2004-2005 spring mt1 - METU Department of Mathematics Group...

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Unformatted text preview: METU Department of Mathematics Group Code Acad. Year Semester Instructor : Math 120 : 2004-2005 : Spring : Date Time Duration Calculus II MidTerm 1 Last Name : : Name Department : : Signature : April 2nd,2005 : 09:30 : 120 minutes 1 2 3 4 List No. Student No. : : Section 5 QUESTIONS ON 6 PAGES TOTAL 50 POINTS 5 SOLUTION KEY Question 1 (10 points) Find the radius of convergence and interval of convergence ∞ (−3)n xn √ . n+1 n=0 of the series Solution: √ | an+1 | 3n+1 | x |n+1 n+1 √ = lim · n = 3 | x | lim n→∞ | an | n→∞ n→∞ 3 | x |n n+2 lim n+1 =3|x|. n+2 So if 3 | x |< 1, by ratio test the given power series is convergent. This gives | x |< 1/3, so R = 1/3. For the interval of convergence, we must check the end points. If x = 1/3: ∞ ∞ (−1)n 3n (−1)n √ √ = . n + 1 3n n+1 n=0 n=0 an = (−1)n √ n+1 is alternating, • √1 n+1 > 0, • limn→∞ an = limn→∞ • |an+1 | |an | = √ √n+1 n+2 = (−1)n √ n+1 n+1 n+2 = 0, < 1, so | an | is decreasing. So by AST, the series is convergent. If x = −1/3: ∞ ∞ √ n=0 ∞ n=1 1 n1/2 1 (−3)n √ = . n n + 1 (−3) n+1 n=0 is divergent by p−test, so the above series is divergent by limit comparison. So the interval of convergence is I = (−1/3, 1/3]. Question 2 (10 points) 1 ; k ≥ 1 an integer. 1 + xk a)Find the power series expansion of f (x) in powers of x. Consider f (x) = Solution: Since ∞ 1 = (−1)n xn 1 + x n=0 we have; for all | x |< 1, ∞ 1 = (−1)n xnk . 1 + xk n=0 b)Use the series in part (a), to expand g(x) = x arctan(x) in powers of x. Solution: x g(x) = x arctan x = x 0 1 du = x 1 + u2 c)Find the expansion of h(x) = of convergence. √ ∞ x ∞ (−1)n u2n du = 0 (−1)n n=0 n=0 x2n+2 2n + 1 for all | x |< 1. √ x + 1 arctan( x + 1) around a = −1. Indicate the interval Solution: h(x) = √ √ x + 1 arctan x + 1 = ∞ (−1)n n=0 (x + 1)1/2 2n + 1 2n+2 ∞ = (−1)n (x + 1)n+1 2n + 1 n=0 for all | x + 1 |< 1. Question 3 (9 points) Determine whether the followings are convergent or divergent. Show tour work. ∞ (n!)2 a) (2n)! n=0 Solution: We apply ratio test, with an = (n!)2 : (2n)! n+1 (n!)2 (n + 1)2 (2n)! 1 an+1 = lim = lim · = < 1. n→∞ 2(2n + 1) n→∞ (2n)!(2n + 1)(2n + 2) (n!)2 n→∞ an 4 lim So by ratio test, the series converges. ∞ b) 1 2 + 3n n=0 Solution: We have; 1 1 1 ≤ n = n 2+3 3 3 0≤ ∞ n=0 1 n 3 n . is a geometric series with r = 1/3, so it is convergent. ∞ 1 n=0 2+3n Therefore, by comparison test, c)Approximate is convergent. 1 1 with an error less than . e 119 Solution: We have; ∞ −x e Then; = (−1)n xn . n! n=0 ∞ 1 (−1)n 1 1 1 1 = e−1 = = 1 − 1 + − + − + ···. e n! 2! 3! 4! 5! n=0 Since the series is alternating; En (1) = 1 1 < , (n + 1)! 119 which gives n + 1 = 5. Therefore; 1 1 1 1 3 ≈1−1+ − + = . e 2! 3! 4! 8 Question 4 (9 points) ∞ a)Prove that for an ≥ 0, if ∞ a2 is also convergent. n an is convergent then n=0 n=0 Solution: ∞ an is convergent, we have lim an = 0. This implies that there exists N ∈ I such N Since n→∞ n=0 that for n ≥ N , we have 0 ≤ an ≤ 1. So for all n ≥ N , 0 ≤ a2 ≤ an . n ∞ Since ∞ a2 is also convergent by comparison test. n an is convergent, n=0 n=0 b)The term of a sequence are defined recursively by the equations 4n + 1 )an . Is (an )∞ convergent? (explain, and if so what is the limit?) a1 = 2, an+1 = ( n=1 5n + 2 Solution: ∞ an . We have an ≥ 0. Apply ratio test to this series; Consider n=1 lim n→∞ 4n + 1 4 an+1 = lim = < 1. n→∞ 5n + 2 an 5 So by ratio test, the series converges. This implies limn→∞ an = 0. c)Give an example of a power series which is convergent in (1, 5), and divergent on (−∞, 1) ∪ (5, ∞). (No condition at x = 1 and x = 5) Solution: ∞ an (x − 3)n ; a power series around 3. We want the radius of convergence to be Consider, n=0 2. So we want lim n→∞ Let an = | an+1 | 1 = . | an | 2 1 , then we get the desired limit and interval of convergence. So 2n ∞ (x − 3)n 2n n=0 is a power series satisfying the conditions we want. Question 5 ( 12 points) → → → → → → → → a) Find all vectors v that satisfy the equation (− i +2 j +3 k )× v = i +5 j −3 k Solution: → Let v = (a, b, c). Then we should have; (−1, 2, 3) × (a, b, c) = i −1 a j k 2 3 b c = (2c − 3b, 3a + c, −2a − b) = (1, 5, −3). This happens if and only if −3b + 2c = 1, 3a + c = 5, and −2a − b = −3 which gives → → → → v = t i +(3 − 2t) j +(5 − 3t) k for all t ∈ I R. b)Show that the four points A(1, 2, −1), B(0, 1, 5), C(−1, 2, 1) and D(2, 1, 3) are coplanar (i.e. are on the same plane). Solution: → → → − − − → → −→ − Let the three vectors u , v and w represent the three segments AB, AC and AD, respectively. → → → Then u = (−1, −1, 6), v = (−2, 0, 2) and w= (1, −1, 4). The four points are coplanar if the volume of the parallelpiped spanned by the three vectors is 0; −1 −1 −2 0 1 −1 6 2 4 =− 0 1 2 4 + −2 2 1 −1 +6 −2 0 1 −1 = −2 − 10 + 12 = 0. c)Find the distance between the lines L1 : intersection of the planes (x + 2y = 3 and y + 2z = 3) and L2 : intersection of the planes (x + y + z = 6 and x − 2z = −5) Solution: L1 contains the points (1, 1, 1), (3, 0, 3/2), so it is parallel to the vector (3, 0, 3/2)−(1, 1, 1) = → (2, −1, 1/2), so it is parallel to the vector u 1 = (4, −2, 1). L2 contains the points (−5, 11, 0), (−1, 5, 2), so it is parallel to the vector (−1, 5, 2) − → (−5, 11, 0) = (4, −6, 2), so it is parallel to the vector u 2 = (2, −3, 1). → → → → Using the vectors r 1 = (1, 1, 1), r 2 = (−1, 5, 2), u 1 = (4, −2, 1) and u 2 = (2, −3, 1) we find the distance: → d= → → → u1 → u2 → | ( r 2 − r 1) · ( u 1 × u 2) | | × | = | (2, −4, −1) · (1, −2, −8) | 18 =√ . | (1, −2, −8) | 69 ...
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