MTH 142 Practice Problems for Exam 3 -Spring 2004Last changed: April 1, 2004, 9:00 a.mSections 9.1, 9.2, 9.3, 9.4, 10.1, 10.2, 10.3, 10.4, 10.51. Obtain the first three nonzero terms of the Taylor series of f(x) =√x about 4.2. Obtain P3(x) = the Taylor polynomial of order 3 of tan x about a = π/43. Calculate the radius of convergence of the power series∞Xn=0n3n(x − 2)n.4. Given that R = 2 is the radius of convergence of the series∞Xn=11n22n(x + 1)n, find theinterval of convergence (include endpoint analysis).5. Calculate the radius of convergence of the power series∞Xn=0n + 1n + 3xn.6. Given that R = 1 is the radius of convergence of the series∞Xn=1nn2+ 1xn, find the intervalof convergence (include endpoint analysis).7. A certain amount of fresh water shrimp is placed in a tank together with 2 lbs. of food,at 12:00 noon on January 1st, and 2 lbs. of food are added to the tank at noontime everyday after Jan 1. After every 24 hours, 15% of the food either decomposes or is eaten.a) How much food is in the tank right after 12:00 noon on January 20 th? Give details,and explain how you arrived at your answer.b) In the long run, how much food is in the tank right after noontime? (Section 9.1)8. The degree 2 Taylor polynomial approximation of f(x) = ln(1 + x) for x near 0 is P2(x) =x −12x2. We wish to approximate ln(1.5) by P2(x).a) What should x be? What is the error? (use your calculator)b) Find a bound for the error using the formula studied in class.9. a) Use the me thod of substitution to find the first 4 nonzero terms of the Taylor series ofthe function f(x) =1√1+x2about x = 0.b) Use series to answer the following question: For values of x that are close to 0, whichfunction is larger, cos(x) or1√1+x2?10. Calculate the order 3 Fourier polynomial of the 2π periodic function given on [−π, π) byf(t) =(−0.5 if − π ≤ t < 01 if 0 ≤ t < π11. Calculate the order 3 Fourier polynomial of the function given on [−π, π) by f(t) = 3t.Use the suggested method to determine if the series converges12.∞Xn=2n21 + n3(a) Comparison Test. (b) Integral Test.13.∞Xn=11 +1n(a) comparison test. (b) integral test14.∞Xn=3(−2)n+1πn(a) Geometric series. (b) alternating series test. (c) ratio test.SOLUTION MTH142 Practice Problems for Exam 21.T (x) = 2 +14(x − 4) −164(x − 4)2+1512(x − 4)3+ ···2.P3(x) = 1 + 2(x −π4) + 2(x −π4)2+83(x −π4)33. Using the ratio test we haveL = limn→∞|an+1||an|= limn→∞(n + 1)3n+1(x − 2)n+1n3n(x − 2)n= limn→∞(n + 1)3|x − 2|n= 3|x − 2|Now L < 1 when 3|x − 2| < 1, that is, |x − 2| <13. Then the radius of convergence is13.4. Given that R = 2 is the radius of convergence, we note that the “base point” is a =−1, hence the interval of convergence goes from -3 to 1. We now analyze convergenceat the endpoints. (a) Substituting x = −3 into∞Xn=01n22n(x + 1)n, we have the series∞Xn=11n22n(−2)n=∞Xn=1(−1)nn2The absolute convergence test guarantees the convergenceof the series provided the seriesP∞n=11n2converges, which does converge (p-series, withp = 2). Hence x = −3 belongs to the interval of convergence. (b) Now substitute x = 1into∞Xn=01n22n(x + 1)n, to obtain the series∞Xn=11n22n(2)n=∞Xn=11n2which converges (p = 2again). Conclusion: the interval of convergence is −3 ≤ x ≤ 1 (or, in interval notation,[−3, 1].)5. Using the ratio test we haveL = limn→∞|an+1||an|= limn→∞n+2n+4xn+1n+1n+3xn= limn→∞(n + 2)(n + 3)|x|(n + 1)(n + 4)= limn→∞(n2+ 5n + 6)|x|n2+ 5n + 4= 1|x|Now L < 1 when |x| < 1. Then the radius of convergence is 1.6. Given that R = 1 is the radius of convergence, we note that the “base point” is a = 0, hencethe interval of convergence goes from -1 to 1. We now analyze convergence at the endpoints.Substituting x = −1 into∞Xn=0nn2+ 1xn, we obtain the series∞Xn=1nn2+ 1(−1)nwhich is analternating series. Since112+1≥222+1≥332+1≥ ··· and also limn→∞nn2+1= 0, the seriesconverges. Therefore x = −1 is part of the interval of convergence. To test the otherendpoint, substitute x = 1 into the power series to obtain∞Xn=1nn2+ 1(1)n=∞Xn=1nn2+ 1This is a divergent series, since f(x) =xx2+1is decreasing and positive, we may apply theintegral test for series: we get:R∞1xx2+1dx = limb→∞Rb1xx2+1dx = limb→∞12ln |x2+ 1|b1=limb→∞12ln |b2+1|−12ln |2| = ∞ Therefore x = 1 is not part of the interval of convergence.We conclude that the interval of converge is −1 ≤ x < 1, or in interval notation, [−1, 1).7. The following table is helpful:day amount right after 12:00 noonJan 1 2Jan 2 2 + (0.85) 2Jan 3 2 + (0.85) 2 + (0.85)22Jan 4 2 + (0.85) 2 + (0.85)22 + (0.85)32......Jan 20 2 + (0.85) 2 + (0.85)22 + ··· + (0.85)192Then, the amount right after noontime on January 20th is (note that there are 20 termsin the left-hand-side of equation):2 + (0.85) 2 + (0.85)22 + ··· + (0.85)192 =2(1 − (0.85)20)1 − 0.85≈ 12.8165In the long run, the amount of food after noontime is21−0.85≈ 13.33338. a) Take x = 0.5, so the error is E = f(0.5) − P2(0.5) = 0.4055 − 0.3750 = 0.0305b) The function |f(3)(t)| = |2/(1 + t)3| attains its maximum in the interval 0 ≤ t ≤ 0.5 att = 0. The maximum is M = |f(3)(0)| = 2. The error bound when approximating f(0.5)by P2(0.5) is |f(0.5 −P2(0.5)| ≤ M(0.5)33!≈ 0.041669. a) Begin from the binomial series with exponent p = −1/2:(1+y)−1/2= 1+−12y+−12(−12− 1)2!y2+−12(−12− 1)(−12− 2)3!y3+··· = 1−y2+3 y28−5 y316+···, −1 < y < 1.Substitute y = t2to obtain1√1 + t2= 1 −t22+3 t48−5 t616+ ···The expansion given above is valid on −1 < t < 1.b) Since cos(t) = 1 −t22+t424−t6720+ ··· and1√1+t2= 1 −t22+3 t48−5 t616+ ··· we see thatcos(t) <1√1+t2. Reason: in the series expansions, the first terms that are different are thecoefficients of t2. Note that the coefficient of t2in the expansion of cos(t) is smaller thanthe coefficient of t2in the expansion of1√1+t2. (For small t, smaller powers dominate).10. Short answer: F3(x) =14+3πsin(t) +1πsin(3t).11. Short answer: F3(t) = 6 sin(t) − 3 sin(2t) + 2 sin(3t)12. (a) Comparison test: Note thatn21+n3behaves like1nwhen n is large, so we suspect thatthe series diverges. The following inequalities are clearly valid:0 ≤n2n3+ n3≤n21 + n3, n = 2, 3, . . .The term in the center simplifies to12n.