MATH 152 Exam III - - Spring 1996
Part I - - Multiple Choice
No calculators. No partial credit.
1. The series infinity ----- n \ (-1) ) ----- / 1/4 ----- n n = 1 is a) harmonic b) divergent c) absolutely convergent d) geometric e) conditionally convergent. SOLUTION: e) The terms 1/n^(1/4) monotonically converge to zero; so this series converges by the alternating series test. This series does not converge absolutely because the exponent 1/4 is less than or equal to one (the integral test implies that the series involving 1/n^p converges for p>1). 2. The curve parametrized by x= e^t y=e^{-t} for 0<t<1, may be written as: a) y = e^x for 0<x<1 b) y = 1/x for 1<x<e c) y= -x for 1<x<e d) y = e^{-x)$ for 1<x<e e) y = 1/x, for 0<x<1 SOLUTION: b) Note that y = e^(-t) = 1/e^t = 1/x and that when t=0, x = 1 and when t = 1, x= e. 3. The function f(x) defined by the series infinity ----- (2 n) \ x f(x) := ) ------ / (2 n)! ----- n = 0 satisfies which of the following differential equations? a) f'' = - 2 f b) f'' = - f c) f'' = 0 d) f'' = f e) f'' = 2 f SOLUTION: d) The second derivative of x^(2n)/(2n)! is (2n)(2n-1)x^(2n-2) / (2n)!. = x^(2n-2)/(2n-2)! Since (2n)! = (2n)(2n-1)(2n-2)! = x^(2m)/(2m)! by letting m = n-1 So the second derivative of f equals infinity ----- (2 m) \ x ) ------ / (2 m)! ----- m = 0 which is another way of writing the function f. 5. The series infinity ----- \ ) (arctan(n + 1) - arctan(n)) / ----- n = 1 equals a) Pi/8 b) Pi/4 c) Pi/2 d) Pi e) does not converge SOLUTION: b) This series telescopes to arctan(infinity) - arctan(1) (or more precisely to the limit of arctan(N) - arctan(1) as N -> infinity). Since arctan(infinity) = Pi/2 and arctan(1) = Pi/4, the answer is Pi/2 - Pi/4 = Pi/4. 6. Find the radius of convergence of the series > Sum((x-4)^n/n^n,n=1..infinity); infinity ----- n \ (x - 4) ) -------- / n ----- n n = 1 a) 0 b) 1/2 c) 1 d) 2 e) infinity SOLUTION: e) Use the root test - - the nth root of the nth term is |x-4|/n which converges to zero for any value of x. Therefore the root test implies that this series converges for any value of x. In other words, the radius of convergence is infinity. 7. Find the coefficient of (x-2)^2 in the Taylor series expansion of g(x) = 1/x about x=2. a) 1/2 b) 1/4 c) 1/8 d) 1/16 e) 1/32 SOLUTION: c) The coefficient is g''(2)/2; since g''(x) = 2/x^3 inserting x = 2 yields g''(2) = 2/8 = 1/4. Therefore g''(2)/2 = 1/8. 8. The line tangent to the curve x = t^3-t y = t^2 at the point (6,4) has slope a) 8/107 b) 4/11 c) 2/3 d) 3/2 e) 47 SOLUTION: b) The point (6,4) corresponds to t = 2. The slope of the tangent at (6,4) is therefore equal to y'(2)/x'(2); since x' = 2t and y' = 3t^2 - 1, the slope is 4/11. 9. Find the coefficient of x^3 in the Maclaurin series expansion (Taylor series about x=0) of > g(x) := (1-x^2/2)*exp(x); 2 g(x) := (1 - 1/2 x ) exp(x) a) -1/3 b) 0 c) 1/2 d) 1 e) 3/2 SOLUTION: a) Expanding e^x, we get g(x) = (1 - x^2/2) (1 + x + x^2/2! + x^3/3! + ...) The x^3 term of this product equals x^3/3! - x^3/2 = -x^3/3. Therefore the coefficient of x^3 is -1/3. 10. On which of the following intervals does the series infinity ----- \ n ) (x - 4) / ----- n = 1 converge? a) (-1,1) b) (0,8) c) (3,5) d) (-1/2, 1/2) e) (-infinity, infinity) SOLUTION: c) This is a geometric series, which converges provided |x-4| < 1 or another words, 3 < x < 5. 11. Suppose that infinity ----- \ ) a_n / ----- n = 1 infinity ----- \ ) b_n / ----- n = 1 are series with positive terms such that a_n/b_n --> 0 ans n - -> infinity. Which statement is true? a) If infinity ----- \ a) If ) b_n converges then / ----- n = 1 infinity ----- \ ) a_n converges / ----- n = 1 infinity ----- \ b) If ) b_n diverges, then / ----- n = 1 infinity ----- \ ) a_n diverges / ----- n = 1 infinity ----- \ c) If ) a_n converges, then / ----- n = 1 infinity ----- \ ) b_n diverges / ----- n = 1 infinity ----- \ d) If ) b_n converges, then / ----- n = 1 infinity ----- \ ) a_n diverges. / ----- n = 1 infinity ----- e) If \ ) a_n converges, then / ----- n = 1 infinity ----- \ ) b_n converges. / ----- n = 1 SOLUTION: a) If a_n/b_n --> 0, then eventually, the a_n are all smaller than the b_n. This means that if the series involving b_n converges, then the series involving a_n must also converge.
Part II - - Work Out Problems
Calculators are permitted. Partial credit may be given.
12. Find the Maclaurin series for 1/(1+x^5). Use this series to approximate .1 / | 66 | ------ dx | 5 / 1 + x 0 SOLUTION: The Maclaurin series is infinity ----- \ n (5 n) ) (-1) x / ----- n = 0 using the formula for a geometric series. To approximate the integral, the idea is to break up this series into a sum of the first N terms and a remainder which is a sum from N+1 to infinity. Then we integrate the first N terms (which is a polynomial). The trick is to find a convenient value of N so that the integral of the remainder is less than 10^(-14). Now the remainder is a geometric series whose value is (5 N + 5) x 66 ---------- 5 1 + x which is less than 66 x^(5N+5). Integrating this over 0 to 0.1, gives 66 10^(-5N-6)/(5N+6). By trial and error, we see that N=2 gives (66/16)(10)^(-16) which is less than 10^(-14). Therefore, the approximate value for the integral is obtained by computing 1/10 / | 5 10 | 66 - 66 x + 66 x dx | / 0 which is 6.59998900006. 13. Find the radius of convergence of the series infinity ----- (2 n) \ x ) ------ / n ----- 2 n = 1 SOLUTION: Write the series as infinity ----- \ 2 n ) (1/2 x ) / ----- n = 1 This is a geometric series which converges when x^2/2 < 1, or x^2 < 2 which is equivalent to -sqrt(2) < x < sqrt(2). So the radius of convergence is sqrt(2). 14. Determine whether the series infinity ----- \ 1 ) ---------- / 2 n ----- n + (-1) n = 2 converges. Justify your answer. SOLUTION: This series converges by comparison with the series infinity ----- \ 1 ) ---- / 2 ----- n n = 2 which converges by the integral test. 15. What is the smallest value of n for which we can be sure that the nth partial sum n ----- \ / 5 \ ) |----| / | 6 | ----- \ k / k = 1 approximates infinity ----- \ / 5 \ ) |----| / | 6 | ----- \ k / k = 1 to within 10^(-10)? HINT: Use the integral test to bound the remainder. SOLUTION: By the integral test, the remainder infinity ----- \ / 5 \ ) |----| / | 6 | ----- \ k / k = n + 1 is less than infinity / | 5 | ---- dx | 6 / x n which is 1 ---- 5 n This remainder is within 10^(-10) provided n is at least 10^2 = 100. 16. The point A=(a,0) is moving on the positive x - axis and the point B = (0,b) is moving on the positive y - axis so that the distance between A and B is always 5. Let C be the midpoint of the line segment from A to B.. Find parametric equations for the curve traced out by C, using b as the parameter. SOLUTION: The midpoint from A to B is x = a/2 , y = b/2. We must describe these equations in terms of the parameter b. From the given information, a^2+b^2 = 25 (Pythagorean Theorem). So 2 1/2 b = (25 - a ) and so the point is 2 1/2 x = 1/2 (25 - a ) y = 1/2 b 17. What is the length of the curve parametrized by x = t^2/2 - t y = (4/3) t^(3/2) between t = 0 and t = 2? SOLUTION: The arclength is 2 / | //d \2 /d \2\1/2 | ||-- x(t)| + |-- y(t)| | dt | \\dt / \dt / / / 0 Inserting x = t^2/2 - t y = (4/3) t^(3/2), the integrand becomes sqrt( (t-1)^2 +4t) = sqrt(t^2+2t+1) = t+1. Therefore the length is 2 / | | 1 + t dt | / 0 which is 4.