Fall 2008 Midterm &Solutions.pdf

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Department
Mathematics & Statistics (Sci)
Course Code
MATH 222
Professor
Christa Scholtz

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McGill University Department of Mathematics and Statistics Calculus 3 (MATH 222) Fall 2008 Instructors: Prof. K. GowriSankaran, N. Sonnerat Midterm Exam Solutions October 22nd 2008, 6pm - 8pm Instructions: This is a 2 hour, closed book/notes examination. No calculators are permitted. To get full marks, you must answer all five questions. Justify all your answers, unless the question instructs you not to. (−1) en 1. (a) Compute lim n→∞ { n! }. Solution: The easiest way would have been to say that you saw in class that xn lim n! = 0 for all values of x, so in particular this is true for x = −e. Alternatively, you can use the ratio test to show that the series X (−1) e n n! converges absolutely. Then the test for divergence implies that the sequence must have limit 0. As a third option, you could have shown that (−1) e n en e2e 0 ≤ | | = ≤ , n! n! 2 n so lim|an| = 0 by the squeeze theorem (see assignment solutions for more details). But you have another theorem that tells you that if |a | tends to 0, then a must n n also tend to 0. (b) Determine whether the series X∞ 2 k! k=0 (k + 2)! converges or diverges. Solution: Either use l’Hopital’s Rule to show that 2 k! 2k lim = lim 6= 0, (k + 2)! (k + 2)(k + 1) which implies that the series is divergent by the test for divergence. Note that x x x−1 d/dx(2 ) = ln2 · 2 , NOT x2 . 1 Alternatively, apply the ratio test to find that |an+1| lim = 2 > 1, |an| so the series must diverge. (c) Determine whether the series √ X j (−1) j+1 j=0 j + 5 converges or diverges. Solution: √ Let a = j. It is obvious that a ≥ 0 for all j. It’s not so obvious that it’s j j+5 j √ decreasing, but if you take the derivative of f(x) , you’ll see that f (x) < 0 x+5 for all x > 5. This implies that for j > 5, the series is decreasing. Finally, √ j 1 lim = lim √ √ = 0 j + 5 j + 5/ j √ √ √ 1 because lim5/ j = 0, so j +5/ j tends to ∞ and thus √j+5/ jtends to 0. So all the conditions of the alternating series test are met, and therefore the series converges. (It does not converge absolutely, but we didn’t ask for that.) (d) Determine whether the series X k lnk (k + 1) k=1 converges or diverges. Solution: It converges, which can be seen either by using the limit comparison test with bk= k −3/2(see assignment solutions for details), or by arguing that k lnk lnk 3 < 2 = ck. (k + 1) k P P Then you can show that ckconverges using the integral test, and so k l3 k k=1(k+1) converges by the comparison test. (Note that all the terms in all the series here are ≥ 0 as long as k ≥ 3, so the comparison test, limit comparison test, and integral test can be applied.) 2 2. (a) Find the Maclaurin series of f(x) = ln(1 + x ), and determine its radius and interval of convergence. Also discuss convergence at the endpoints of the interval. 2 Solution: Take the derivative to find f (x) =2x2. Using the formula for the 1+x geometric series, ∞ ∞ 2x 2x X 2 n X n 2n+1 2 = 2 = 2x (−x ) = 2 (−1) x , 1 + x 1 − (−x ) 0 0 provided that |x| < 1. You know that power series can be integrated term by term, so X∞ x2n+2 X∞ x2n+2 ln(1 + x ) = C + 2 (−1)n = C + (−1)n . 0 2n + 2 0 n + 1 2 Since ln(1 + 0 ) = ln1 = 0, you get that C must be 0 (because all the terms of the series are 0 if x = 0). Thus X x2n+2 ln(1 + x ) = (−1) n . n + 1 0 Alternatively, if you happened to know the series for ln(1−t) by heart, you could have used that and substituted −x for t to get the same answer. Now the ratio test tells you that the radius of convergence is 1. You can also argue this by saying that the radius of convergence doesn’t change when you integrate or differentiate term by term, and you know it’s 1 for the geometric series. Finally, when x = 1 or x = −1, you get the convergent alternating series ∞ n X (−1) , 0 n + 1 so the interval of convergence is the closed interval [−1,1]. (b) State the Maclaurin series for f(x) = sinx. Also state its radius of covergence. (“state” means that no justification is required.) Solution: X 2n+1 n x sinx = (−1) (2n + 1)! 0 which converges for all values of x (i.e. R = ∞). (c) Approximate the value of Z 0.5 2 sin(t )dt 0 correct to 6 decimal places (i.e. |error| < 0.000001). Justify your answer. (You (0.5) may use the fact that ≤ 0.0000004.) 1320 Solution: Using the series for sint, we get that ∞ ∞ 2 X n (t )n+1 X n t4n+2 sin(t ) = (−1) = (−1) . 0 (2n + 1)! 0 (2n + 1)! 3 Integrating term by term, you get that Z 0.5 X∞ Z 0.5 t4n+2 X∞ t4n+3 sin(t )dt = (−1) n dt = [ (−1) n ]0.5 0 0 (2n + 1)! (4n + 3)(2n + 1)! 0 0 0 which equals X∞ 4n+3 3 7
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