Math 1B: Discussion ExercisesGSI: Theo Johnson-Freydhttp://math.berkeley.edu/~theojf/09Summer1B/Find two or three classmates and a few feet of chalkboard. As a group, try your hand at thefollowing exercises. Be sure to discuss how to solve the exercises — how you get the solution ismuch more important than whether you get the solution. If as a group you agree that you allunderstand a certain type of exercise, move on to later problems. You are not expected to solve allthe exercises: some are very hard.Exercises marked with an § are from Single Variable Calculus: Early Transcendentals for UCBerkeley by James Stewart. Others are my own or are independently marked.Infinite SeriesA series is an infinite sum of numbers. There are two sequences associated to each series. Firstof all, there’s the sequence of terms in the sum (a.k.a. “summands”):1 +12+14+18+ ··· 1,12,14,18, . . .Second, there’s the sequence of partial sums:1 +12+14+18+ ··· 1,32,74,158, . . .We normally write the former explicitly:1 +12+14+18+ ··· =∞Xn=012nOn the other hand, a series converges exactly if the sequence of partial sums converges, and thelimit of the sequence of partial sums is the “value” of the series. Standard notation: the sequenceof summands is an, the sequence of partial sums is sn=Pnk=0ak, and if {sn} converges, thenP∞n=0an= limn→∞sn.Here are some facts about series:• A series cannot converge unless the sequence of summands tends to 0. But the sequence ofsummands can go to 0 without the series converging.• The geometric series a + ar + ar2+ ar3+ ··· =P∞n=0arnconverges if and only if |r| < 1. If itconverges, then it converges to a/(1 −r). (A proof and generalization of this is in exercise 3.)• IfPan= A andPbn= B, thenP(an+ bn) = A + B. Let cn=Pnk=0akbn−k. ThenPcn= AB. This is called the “Cauchy product” or “discrete convolution” of the two series.• This one isn’t so much a fact as a technique. If we can write each anas a difference an=bn− bn+1, then sn= b0− bn+1, so the sumPanconverges if the sequence {bn} converges.Partial fractions are a good way to find such decompositions of terms as differences.1. What is the sum∞Xn=112n=12+14+18+116+ . . . ?1How about∞Xn=113n=13+19+127+181+ . . . ?If k is some number strictly greater than 1, what is∞Xn=11kn=1k+1k2+1k3+1k4+ . . . ?2. § Determine whether the following series are convergent or divergent. If convergent, find thesums:(a)∞Xn=1(−3)n−14n(b)∞Xn=01√2n(c)∞Xn=1πn3n+1(d)∞Xn=01 + 3n2n(e)∞Xn=0[(.8)n− (.3)n] (f)∞Xn=0(cos 1)n(g)∞Xn=03n(n + 3)(h)∞Xn=1arctan n (i)∞Xn=1lnnn + 13. Remember how to evaluate the geometric series: if |r| < 1, then we can evaluate S =P∞n=0arn= a + ar + ar2+ ar3+ . . . by mutiplying by r and subtracting:S = a + ar + ar2+ ar3+ . . .− ( r · S = ar + ar2+ ar3+ . . . )S − rS = athus S =a1 − r.Use this method to compute the following sums:(a) 1 +23+39+427+581+6243+ . . . (b) 1 +42+94+168+2532+3664+ . . .4. § When money is spent on goods and services, those who receive the money also spend someof it. The people receiving some of the twice-spent money will spend some of that, and soon; this chain reaction is called the multiplier effect. In a hypothetical isolated community,the local government begins the process by spending D dollars. Suppose that each recipientof spent money spends 100c% and saves 100s% of the money she or he receives. The valuesc and s are called the marginal propensity to consume and the marginal propensity to saveand, of course, c + s = 1.(a) Let Snbe the total spending that has been generated after n transactions. Find anequation for Sn.(b) Show that limn→∞Sn= kD, there k = 1/s is the multiplier. What is the multiplier ifthe marginal propensity to consume is 80%?(c) In fact, the marginal propensities to save and to consume vary by socioeconomic class(among other things): poor people spend a larger proportion of the money they receive,and rich people save a larger proportion. If the government is trying to stimulate theeconomy, to whom should it give its
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