see that n 5 + 1) 5 2. limit = 2 3. limit doesnt exist ( n an = 4+n 12 4. limit = 5 But 5 5. limit = 2 n (1) , . 3 3 0 n n 5 as n , so an 4 as n . Consequently, the sequence converges and has Explanation: By properties of limits limit = 5 4 lim 4anbn n2 = 4 lim a n correct 4 3an bn a = 6, n 2. converges with limit = 4a n b n lim n = 48 b n . n 005 10.0 points while (3a b ) lim n =3 Determine if the sequence {a n} converges 2 n (2n 1)! , when an = n n 5 n lim a lim n n b n = 20 6= 0. Thus, by properties of limits again, lim n 4anbn 3an bn = 1. converges with limit = 12 5 (2n + 1)! and if it converges, find the limit. 5 4 . 2. converges with limit = 004 25 3. converges with limit = 5 10.0 points Determine if the sequence {an} converges, and if it does, find its limit when n an = 5n + (1) . 4n + 3 1. converges with limit = 1 2 4. converges with limit = 4 5 5. does not converge Explanation: correct HW02 gilbert (55035) 3 But By definition, m! is the product ln 2 m! = 1 2 3 . . . m of the first m positive integers. When m = 2n 1, therefore, (2n 1)! = 1 2 3 . . . (2n 1) , ln(n) + 1, ln 5 ln(n) +1 1 as n . Consequently, the given sequence converges and has while limit = 1 . (2n + 1)! = 1 2 3 . . . (2n 1)2n(2n + 1) . 007 when m = 2n + 1. But then, 2 2 5n (2n 1)! = 5n (2n + 1)! Determine if the sequence 5 2n(2n + 1) {2, 0, 2, 0, 0, 2, 0, 0, 0, 2, . . . } 4 as n . Consequently, the given sequence converges with limit = 006 5 4 10.0 points . 10.0 points Determine whether the sequence {an} cov-erges or diverges when converges and, if it does, find its limit. 1. limit = 4 2. limit = 0 3. sequence diverges correct 4. limit = 1 ln (2n) an = 5. limit = 2 ln (5n) , and if it converges, find its limit. 1. converges with limit = 1 correct Explanation: Since a sequence {an}n converges and has limit L precisely when ALL the values of an are as close as we please to L for it all suciently large n, the 2. sequence diverges sequence diverges 3. converges with limit = 0 4. converges with limit = ln 2 ln 5 because the values of an oscillate from 0 to 2 and back as n . 008 5. converges with limit = 25 Rewrite the finite sum Explanation: Since ln (2n) ln (5n) 10.0 points 9 12 6 + + 2+4 3+4 4+4 = ln 2 + ln(n) = ln 5 + ln(n) ln 2 +1 ln(n) . ln 5 ln(n) + 1 21 + 15 + . . . + 5+4 7+4 using summation notation. HW02 gilbert (55035) 5 X 1. k =0 3k k2 (1) k+4 8 X 2. k =3 1. convergent with sum 2. convergent with sum 3k k2 (1) k+4 X 5 16 16 correct 5 3. convergent with sum 5 3. 4 3k 16 3 3 4. convergent with sum 16 k =0 k + 4 7 5. divergent X 4. k =2 k 3+k 4 correct Explanation: The infinite series 8 X 5. k =3 k 3+k 4 7 X 6. k =2 1 k2 (1) k+4 The numerators form a sequence 9, 12 , 1 X a rn1 n=1 a 2+4, 3+4, 4+4, 5+4, ... , 7+4. Thus the general term in the series is of the form (i) converges with sum 1 r when |r| < 1, (ii) and diverges when |r| 1 . Consequently, the given series is 3k k +4 k where the sum ranges from k = 2 to k = 7. Consequently, the series becomes 16 convergent with sum 7 X 3k 010 k+ 4 10.0 points Determine whether the series k=2 in summation notation. X n =1 009 10.0 points Determine whether the infinite series 41+ 1 rn1 15 , . . . , 21 , while the denominators form a sequence a= X is an infinite geometric series with a = 4 1 and r = 4 . But a geometric series Explanation: 6, 1 4 1 + 4 16 + 64 = n =1 a 3k 1 4 16 + 64 1 is convergent or divergent, and if convergent, find its sum. 2 5n +n 4 is convergent or divergent, and if convergent, find its sum. 1. convergent with sum = 2. divergent correct 9 2 5 . HW02 gilbert (55035) 5 25 5. converges with sum 7 = 12 4. convergent with sum = 52 6. converges with sum = 23 3. convergent with sum = 2 Explanation: An infinite geometric series 5. convergent with sum = 9 r< Explanation: X an is divergent when The infinite series (ii) diverges when |r| 1 . n Now Note for the given series, 2n a = n 2 5n + 4 = 5+n 4, so lim n sum = 1 r , while it lim an 6= 0 . a = lim n n 2n X k =1 1 5 k X 1 k1 = k =1 5 5 1 1 is a geometric series with a = r = 5 < 1. Thus it converges with 2 = 6= 0 . 5n + 4 5 Thus the given series is divergent . 011 10.0 points sum = while X 1 X 2k = k =1 k =1 5 k 2 k1 5 52 2 is a geometric series with a = r = 5 < 1. Thus it too converges, and it has sum = 2 k =1 3. Consequently, being the sum of two conver-gent series, the given series 1 + 2k 5k converges or diverges, and if it converges, find its sum. converges with sum = 1. series diverges 2. 4, Determine if the series X n1 n =1 a r (i) converges when | | 1 and has a n=1 lim an exists but n P 1 4+ 2 3 11 = 12 012 10.0 points 11 12 correct converges with sum = Find the sum of the infinite series X 2k k =1 (cos ) , (0 < 2) , 3 3. converges with sum = 4 5 4. converges with sum = 6 whenever the series converges. 2 2 1. sum = sin cos . HW02 gilbert (55035) 2 2. sum = csc 2 3. sum = tan 014 (part 2 of 3) 10.0 points (ii) is What an for n > 1? 4 a 2 4. sum = cot correct a Explanation: For general the series 2. k=1 (cos2 )k a 4. is an infinite geometric series with common 2 ratio cos . Since the series starts at k = 1, its sum is thus given by cos 2 2 1 cos Consequently cos 2 2. = sin P a 5n 1 = n 1. a1 = S n S n1 . n1 , 5(n + 1) 6 = n +1 n +1 6 = 5n+ 1. =5 4. a1 5 Sn = = 2 correct 3. a1 correct Consequently, = 2 2. a1 n(n + 1) n But is n +1 (i) what is a1? 6 = an n=1 S a n2 Since Sn = a1 + a2 + + an , we see that . n = n(n + 1) 4 = n Explanation: 013 (part 1 of 3) 10.0 points partial sum of 4 5 . an 6. 2 sum = cot th = n(n 1) 6 =2 n n 3. a n X = n(n 1) 6 n 1. 2 5. sum = sec If the n 6 =3 5. a1 an = 6 n 6 = n +1 6 n(n + 1) for all n > 1. 015 (part 3 of 3) 10.0 points P (iii) What is the sum 1. sum = 5 correct = 3 2. sum = 6 Explanation: Since a1 = S1, a1 =2 . 3. sum = 4 n =1 an? 4. sum = 3 moreno (am49869) HW02 gilbert (55035) series X 5.sum=2 Explanation: By definition a, n n =3 sum = lim Sn = n 5 n 1 . lim an = f (n) . Clearly from this figure we see that a3 = f (3) n+1 n 7 < 3 2 f (x) dx, = f (4) < 4 3 f (x) dx , Let f be a continuous, positive, decreasing function on [2, ). Compare the values of the series a5 = f (5) < 5 4 f (x) dx, 17 a6 = f (6) < 6 5 f (x) dx , Thus a sum = 5 . 016 4 10.0 points X A= while f (n) n=3 and so on. Consequently, and the integral 17 B= A<B f (x) dx . . 2 1. A = B keywords: 2. A < B correct 017 10.0 points Determine whether the series 3. A > B Explanation: X In the figure k=1 4 k ln(5k) is convergent or divergent. 1. series diverges correct 2. series converges Explanation: The function a a 3 2 a 4 3 a 5 4 f (x ) = ... 6 5 6 the bold line is the graph of f on [2, ) and the areas of the rectangles the terms in the 4 x ln(5x) is continous, positive and decreasing on [1, ). By the Integral Test, therefore, the series 4 X k =1 k ln(5k) moreno (am49869) HW02 gilbert (55035) converges if and only if the improper integral 4 f (x) dx = 1 x ln(5x) 1 2. r = exists. To evaluate this last integral, set u = ln(5x). Then x 1 2.47474747 . 1. r = lim dx t 1 x ln(5x) du = Use this idea to find the rational number r whose repeating decimal expansion is dx converges, I.E., if and only if t 4 3. r = dx , 4. r = in which case, 4 dx = t 1 x ln(5 x) 4 LN(5t) LN 5 u 5. r = du = 4 ln(ln(5t)) ln(ln 5))o . But Consequently, 4 lim t dx 1 245 250 250 49 101 99 correct 101 99 20 Explanation: The decimal expansion 2.47474747 x ln(5x) does not exist. The Integral Test thus ensures that the given series diverges 18 247 can be interpreted as the infinite geometric series 47 47 47 + + + ... 2+ 2 4 6 10 10 10 lim ln(ln(5t)) = . t t 8 = 2+ 245 r= 019 10.0 points 99 . 10.0 points Which of the following series are convergent: x = 5.78787878 can be interpreted as the infinite geometric series 78 78 78 x = 5 ++ ++ . . . . 2 4 6 10 10 10 By summing this infinite geometric series we thus get the rational number having the given repeating decimal expansion. 47 X 1 2 k =0 10 2 k. Consequently, . A repeating decimal expansion such as 10 X A. 2 n 2 /3 n =1 B. X 3 n =1 n + 1 1 C. 1 1 1 1+ + +2 + +... 2 3 5 moreno (am49869) HW02 gilbert (55035) 9 is divergent. 1. A only keywords: convergent, Integral test, 2. A and B only 020 3. all of them 10.0 points Determine the convergence or divergence of the series 4. C only (A) 5. none of them correct 1+ 1 1 1 1 4 + 9 + 16 + 25 + ... , and 6. B only X 7. B and C only 8. A and C only m =1 (B) 2 mem . 1. A convergent, B divergent Explanation: By the Integral test, if f (x) is a positive, decreasing function, then the infinite series 2. both series convergent correct X f (n) 3. A divergent, B convergent n=1 converges if and only if the improper integral 4. both series divergent f (x) dx 1 converges. Thus for the three given series we have to use an appropriate choice of f . 2 A. Use f (x) = x2/3 . Then f (x) dx 1 Explanation: (A) The given series has the form 1 1 1 1 X 1 + 4 + 9 + 16 + 25 + ... = n=1 1 n (B) The given series has the form X f (m) Then f (x) dx m=1 with f defined by 1 is divergent (log integral). 1 C. Use f (x) = x1/2 . Then f (x) dx 1 . This is a p-series with p = 2 > 1, so the series converges. is divergent. 3 B. Use f (x) = x + 1. 2 f (x) = xe x2 . Note first that f is continuous and positive on [1, ); in addition, since x 2 f (x) = e 2 (1 2x ) < 0 moreno (am49869) HW02 gilbert (55035) for x > 1, f is decreasing on [1, ). Thus we can use the Integral Test. Now, by substitu-tion, 12 x dx = 2 ex xe 2 t 1 and so xex dx 2 1 = converges if and only if the infinite series X 1 2 e. n=1 converges. In the given example 1f (x) = xp , Since the integral converges, the series con-verges. 021 10.0 points If the improper integral 1 (A) (B) (C) (D) (E) n X n 1 p f (x) = xp +1 , dx converges, which of the following statements is (are) always true? X which is a function both continuous and pos-itive on [1, ). It will also be decreasing on [1, ) if f (x) < 0 for all x > 1. But xp 1 f (n) t , 1 10 so f will be decreasing provided p > 0. On the other hand, the improper integral p converges; 1 np +1 diverges; 1 X n np 1 converges; 1 X n np 1 diverges; 1 X n np +1 converges. n 1 x 1 dx p converges if and only if 1 n nlim x 1 p dx . exists. But n 1 x 1 1 n p dx = px p1 = p 1 1 1 n p 1 . 1 Consequently, the improper integral 1. A, D and E 2. A and E only correct x 1 3. B and D only 4. A, C and E only 1 dx p converges if and only if p > 1. Hence by the Integral test, the infinite series X 5. A only 1 p n=1 x Explanation: To apply the Integral test we need to start with a function f which is positive, continuous and decreasing on [1, ). Then the integral test says that the improper integral f (x) dx 1 converges if and only if p > 1. Now we can check which of the statements is (are) always true. (A) This is always true because of the Inte-gral test. moreno (am49869) HW02 gilbert (55035) (B), (E) Since Pn >1 1 = p + 1 > 1, the pn Integral test ensures that converges. P+1 Thus (B) is false and (E) is true. P 1 (C), (D) The series nn converges if 1 P and only if p 1 > 1, I.E., when p > 2. Since the convergence of the improper integral 1 1 xp dx guarantees only that p > 1, we see that state-ments (C) and (D) are true for some values of p and false for others. Consequently, of the statements, only A and E are always true. PDF to Word 11
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University of Texas - M 408D - 408D
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University of Texas - M 408D - 408D
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University of Texas - M 408D - 408D
HW06a gilbert (55035)This print-out should have 10 questions.Multiple-choice questions may continueon the next column or page find allchoices before answering.110.0 pointsA rectangular box is constructed in 3space with one corner at the origin and
University of Texas - M 408D - 408D
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University of Texas - M 408D - 408D
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University of Texas - M 408D - 408D
HW08 gilbert (55035)This print-out should have 21 questions.Multiple-choice questions may continueon the next column or page find allchoices before answering.001100210.0 pointsIdentify the quadric surface10.0 pointsClassify the quadric surface
University of Texas - M 408D - 408D
HW09 gilbert (55035)This print-out should have 18 questions.Multiple-choice questions may continueon the next column or page find allchoices before answering.00110.0 pointsFind an equation for the plane passingthrough the origin that is parallel t
University of Texas - M 408D - 408D
HW10 gilbert (55035)This print-out should have 16 questions.Multiple-choice questions may continueon the next column or page find allchoices before answering.00110.0 pointsIn the contour map below identify thepoints P, Q, and R as local minima, lo
University of Texas - M 408D - 408D
DiscGQ02 gilbert (55035)This print-out should have 4 questions.Multiple-choice questions may continueon the next column or page find allchoices before answering.00110.0 points1and the series11B=F (N) .N=61. A &lt; BDetermine whether the series
University of Texas - M 408D - 408D
EXAM 02 gilbert (55035)This print-out should have 14 questions.Multiple-choice questions may continueon the next column or page find allchoices before answering.0011I.E., y = x. Thus by the constraint equation,g (x, x) = 2x + x 6 = 0 ,I.E., x = 2
University of Texas - M 408D - 408D
DiscGQ08 gilbert (55035)This print-out should have 4 questions.Multiple-choice questions may continue onthe next column or page find all choicesbefore answering.12.001 10.0 pointsWhen z = P (x, y) is the pressure at apoint (x, y) in the xy-plane,
University of Texas - M 408D - 408D
DiscGQ07 gilbert (55035)1This print-out should have 4 questions.Multiple-choice questions may continue onthe next column or page find all choicesbefore answering.z001 10.0 pointsWhich one of the following equations hasgraphxBut this circle has
University of Texas - M 408D - 408D
DiscGQ06 gilbert (55035)This print-out should have 4 questions.Multiple-choice questions may continueon the next column or page find allchoices before answering.1Which of the following expressions arewell-defined for all vectors a, b, c, and d?10.
University of Texas - M 408D - 408D
DiscGQ05 gilbert (55035)This print-out should have 4 questions.Multiple-choice questions may continue onthe next column or page find all choicesExplanation:After completion of squares222x + y + z 6x + 2y 8z + 17 = 0before answering.001110.0 p
University of Texas - M 408D - 408D
EXAM 01 gilbert (55035)This print-out should have 14 questions.Multiple-choice questions may continueon the next column or page find allchoices before answering.00141. I =9= 29 correct3. IDetermine whether the series= 894. integral doesnt con
University of Texas - M 408D - 408D
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University of Texas - M 408D - 408D
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Summary of Key Learning Objectives forClass 2: Bridgeton CaseThe Bridgeton Case introduces cost accounting by providing a simple example of the design andfunctioning of traditional product cost systems. The case setting and managerial decisions covered
UC Davis - MGT ` - 170
MGT 170: Class 1Summary of Key Learning Objectives forP&amp;G Polska CaseThe P&amp;G Polska case introduces the topic of management accounting, including both costmanagement and management control (i.e., performance measurement and evaluation). The caseillus
UC Davis - MGT ` - 170
5-10 Year Plan DevelopmentJanF ebM arA prM ayJ unJ ulA ugS epO ct2 pageB u r k e L e t t erCE O J &amp; JDis c u s sw it hE xec .C om mStr ateg yReviewExec . C om mStolz erDeb atean dA dju s tCod m anBoar dFunc . D epts .P r ior Yr
UC Davis - MGT ` - 170
JanFebMarAprMayJunJulAugSepO ctCEO J&amp;JExec. CommStolzerDebateandAdjustCodmanBoardFunc. Depts.Prior Yr2nd YearFcstDebateandAdjustMissionState.5/10 Plan-by area-bus. seg-strategies-mktg plan-other deptsInitialPrepApproved
NUCES - FINANCIAL - 101
CHAPTER 10ACQUISITION AND DISPOSITION OFPROPERTY, PLANT, AND EQUIPMENTIFRS questions are available at the end of this chapter.TRUE-FALSEConceptualAnswerFTFTFTFFFTTTTFFTTFFTNo.Description1.2.3.4.5.6.7.8.9.10.11.12.13
McMaster - CELL BIO - 2b03
Cancer biology: tumorigenesis can stem from errors in signaling pathways and cell cycle regulationmechanisms Cancer is result of uncontrolled cell divisiono Leading cause of premature death in Canadao In young animal, cell proliferation exceeds cell
McMaster - CELL BIO - 2b03
McMaster - CELL BIO - 2b03
Final Exam Practice QuestionsThese questions can be used as a guide to help you focus your studying to cover key points that willlikely be on the final exam. Two disclaimers: 1) This guide should not be used as a substitute for thelecture notes, althou
Utah Valley University - HIST - 2710
AmistadAmistad We are all born into something, a certain belief or ideology. Others create their ownideologies, either out of conviction or for the purpose of convenience. Our lives revolve aroundthese ideas, these ideologies, for it is through them th
Utah Valley University - HIST - 2710
Brazil TourismSubject: Tourism in Brazil Statement: A brief overview of Brazil. Introduction: With its 8,512thousands km2, Brazil is the fifth largest country in the world and it is today America Latin'smost important country, if considering the produc
Utah Valley University - HIST - 2710
Brazil, Japan, UkExecutive Summary A cross analysis between three countries around the world were arrangedto determine which of those countries present the best investment opportunities. An evaluationamong the countries was structured through the compa
Utah Valley University - HIST - 2710
BRAZIL For years Brazil has been a featured country in South America. Not only is Brazil avery pretty country, it is filled with an abundance of natural resources. For three centuries Brazilwas ruled by Portugal. The country finally gained its independe
Utah Valley University - HIST - 2710
Brazilians and North AmericansBrazilians have a strikingly distinct culture which favors physical contact to show love andaffection compared to the North American culture. The personal space that Brazilians have isnot as great as the North Americans ha
Utah Valley University - HIST - 2710
The world's most luxurious hotel, Burj Al-Arab or the Arabian Tower, is the second modernwonder of the world. Ever since its completion in November 1997, the hotel has become an iconfor the technologically expanding city of Dubai. The jewel of the Middl
Utah Valley University - HIST - 2710
Child Prostitution EssayThe word &quot;prostitution&quot; dates back to the year 1553. Webster's dictionary defines it as the act orpractice of indulging in promiscuous sexual relations especially for money or the state of beingprostituted. Prostitution probably