MATH 152, SPRING 2012 COMMON EXAM III - VERSION A Solutions Last Name: First Name: Signature: Section No: PART I: Multiple Choice (4 pts each) 1. What is the intersection of the sphere (x − 2)2 + (y − 1)2 + (z − 3)2 = 16 with the xy-plane? (a) The circle (x − 2)2 + (y − 1)2 = 16. (b) The circle (x − 2)2 + (y − 1)2 = 1. (c) The point (2, 1, 0). (d) The circle (x − 2)2 + (y − 1)2 = 7. Correct Answer (e) The sphere does not intersect the xy-plane. Solution: The sphere intersects the xy-plane when z = 0. Substituting z = 0 into the equation (x − 2)2 + (y − 1)2 + (z − 3)2 = 16 yields (x − 2)2 + (y − 1)2 + (0 − 3)2 = 16, hence (x − 2)2 + (y − 1)2 = 7. 2. If we apply the Ratio Test to the series ∞ (−4)n n2 P , which of the following statements is true? 5n+1 n=1 an+1 = − 4 < 1, therefore the series converges. (a) lim n→∞ an 5 an+1 4 = < 1, therefore the series converges. Correct Answer (b) lim n→∞ an 5 an+1 = 4 < 1, therefore the series converges. (c) lim n→∞ an 25 an+1 = ∞ > 1, therefore the series diverges. (d) lim n→∞ an an+1 = 0 < 1, therefore the series converges. (e) lim n→∞ an n+1 an+1 (−4)n (−4)(n + 1)2 5n+1 (n + 1)2 5n+1 = lim (−4) Solution: lim = lim n→∞ an n→∞ 5n+2 (−4)n (n2 ) n→∞ 5n+1 (5) (−4)n (n2 ) (−4)(n + 1)2 4 = < 1, therefore the series converges absolutely, therefore converges. = lim 5 n→∞ 5n2 3. What does it mean to say ∞ P an is convergent? n=1 (a) lim an = 0. n→∞ (b) lim an exists. n→∞ (c) There are numbers A and A′ such that A ≤ an ≤ A′ for all n. (d) There are numbers B and B ′ such that B ≤ a1 + a2 + ... + an ≤ B ′ for all n. (e) lim (a1 + a2 + ... + an ) exists. Correct Answer n→∞ Solution: The partial sum is sn = a1 + a2 + ... + an and ∞ P n=1 1 an = lim sn . n→∞ 4. Which of the following is a unit vector in the direction of b − 3a, where a = h1, −2, 2i and b = h5, 0, −3i? 6 9 2 (a) − , − , 11 11 11 2 6 9 Correct Answer (b) , ,− 11 11 11 (c) h2, 6, −9i 14 2 18 (d) − √ , −√ ,√ 524 524 524 2 18 14 ,√ , −√ (e) √ 524 524 524 Solution: b − 3a = h5, 0, −3i − 3 h1, −2, 2i = h2, 6, −9i. Divide by the magnitude to make this vector a unit vector. √ 9 2 6 h2, 6, −9i . = , ,− Since | h2, 6, −9i | = 4 + 36 + 81 = 11, u = 11 11 11 11 5. Represent 1 as a power series about 0 and give the interval of convergence. 1 + 4x2 (a) ∞ X 1 1 = (−4)n x2n , where |x| < . Correct Answer 1 + 4x2 2 n=0 (b) ∞ X 1 1 = (−4)n x2n , where |x| < . 1 + 4x2 4 n=0 (c) ∞ X 1 = (−4)n x2n , where |x| < 4. 1 + 4x2 n=0 (d) ∞ X 1 1 = 4n x2n , where |x| < . 1 + 4x2 2 n=0 (e) ∞ X 1 1 = 4n x2n , where |x| < . 1 + 4x2 4 n=0 Solution: ∞ ∞ X X 1 1 2 n = (−4x ) = (−4)n x2n , where | − 4x2 | < 1 , hence |x| < . 1 + 4x2 2 n=0 n=0 6. Find the cosine of the angle, θ, between the vectors a = h2, 0, 1i and b = h1, −2, 1i. 4 (a) cos(θ) = √ 30 5 (b) cos(θ) = √ 30 3 (c) cos(θ) = √ 12 5 (d) cos(θ) = √ 12 3 (e) cos(θ) = √ Correct Answer 30 Solution: cos(θ) = h2, 0, 1i · h1, −2, 1i 3 a·b = =√ |a||b| | h2, 0, 1i || h1, −2, 1i | 30 2 7. Find the 15th derivative at x = 0 for f (x) = (−1)n xn . n n=0 7 (n + 5) ∞ P (a) f (15) (0) = 0 15! 715 (20) 15! Correct Answer (c) f (15) (0) = − 15 7 (20) 15 (d) f (15) (0) = − 15 7 (20) 15 (e) f (15) (0) = 15 7 (20) (b) f (15) (0) = (−1)n 1 1 1 1 1 1 xn = − x + 2 x2 − 3 x3 + ... − 15 x15 + 16 x16 + .... Now, if n (n + 5) 7 5 7(6) 7 (7) 7 (8) 7 (20) 7 (21) n=0 the degree on x is less than 15, the 15th derivative of these terms will be zero. If the degree on x is greater than 15, the 15th derivative of these terms will still contain a power of x, and hence will equal 0 when x = 0. If the degree 15! on x is 15, then f (15) (x15 ) = 15! Thus f 15 (0) = − 15 7 (20) Solution: f (x) = ∞ P 8. Find the Taylor Polynomial of degree 2, T2 (x), for f (x) = 1 (x − 9)2 216 1 T2 (x) = 3 + (x − 9) − 6 1 T2 (x) = 3 + (x − 9) − 6 1 (x − 9)2 T2 (x) = − 108 None of these. √ x centered at a = 9. (a) T2 (x) = − (b) (c) (d) (e) 1 (x − 9)2 Correct Answer 216 1 (x − 9)2 108 Solution: T2 (x) = f (9) + f ′ (9)(x − 9) + f ′′ (9) (x − 9)2 . 2 1 1 (x − 9)2 . T2 (x) = 3 + (x − 9) − 6 216 9. The series ∞ P 2 ne−n n=1 (a) converges by the Integral Test. Correct Answer (b) diverges by the Ratio Test. (c) diverges by the Test for Divergence. (d) diverges by the p-series test. (e) converges by the Comparison Test with ∞ P 2 e−n . n=1 2 Solution: Since xe−x is a continuous, positive, decreasing function on [1, ∞), we may apply the integral test: Z ∞ ∞ 2 2 1 1 xe−x dx = − e−x = . Since the improper integral converges, so does the series. 2 2e 1 1 3 10. In three-dimensional space, R3 , the equation x2 + y 2 = 9 describes (a) a parabola. (b) a sphere. (c) a cylinder. Correct Answer (d) a circle. (e) a plane. Solution: In two-dimension, x2 + y 2 = 9 describes a circle. Thus, in three-dimension, as z varies, x2 + y 2 = 9 describes a cylinder. 11. The series ∞ (−1)n n P 2 n=1 n + 1 (a) is absolutely convergent. (b) is convergent but not absolutely convergent. Correct Answer (c) is divergent. (d) is absolutely divergent but not divergent. (e) converges absolutely by the ratio test. ∞ (−1)n n P n converges by the Alternating Series Test since the sequence decreases Solution: The series 2 n2 + 1 n=1 n + 1 ∞ P n for convergence. Using the Limit Comparison to zero. To test for absolute convergence, test the series 2 n=1 n + 1 n ∞ 1 ∞ ∞ (−1)n n 2+1 P P P 1 n an n = 1 > 0. Since Test, with bn = , lim = lim diverges, so does . Hence 2 2 1 n→∞ n n→∞ bn n=1 n n=1 n + 1 n=1 n + 1 n x is continuous, positive and decreasing is converget but not absolutely convergent. Another solution: Since 2 x +1 ∞ P n diverges: on [1, ∞), we may use the integral test to show 2+1 n n=1 Z ∞ ∞ 1 x 2 = ∞. dx = ln(x + 1) 1 x2 + 1 2 1 12. The series 1 √ is n+ e n n=1 ∞ P ∞ 1 P 1 1 √ and is a divergent geometric series. > n en + n en n=1 e ∞ 1 P 1 1 is a convergent geometric series. Correct Answer (b) convergent because n √ < n and n e + n e n=1 e ∞ P 1 1 1 √ is a convergent p series. (c) convergent because n √ < √ and e + n n n n=1 ∞ P 1 1 1 √ is a divergent p series. (d) divergent because n √ < √ and e + n n n n=1 ∞ P 1 1 1 √ is a divergent p series. (e) divergent because n √ > √ and e + n n n n=1 ∞ n X ∞ 1 P 1 1 1 1 = is a convergent geometric series (r = ), and n √ < n , it follows Solution: Since n e e e e e + n n=1 n=1 (a) divergent because that ∞ X 1 1 √ , thus by the comparison test, since the larger series converges, so does the smaller series. < n n n=1 en n=1 e + ∞ P 4 ∞ (−2)n P . n! n=0 13. Find the sum of the series (a) e2 (b) cos(−2) (c) sin(−2) (d) 0 (e) e−2 Correct Answer Solution: Since ex = ∞ (−2)n ∞ xn P P , it follows that e−2 = . n! n=0 n=0 n! PART II: WORK OUT (48 points total) Directions: Present your solutions in the space provided. Show all your work neatly and concisely and box your final answer. You will be graded not merely on the final answer, but also on the quality and correctness of the work leading up to it. 14. a.) (5 pts) Using the known Maclaurin series for sin x, find the Maclaurin series for sin(x3 ). ∞ ∞ X ∞ (−1)n x2n+1 P P (−1)n x6n+3 , thus sin(x3 ) = . Solution: sin x = n=0 n=0 (2n + 1)! n=0 (2n + 1)! sin(x3 ) − x3 . x→0 x9 b.) (5 pts) Using the Maclaurin series for sin(x3 ), compute lim (−1)n x6n+3 x9 x15 x21 3 − x3 x − + − + ... − x3 sin(x ) − x (2n + 1)! 3! 5! 7! = lim = lim Solution: lim x→0 x→0 x→0 x9 x9 x9 x6 x12 1 − + ... x9 − + 1 1 3! 5! 7! = lim =− =− x→0 x9 3! 6 3 3 2 15. (5 pts) Using the known Maclaurin series for ex , find a Maclaurin series for e−x . ∞ xn ∞ (−1)n x2n P P 2 Solution: ex = , thus e−x = . n! n=0 n! n=0 2 b.) (5 pts) Expand the Maclaurin series for e−x found above to degree 4. Use this 4th degree Maclaurin polynomial Z 1 2 e−x dx. Do not simplify. to estimate 0 ∞ (−1)n x2n P x4 ≈ 1 − x2 + . n! 2 n=0 1 Z 1 Z 1 x3 1 x5 1 23 x4 2 −x2 dx = x − + 1−x + Thus e dx ≈ = 1 − 3 + 10 = 30 . 2 3 10 0 0 0 2 Solution: e−x = 16. (10 pts) Find a power series about 0 for f (x) = ln(6x + 5). What is the associated radius of convergence? n+1 n X ∞ ∞ 6x 6x 6 d 6 6 6X = − = (−1)n xn , where < 1. Solution: (ln(6x + 5)) = = 6x dx 6x + 5 5 n=0 5 5 5 n=0 5 1+ 5 n+1 n+1 n+1 Z X ∞ ∞ X 6 6 x 5 (−1)n Thus, ln(6x + 5) = xn dx = C + (−1)n , where |x| < . Choosing x = 0 to 5 5 n + 1 6 n=0 n=0 n+1 n+1 ∞ P x 5 6 solve for C yields C = ln 5. Thus ln(6x + 5) = ln 5 + ,R= . (−1)n 5 n + 1 6 n=0 5 ∞ (2x − 1)n P . Be sure to test the endpoints for convergence. n4n n=1 an+1 (2x − 1)(n) 2x − 1 (2x − 1)n+1 n4n . Solution: Using the Ratio Test, lim = lim = lim = n→∞ an n→∞ (n + 1)4n+1 (2x − 1)n n→∞ (n + 1)4 4 ∞ an+1 P < 1, then the series The Ratio Test states if lim an converges. Hence n→∞ an n=1 2x − 1 < 1, so |2x − 1| < 4, thus x − 1 < 2. Solve this inequality for x: − 3 < x < 5 . 4 2 2 2 Test the endpoints of the interval for convergence : ∞ X ∞ (−4)n (−1)n 3 P = which converges by the Alternating Series Test since the sequence Test x = − : 2 n=1 n4n n n=1 1 decreases to zero. n ∞ X ∞ (4)n 1 5 P = which diverges by the p-series test. Thus the interval of convergence is Test x = : 2 n=1 n4n n n=1 17. (10 pts) Find the interval of convergence of the series − 3 5 ≤x< . 2 2 18. Consider the series ∞ (−1)n P . 2 n=1 2n + 1 a.) (4 pts) Prove the series is absolutely convergent. ∞ ∞ ∞ X ∞ 1 ∞ (−1)n P P 1 1 1X 1 = P = . Since is a convergent p-series, it follows Solution: < 2n2 + 1 2 2 2 2 2n 2 n=1 n n=1 2n + 1 n=1 n n=1 n=1 ∞ ∞ (−1)n ∞ P P P 1 1 converges, thus also converges by the comparison test. Hence is absolutely that 2 2 2 n=1 2n + 1 n=1 2n + 1 n=1 2n convergent. ∞ (−1)n P b.) (2 pts) Find s2 , the sum of the first 2 terms, to approximate . 2 n=1 2n + 1 Solution: S2 = 2 (−1)i P 1 1 2 =− + =− 2 3 9 9 i=1 2i + 1 c.) (2 pts) Find an upper bound on |R2 |, the absolute value of the remainder, in using s2 to approximate Solution: |R2 | ≤ |a3 |, where an = (−1)n 1 . Thus |R2 | ≤ . 2n2 + 1 19 6 ∞ (−1)n P . 2 n=1 2n + 1