Math 1152 Sample Midterm 3 AU24 PDF

Math 1152 - Sample Midterm 3 - AU24 - Page 2 1. Directions: Fill in the circle next to the correct response(s). 8x Q1: Multiple Choice The radius of convergence for the Taylor ser...

Math 1152 - Sample Midterm 3 - AU24 - Page 2 1. Directions: Fill in the circle next to the correct response(s). 8x Q1: Multiple Choice The radius of convergence for the Taylor series centered at x = 0 for f (x) = is: 2+x 1 1 A. 8 B. 4 C. 2 D. E. F. ∞ G. None of these 2 8 Q2: Multiple Choice The third degree Taylor polynomial centered at x = 0 for f (x) = x cos(2x) is: 2 1 5 A. x − 2x3 B. 2x − x3 C. x − 2x3 + x5 D. 2x − x3 + x E. None of these 3 12 Q3: MultiSelect Fill in the circle next to each differential equation below which is separable: y′ A y ′ = e2x−3y B = 4x2 C y ′ = y 2 + 2x3 D y ′ − x ln(x) = yy ′ E xy ′ + xy = ex y ∞ X f (x) Q4: Multiple Choice If f (x) = kxk , find lim. x→0 2x + 3x2 k=2 1 1 2 A. 0 B. C. D. E. 1 F. The limit does not exist. G. None of these 3 2 3  Math 1152 - Sample Midterm 3 - AU24 - Page 3 Q5 :Multiple Choice Direction fields A, B, C and D are shown below. dy (i) Which of these is the direction field for the differential equation = x(1 − y 2 ) ? dx A. Field A B. Field B C Field C D Field D E None of these dy (x + y + 2) (ii) Which of these is the direction field for the differential equation = ? dx 2 A. Field A B. Field B C Field C D Field D E None of these Q6 : Multiple Choice Fill in the circle next to the best response. Consider the initial value problem y ′ (x) = y 2 esin(2x) , y(0) = 2. Calculate y ′′ (0). A. 8 B. 12 C 16 D 20 E 24 F. 28 G. None of these  Math 1152 - Sample Midterm 3 - AU24 - Page 4 Q7: True or False Determine whether each statement below is true or false. T F Differentiating a Taylor series could change its interval of convergence but not its radius of convergence. ∞ ∞ 1 X 1 X T F Since = xk , it follows that = 3k. 1−x 1−3 k=0 k=0 T F If p4 (x) = 2(x − 1)2 + 4(x − 1)4 is the fourth degree Taylor polynomial centered at x = 1 for an infinitely differentiable function f (x), then f ′′′ (1) = 0. ∞ ∞ " # d 2 X xk+1 X T F x = 2x xk dx k+1 k=0 k=0 e T F The second degree Taylor polynomial centered at x = 1 for ex is e + e(x − 1) + (x − 1)2 2 ∞ X (−1)k x2k T F Since cos(x) = , we find the Taylor series centered at x = 0 for x cos(2x) as follows: (2k)! k=0 ∞ ∞ ∞ X (−1)k X (−1)k X (−1)k x cos(2x) = x · (2x)2k = (2x2 )2k = 2k x2k (2k)! (2k)! (2k)! k=0 k=0 k=0 ∞ X x3k+1 T F If f (x) = , then f (82) (0) = 0 2k + 7 k=0  Math 1152 - Sample Midterm 3 - AU24 - Page 5 ∞ X (−1)k 2. [15 pts] Consider the series √. k=1 k ∞ X (−1)k A. Explain why √ converges. State any convergence test(s) you use, explain why they apply, and show any k=1 k necessary computations. N X (−1)k B. Use the bound from the Alternating Series Test to find the smallest integer N needed to ensure that √ is k=1 k ∞ X (−1)k within.01 of the exact value for √. Explain your response! k=1 k  Math 1152 - Sample Midterm 3 - AU24 - Page 6 √ 3. Find the third degree Taylor Polynomial centered at x = 1 for f (x) = 2x − 1. ∞ X ∞ X 4. Suppose that f (x) = ak (x + 4)k and it is known that 3k ak converges but the series represented by f (8) k=0 k=0 diverges. A. [6 pts] Indicate whether the following series must converge, must diverge, or could converge or diverge by filling in the bubble to the immediate left of your choice. No justification is necessary. i The series represented by f (3): must converge must diverge could converge or diverge. ∞ X ii The series 4k ak : must converge must diverge could converge or diverge. k=0 iii The series represented by f (−2): must converge must diverge could converge or diverge. ∞ X B. [3 pts] State maximum possible interval of convergence for the series f (x) = ak (x + 4)k. k=0  Math 1152 - Sample Midterm 3 - AU24 - Page 7 ∞ X (−1)k 5. Let f (x) = (x − 3)2k k · 9k k=1 A. Determine the radius of convergence and determine the interval of convergence. Remember to check the endpoints. Show your work. B. Does f ′′ (5) converge? Justify your answer. C. Write out the 6th Taylor polynomial for f centered at x = 3. Use this to determine f ′ (3), and f ′′ (3) D. Determine f (26) (3).  Math 1152 - Sample Midterm 3 - AU24 - Page 8 ∞ X (−1)k 2k+1 6. Given that the Taylor series centered at x = 0 for arctan(x) is x , answer the following. 2k + 1 k=0 10x arctan(x) A. Calculate lim or explain why the limit does not exist x→0 2x2 + 3x3  x B. Let f (x) = 3t3 arctan(t) dt. Calculate the seventh degree Taylor polynomial for f (x) centered at x = 0 and 0 simplify your final answer.  Math 1152 - Sample Midterm 3 - AU24 - Page 9 ∞ 1 X 7. Given that the Taylor series for centered at x = 0 is (k + 1)xk , find the fifth degree Taylor polynomial (1 − x)2 k=0 3x3 centered at x = 0 for. Simplify your final answer. (1 + 2x)3  Math 1152 - Sample Midterm 3 - AU24 - Page 10 dy 6y 8. Consider the initial value problem: = 2 , y(0) = b. dx x − 2x − 8 a. Sketch the solutions to the initial value problem when b = 1/2 and b = −1 on the direction field below. From the direction field, does it appear that lim y(x) depends on b? YES NO x→−2+ dy 6y b. Find the solution to the initial value problem = 2 , y(0) = b. In your final answer, dx x − 2x − 8 explicitly solve for y as a function of x and leave your answer in terms of b.  Math 1152 - Sample Midterm 3 - AU24 - Page 11 Problem 8 continued... c. To verify your conjecture from Part I, calculate lim y(x) and explain why the limit either does or does not x→−2+ depend on b. Math 1152 - Sample Midterm 3 - AU24 - Page 12 - - - - Extra Workspace - - - -