MAT 127: Calculus C, Fall 2009Solutions to Mini-Quiz 1:qualitative properties of 1st-order differential equationsProblem AConsider the four differential equations for y = y(x):(a) y′= 2 − y, (b) y′= x(2 − y), (c) y′= x + y −1 (d) y′= sin x sin y.Each of the four diagrams be low shows three solution curves for one of these equations:Iyx11IIyx11IIIyx11IVyx11Match each of the diagrams to the corresponding differential equation (the match is one-to-one)and explain your reasoning:diagram I II III IVequation b c d a• I and IV are not (c) or (d) because the constant function y = 2 is not a solution of (c) or (d)• I is not (a) because some solution curves in I descend when y < 2 (and x < 0); IV is not (b)because one of the solution curves in IV ascends when y < 2 and x < 0• III is not (a), (b), or (c) because the constant function y = 0 is not a solution of (a), (b),or (c)• II is not (a), (b), or (d) because the function y =−x is not a solution of (a), (b), or (d); alsothe curve s in II do not have zero slope for y =2 or y =0Problem B: 7.2 3-6, p511diagram I II III IVequation 4 6 3 5• I is not 3 because the slopes in I depend on x (change under horizontal shifts); I is not 5 or6 because the slopes in I are horizontal for y =2• II is not 3, 4, or 5 because the slopes in II are horizontal for y = 0; II is not 3 also becausethe slopes in II depend on x• III is not 4, 5, or 6 because the slopes in III do not depend on x (do not change underhorizontal shifts)• IV is not 3, 4, or 6 because the slopes are not 0 for either y =2 or y =0 (at least on the y-axis,where x=0)Remark: The above justifications, for Problems A and B, contain 12 elimination statements: eachof the 4 diagrams is shown to be incompatible with 3 of the equations. Since you know that thematch is one-to-one, it is possible to fully justify the answer with just 6 elimination statements,provi ded they are chosen properly. For example, after you match diagram I with equation 4 inProblem B, you can forget about equation 4 when considering the remaining 3 diagrams.Solutions to Mini-Quiz 2:convergence/divergence of sequences/seriesDetermine whether each of the following sequences or series converges or not. In each case, clearlycircle either YES or NO, but not both.(a) the sequence an=(−1)n−1nn2+ 1YES NOSince an=(−1)nn·11 + 1/n2, this sequence conve rges to 0.(b) the sequence an= 1 + cos(2/n)YES NOSince cos(2/n) −→ cos 0 = 1, an−→ 2(c) the sequence an= n cos n YESNOSince cos n does not approach 0 as n −→0, |an| takes arbitrarily large values(d) the sequence an= (−1)nnn + 1YESNOSince an= (−1)n11+1/n, the odd terms approach -1, while the even terms approach 1.(e) the sequence an=sin 2n1 +√nYES NOSince |sin 2n| ≤ 1, while√n −→ ∞, an−→ 0(f) the series∞Xn=1n√n2+ 1YESNOn√n2+ 1looks liken√n2= 1:n√n2+ 1=1√n2+ 1/√n2=1p1 + 1/n2−→ 1 6= 0; so the seriesdiverges by the most important divergence test for series(g) the series∞Xn=14 + 3n2nYESNO4 + 3n2nlooks like3n2n=32nand the geometric series∞Xn=132ndiverges:(4 + 3n)/2n3n/2n=4 + 3n3n= 4/3n+ 1 −→ 1.Alternative ly,∞Xn=14 + 3n2n=∞Xn=142n+∞Xn=13n2nand the 1st series on RHS converges, while the seconddiverges, and so the series on LHS also diverges (but diverges+diverges would imply nothing!).Also, 0 ≤3n2n≤4 + 3n2nand the “smaller” series∞Xn=132ndiverges(h) the series∞Xn=17nn!YES NOThe power series∞Xn=1xnn!converges to exfor all x.You can also use the Ratio Test:|an+1||an|=7n+1/(n+1)7n/n!=7n+17n·n!(n+1)!= 7 ·1n+1−→ 0 < 1.(i) the series∞Xn=1sin n2nYES NOThe series∞Xn=112nconverges; since 0 ≤ |sin n|/2n≤ 1/2n, so does the series∞Xn=1|sin n|2nby Compar-ison Test, and thus so does the serie s∞Xn=1sin n2nby the Absolute Convergence Test.(j) the series∞Xn=1(−1)nnn2+ 1YES NOnn2+ 1=1n + 1/n−→ 0,1n + 1/n>1n+1 + 1/(n+1), becausen+1 + 1/(n+1) −n + 1/n> 1 − 1/n ≥ 0 ,and the series is alternating.Solutions to Mini-Quiz 3:approximating sums of infinite seriesExplain why each of the following series converges. Then estimate its sum to within 2−10using theminimal possible number of terms, justifying your estimate; leave your answer as a simple fractionp/q for some integers p and q with no common factor. Is your estimate an under- or over-estimatefor the sum? Explain why.(a)∞Xn=11n5converges by the p-Series Test, since here p = 5 > 1.All terms in this series are positive, and the Remainder Estimate for the Integral Test Theorem,on p583, should be used to estimate its sum with the required precision after all assumptions arechecked. Since f(x)=1/x5>0 is continuous and decreasing on [1, ∞),Z∞m+11x5dx <∞Xn=11n5−n=mXn=11n5=∞Xn=m+11n5<Z∞m1x5dxSinceZ∞m1x5dx =Z∞mx−5dx =1−4x−4∞m=14m−4=14m4,we find that14(m+1)4<∞Xn=m+11n5<14m4.Since we need the middle term to be at most 1/210= 1/45, by the second inequality m=4 works;by the first inequality m=3 would not work (with m = 3, the middle term above is strictly greaterthan 1/45=2−10). So the required estimate isn=mXn=11n5=n=4Xn=11n5=115+125+135+145=35· 45+ 35· 25+ 45+ 3535· 45=243 · 1024 + 243 · 32 + 1024 + 243243 · 1024=257, 875248, 832This is an under-estimate for the infinite sum, because the finite-sum estimate is obtained bydropping only positive terms from the infinite sum.Remark 1: You can justify convergence using the Integral Test, after checking the required assump-tions: since f(x)=1/x5>0 is continuous and decreasing on [1, ∞), the infinite series convergesbecauseZ∞11x5dx =Z∞1x−5dx =1−4x−4∞1=14converges.With this approach, you do not need to re-check the three assumptions before applying the Re-mainder Estimate for the Integral Test Theorem.Remark 2: If you were asked to estimate the sum within 1/1000, according to the book’s recipe youwould still need to take m = 4. This is the smallest value of m for which the book’s upper-boundon the remainder of the infinite series is no greater than the required precision (with m = 3, theupper-bound is 1/(4 ·34)=1/324 > 1/1000). In principle, the remainder is smaller than the upperbound, so that a smaller m could still work. But if the estimate is to be within 1/1000, m=3 stillcannot work because the lower bound for the estimate can be improved toam+1+Z∞m+21x5dx =1(m+1)5+14(m+2)4.For m=3, this is 881/640, 000 > 1/1000.(b)∞Xn=1(−1)nn5converges because this is an alternating