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Quiz 7 for MATH 105 SECTION 205 March 30, 2015 Family Name Given Name Student Number 1. (1 point) Is the series ∞ X k 10 10k (k!)2 (2k)! k=1 convergent or divergent? divergent 1. Answer. Let ak = k 10 10k (k!)2 , then (2k)! ak+1 ak = = (k+1)10 10k+1 [(k+1)!]2 [2(k+1)]! k10 10k (k!)2 (2k)! 10 (k + 1) 10k+1 [(k + 1)!]2 [2(k + 1)]! · (2k)! k 10 10k (k!)2 (k + 10k+1 (k + 1)! 2 (2k)! = · · · 10 k k k! (2k + 2)! 10 10 1 1 = 1+ · 10 · (k + 1)2 · k (2k + 2)(2k + 1) 10 (k + 1)2 1 · = 10 1 + k (2k + 2)(2k + 1) 5 1 → 10 · 1 · = , as k → ∞. 4 2 1)10 By the ratio test, we know that the series ∞ X k 10 10k (k!)2 k=1 2. Consider the power series ∞ X 2k (x − 3)k k=1 k+1 (2k)! diverges. . (a) (1 point) Find the radius of convergence of this series. (a) Answer. Let ck = 1 2 2k , then k+1 ck+1 2k+1 k + 1 2k+1 k + 1 k+1 = · k = k · =2· → 2, ck k+2 k+2 k+2 2 2 as k → ∞. 1 So the radius of convergence is R = . 2 (b) (1 point) Find the interval of convergence of this series. (b) (5/2, 7/2) 1 and center is a = 3, then the interval of convergence 2 is (a − R, a + R) = (3 − 1/2, 3 + 1/2) = (5/2, 7/2). Answer. Since the radius of convergence is R = 3. (1 point) Find the value of x such that ∞ X ekx = 1. k=1 − ln 2 3. Answer. ∞ ∞ X X x k Since (e ) = ekx = 1, then |ex | < 1 and k=1 k=1 ∞ X ekx = k=1 ex = 1. 1 − ex 1 So we get ex = 1 − ex , that is, 2ex = 1. So we get ex = , that is, 2 x = ln 4. (1 point) Find the power series of f (x) = 1 = − ln 2. 2 4 centered at 0. 4 + x2 4. ∞ X 1 k 2k − x 4 k=0 Answer. In fact, we have f (x) = = 4 4 + x2 4 1 · 4 1 + x2 4 = 1 1+ x2 4 1 2 1 − − x4 k ∞ X x2 x2 = − if − < 1, that is, |x| < 2 4 4 k=0 ∞ X 1 k 2k = − x . 4 = k=0 5. (2 points) Find the power series of f (x) = 1 centered at 1. 4 − 2x 5. ∞ X (x − 1)k k=0 2 Answer. In fact, we have 1 4 − 2x f (x) = 1 4 − 2(x − 1) − 2 1 2 − 2(x − 1) 1 1 · 2 1 − (x − 1) ∞ 1X (x − 1)k if |x − 1| < 1 2 = = = = k=0 ∞ X = k=0 (x − 1)k . 2 6. (1 point) Find the function represented by the series k ∞ 2 X x −1 k=0 3 . 6. Answer. In fact, we have k ∞ 2 X x −1 = 3 k=0 = = 1 1− x2 − 1 < 1, that is, |x2 − 1| < 3, then |x| < 2 3 if x2 −1 3 3 3 − x2 + 1 3 . 4 − x2 7. (2 points) *Find the function represented by the series ∞ X x2k k=1 4k . 7. Answer. 3 4 − x2 Approach I: Let f (x) = ∞ X x2k k=1 4k , then f 0 (x) = = d dx ∞ X x2k k=1 ∞ X x2k−1 k=1 x 2 ! 4k 2 if |x2 | < 1 1 − x2 1 x = − · 2 . 2 x −1 = So we get 1 f (x) = − 2 Z x2 x 1 dx = − ln |1 − x2 | + C. −1 4 1 − ln(1 − x2 ) 4 Taking x = 0 in the above identity, we get C = 0. So we get ∞ X x2k k=1 1 = − ln(1 − x2 ). 4k 4 Approach II: In fact, we have ∞ X x2k k=1 4k ∞ = = = = = = 1 X yk Let y = x2 4 k k=1 ∞ Z X 1 y k−1 dy 4 k=1 ! Z X ∞ 1 y k−1 dy 4 Z k=1 1 1 dy 4 1−y 1 − ln |1 − y| + C if |y| < 1, that is, |x| < 1 4 1 − ln(1 − x2 ) + C. 4 Taking x = 0 in the above identity, we get C = 0. So we get ∞ X x2k k=1 8. *Consider the power series ∞ X 4k 1 = − ln(1 − x2 ). 4 (−1)k 32k+1 x3k+1 . k=0 (a) (1 point) Find the radius of convergence of this series. 2 3− 3 (a) Answer. Notice that ∞ X (−1)k 32k+1 x3k+1 = k=0 ∞ ∞ X X (−1)k 32k+1 x3k · x = x (−1)k 32k+1 x3k . k=0 Let y = x3 , then k=0 ∞ ∞ X X (−1)k 32k+1 x3k+1 = x (−1)k 32k+1 y k . k=0 k=0 Let ck = (−1)k 32k+1 , then |ck+1 | 32k+3 = 2k+1 = 9, |ck | 3 Then the radius of convergence of the series ∞ X (−1)k 32k+1 y k is k=0 r ∞ X convergence of the series (−1)k 32k+1 x3k+1 is k=0 for all k ≥ 0. 3 2 1 = 3− 3 . 9 1 . Since y = x3 , then the radius of 9 (b) (1 point) Evaluate the sum of this series. 3x 1 + 9x3 (b) Answer. In fact, we have ∞ X k 2k+1 3k+1 (−1) 3 x ∞ X = (−1)k 32k · 3 · x3k · x k=0 k=0 = 3x ∞ X (−1 · 32 · x3 )k k=0 = 3x ∞ X (−9x3 )k k=0 = 3x 1 + 9x3 if | − 9x3 | < 1. Your Score: /12