Student s Printed Name: Instructor: CUID: Section: Instructions: You are not permitted to use a calculator on any portion of this test. You are not allowed to use any textbook, notes, cell phone, laptop, tablet, SMART watch, or any technology on any portion of this test. All devices must be turned off while you are in the testing room. During this test, any communication with any person (other than the instructor or a designated proctor) in any form, including written, signed, verbal, or digital, is understood to be a violation of academic integrity. No part of this test may be removed from the examination room. Read each question carefully. In order to receive full credit for the free response portion of the test, you must:. Show legible and logical (relevant) justification that supports your final answer. 2. Use complete and correct mathematical notation. 3. Include proper units, if necessary. 4. Give exact numerical values whenever possible. You have 90 minutes to complete the entire test. On my honor, I have neither given nor received inappropriate or unauthorized information at any time before or during this test. Student s Signature: Do not write below this line. Free Response Possible Points Free Response Possible Points Problem Points Earned Problem Points Earned. 5(a). 5 2. 5(b). 5 3. 6. 4(a). 9 Free Response 69 4(b). 6 Multiple Choice 3 Test Total 00 - Page of 6
Multiple Choice: There are 3 multiple choice questions. They do NOT all have the same point value. Each question has one correct answer. The multiple choice problems will count for 3% of the total grade. Use a number 2 pencil and bubble in the letter of your response on the scantron sheet for problems - 3. For your own record, also circle your choice on your test since the scantron will not be returned to you. Only the responses recorded on your scantron sheet will be graded.. (3 pts.) Find the power series representation for f (x) if f(x) = ( ) n x 4n. (2n)! (a) f (x) = ( ) n+ x 4n+ (2n)!(4n + )(n + ) (c) f (x) = ( ) n x 4n+ (2n)!(4n + ) (b) f ( ) n 2nx 4n (x) = (2n )! (d) f (x) = ( ) n 2x 4n (2n )! Answer: (d) 2. (3 pts.) Determine the third degree term in the Taylor series for f(x) = /x 2 centered at a =. (a) 24(x + ) 3 (c) 4(x + ) 3 (b) 6(x + ) 3 (d) 8(x + ) 3 Answer: (c) - Page 2 of 6
3. (3 pts.) The Maclaurin series for f(x) = arctan x is arctan x = ( ) n x 2n+, for x. 2n + Use the Maclaurin series for arctan x to determine the Maclaurin series for x 3 arctan(x 2 ). (a) ( ) n x 4n+5 2n + (c) ( ) n x 2n+4 2n + (b) ( ) n x 4n+8 2n + (d) ( ) n x 4n+4 2n + Answer: (a) 4. ( pt.) Determine whether the series is absolutely convergent, conditionally convergent, or divergent. (a) The series is absolutely convergent. ( ) n n 2/3 (c) The series is divergent. (b) The series is conditionally convergent. Answer: (b) - Page 3 of 6
5. (3 pts.) The Maclaurin series for f(x) = xe x3 is ( ) n x 3n+ xe x3 =, for < x <. n! Use the Maclaurin series for xe x3 to evaluate xe x3 dx as a power series. (a) C + ( ) n x 3n+2 n!(3n + 2) (c) C + ( ) n+ x 3n+2 (n + )!(3n + 2) ( ) n (3n + )x 3n (b) C + n!(3n) ( ) n (3n + )x 3n (d) C + n! Answer: (a) 6. ( pt.) Determine whether the series convergent, or divergent. ( ) n+ n 2 3n 2 + is absolutely convergent, conditionally (a) The series is absolutely convergent. (c) The series is divergent. (b) The series is conditionally convergent. Answer: (c) - Page 4 of 6
7. (3 pts.) Using the Comparison Test, (a) Diverges because 5 n 3 n 4 : 5 n 3 n 4 < 5n 3 n and 5 n 3 n diverges. (b) Diverges because 5 n 3 n 4 > 5n 3 n and 5 n 3 n diverges. (c) Converges because 5 n 3 n 4 < 5n 3 n and 5 n 3 n converges. (d) Converges because Answer: (b) 5 n 3 n 4 > 5n 3 n and 5 n 3 n converges. 8. ( pt.) Determine whether the series convergent, or divergent. (a) The series is absolutely convergent. ( 3 n ( ) 4) n is absolutely convergent, conditionally (c) The series is divergent. (b) The series is conditionally convergent. Answer: (a) - Page 5 of 6
9. (3 pts.) Consider the series series, we find that: ( 3n ( ) n+ 2 ) n + 4n 2n 2. By applying the Root Test to this + (a) The series converges absolutely. (c) The Root Test is inconclusive. (b) The series converges conditionally. (d) The series diverges. Answer: (d) ( ) n+ 0. (3 pts.) Consider the convergent series n 2. If the first five terms are used to estimate + 3 the value of the sum, then the bound on the absolute error given by the Alternating Series Estimation Theorem is: (a) 52 (c) 39 (b) 28 (d) 9 Answer: (c) - Page 6 of 6
. (3 pts.) Which of the following is the graph of the curve given by y 2 4x 2 + 24x + 2y = 39? (a) (c) (b) (d) Answer: (b) - Page 7 of 6
2. ( pt.) Determine whether the series sin(2n) is absolutely convergent, conditionally con- + n3 vergent, or divergent. (a) The series is absolutely convergent. (c) The series is divergent. (b) The series is conditionally convergent. Answer: (a) 3. (3 pts.) If 0 x 2π/3, use Taylor s Inequality to find the best estimate of the maximum 3 error in the approximation sin x T 2 (x), where T 2 (x) = 2 + ( x π ) 3 ( x π ) 2 is the 2 3 4 3 second-order Taylor polynomial of sin x at π/3. (a) π2 9 (c) π3 27 (b) π2 36 Answer: (d) (d) π 3 62 - Page 8 of 6
Free Response. The Free Response questions will count for 69% of the total grade. Read each question carefully. To receive full credit, you must show legible, logical, and relevant justification which supports your final answer. For problems involving analysis of convergence or divergence of series, remember to include: the name of the convergence test used, work to show that the test conditions have been met, and a conclusion statement about convergence or divergence of the series.. ( pts.) Determine whether the following series converges or diverges: Solution: ( ) n+ n 2n 3/2 + is an alternating series with b n = We try the Alternating Series Test. n 2n 3/2 + > 0. ( ) n+ n 2n 3/2 +. Let f(x) = x 2x 3/2 +. Then f (x) = (2x3/2 + )() x(3x /2 ) (2x 3/2 + ) 2 = 2x3/2 + 3x 3/2 (2x 3/2 + ) 2 = x3/2 (2x 3/2 + ) 2. f (x) = 0 when x =, and f (x) < 0 for x >. So f(x) is decreasing for x >, meaning {b n } is decreasing for n >. 2. lim n So n 2n 3/2 + = lim n ( ) n+ n 2n 3/2 + n /2 2 + n 3/2 = 0 converges by the Alternating Series Test. Work on Problem: Defines f(x) Finds f (x) States f (x) < 0 for x > Concludes b n is decreasing for n > Shows the limit of b n as n is 0 for setting up the limit of correct b n 3 points for evaluating the limit with supporting work Concludes the series converges by Alternating Series Test ( for test name, for converges) Notes: Deduct 0.5 points for algebra or notation errors. (with a max. deduction of for notation errors) Points 0.5 points.5 points 4 points - Page 9 of 6
2. ( pts.) Determine whether the following series converges or diverges: Solution: lim 4 n+ (n + )! n 2 (n + ) 2 4 n n! = lim n n = lim n = lim n Since lim a n+ n a n >, the series diverges by the Ratio Test. 4 4 n (n + ) n!n 2 (n + ) (n + ) 4 n n! 4n 2 n + 4n + /n = > 4 n n! n 2 Work on Problem: Considers lim a n+ n a n for the a n in the problem Simplifies the limit Evaluates the limit Notes that the value of the limit is > Concludes that the series diverges by the Ratio Test ( for test name, for diverges) Notes: Deduct 0.5 points for each notation error. (with a max. deduction of for notation errors) Points 4 points 3. ( pts.) Determine whether the following series converges or diverges: Solution: Since the terms of the series are positive, we try a comparison test. (n + ) 2 Note that 4n 5 + = n 2 + 2n + 4n 5. + We try the Limit Comparison Test and compare the given series to the p-series test since p = 3 >. So the series lim n n 2 + 2n + 4n 5 + n 3 n 5 + 2n 4 + n 3 = lim n 4n 5 + + 2/n + /n 2 = lim n 4 + /n 5 =, where 0 < /4 < 4 (n + ) 2 4n 5 +, which converges by n3 (n + ) 2 4n 5 + converges by the Limit Comparison Test. - Page 0 of 6
Work on Problem: States an appropriate comparison series States the comparison series converges with justification Sets up the limit for the Limit Comparison Test Simplifies the limit Evaluates the limit Observes that the limit is finite and > 0 Concludes that the series converges by the Limit Comparison Test ( for test name, for converges) Notes: Deduct 0.5 points for algebra or notation errors. Points - Page of 6
4. Consider the power series ( ) n (x 5) n n3 n. (a) (9 pts.) Determine the radius of convergence, R, for the power series. Solution: We use the Ratio Test. (x 5) n+ (n + )3 lim n+ n (x 5) n = lim (x 5) n+ n3 n n (n + )3 n+ (x 5) n n3 n = lim (x 5) (x 5) n n 3 n n (n + )3 3 n (x 5) n By the Ratio Test, the series converges when So R = 3. Work on Problem: Considers lim a n+ n a n Simplifies the limit Evaluates the limit n = lim x 5 n 3n + 3 x 5 = 3 x 5 3 <, which implies x 5 < 3. Concludes that the series converges when limit value is < by the Ratio Test States the radius of convergence Notes: Deduct 0.5 points for algebra or notation errors. (with a max. deduction of for notation errors) Points 3 points - Page 2 of 6
(b) (6 pts.) Determine the interval of convergence for the power series. Hint: Don t forget to check for convergence at the endpoints. Solution: We need to check convergence of the series at x = 5 3 = 2 and at x = 5 + 3 = 8. When x = 2, the series is ( ) n ( 3) n n3 n = ( ) 2n 3 n n 3 n = This series diverges by the p-series test since p = (or diverges because it is the harmonic series). So x = 2 is NOT included in the interval of convergence. When x = 8, the series is ( ) n (3) n n3 n = ( ) n 3 n n 3 n = n ( ) n This is an alternating series with b n = /n > 0 so we try the Alternating Series Test.. For n, n + > n = /(n + ) < /n. So b n+ b n. 2. lim n n = 0. ( ) n Since the conditions for the Alternating Series Test are met, converges by the n ( ) n Alternating Series Test. (or we can say converges because it is the alternating n harmonic series). So x = 8 is included in the interval of convergence. So the interval of convergence is (2, 8]. n Work on Problem: Considers the series at x = 5 R States series diverges, with justification Considers the series at x = 5 + R States series converges, with justification States the interval of convergence Notes: Deduct 0.5 points for algebra or notation errors. (with a max. deduction of for notation errors) Points.5 points.5 points - Page 3 of 6
5. Consider the curve given by the following parametric equations: x = 2 sin t, y = 2 cos t, 0 t π/2. (a) (5 pts.) Eliminate the parameter to find a Cartesian equation of the curve. Be sure to include any necessary restrictions. Solution: ( x ) 2 ( y ) 2 We know sin 2 t + cos 2 t =. Since x/2 = sin t and y/2 = cos t, we have + =, 2 2 or x 2 + y 2 = 4. But note that for 0 t π/2, 0 sin t and 0 cos t so the circle will only include the portion with 0 x 2 and 0 y 2. So a Cartesian equation of the curve is x 2 + y 2 = 4, for 0 x 2 and 0 y 2. Work on Problem: Uses sin 2 t + cos 2 t = States correct Cartesian equation Includes appropriate interval restriction Notes: Deduct max of 0.5 points for notation errors. Points 3 points (b) (5 pts.) Sketch the curve and indicate with an arrow the direction in which the curve is traced as the parameter increases. Include the coordinates of the initial and terminal points with the appropriate value of t in the blanks below. Coordinates of initial point: (0, 2) at t = 0 Coordinates of terminal point: (2, 0) at t = π/2 Work on Problem: Graph is correct portion of the circle Arrows indicating direction Coordinates of initial and terminal points Points 3 points - Page 4 of 6
6. ( pts.) Find the 3rd-order Taylor polynomial, T 3 (x), of the function f(x) = ln(x 2) centered at a = 3. Solution: n f (n) (x) f (n) (3) 0 f(x) = ln x 2 f(3) = 0 f (x) = x 2 f (3) = 2 f (x) = (x 2) 2 f (3) = 3 f (x) = 2 (x 2) 3 f (3) = 2 T 3 (x) = (x 3) + 2! (x 3)2 + 2 (x 3)3 3! Work on Problem: Finds f (x), f (x), f (x) Evaluates f(x), f (x), f (x), f (x) at x = 3 States T 3 (x) (can be unsimplified) per term in T 3 (x) Notes: Deduct 0.5 points for algebra or notation errors. (with a max. deduction of for notation errors) Points 3 points 6 points - Page 5 of 6
Scantron: Check to make sure your Scantron form meets the following criteria: My Scantron: is bubbled with firm marks so that the form can be machine read; is not damaged and has no stray marks (the form can be machine read); has 3 bubbled in answers; has MATH 080 and my Section number written at the top; has my Instructor s last name written at the top; has Test No. 3 written at the top; has the correct test version written at the top and bubbled in below my XID; shows my correct XID both written and bubbled in. **Bubble a zero for the leading C in your XID**. - Page 6 of 6