Homework 13 - tgo72 – Homework 13 – Gompf – (58370)...

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Unformatted text preview: tgo72 – Homework 13 – Gompf – (58370) This print-out should have 19 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. 001 10.0 points Thus the given series ∞ 1 sin k=1 1 k converges if and only if the series ∞ Determine whether the series ∞ k=1 (−1)k−1 sin 1 k k=1 1 k is absolutely convergent, conditionally convergent or divergent. 1. absolutely convergent 2. series is divergent converges. But by the p-series test with p = 1 (or because the harmonic series is divergent), this last series is divergent. Thus the given series is not absolutely convergent. On the other hand, since sin 1 k > sin 1 k+1 3. conditionally convergent correct Explanation: Th given series can be written as an alternating series ∞ k=1 (−1)k ak 1 = 0, k→∞ k the Alternating series test applies, showing that the series lim sin ∞ while 1 > 0. k To test for absolute convergence, i.e., convergence of the series ak = sin ∞ with k=1 (−1)n−1 sin 1 k is convergent. Consequently, the given series is conditionally convergent . 002 10.0 points ak , k=1 we use the Limit Comparison test with ak 1 = sin k , bk 1 =. k Determine whether the series ∞ m=1 (−6)m+1 54m For then ak and positive terms and bk are series with is absolutely convergent, conditionally convergent or divergent. 1. absolutely convergent correct 2. divergent 3. conditionally convergent Explanation: 1 sin ak k = lim lim 1 k→∞ k → ∞ bk k = lim θ→0 sin θ = 1 > 0. θ tgo72 – Homework 13 – Gompf – (58370) The given series can be written in the form ∞ 2 m=1 (−6)m+1 = 54m ∞ is absolutely convergent, conditionally convergent, or divergent. 1. conditionally convergent 2. divergent correct an m=1 with am = (−1)m+1 But then 6 6m+1 = (−1)m+16 4 4m 5 5 = 6 ; 54 m . 3. absolutely convergent Explanation: The given series has the form ∞ am+1 am am+1 am in particular, m→∞ an n=1 lim 6 = 4 < 1; 5 where an = But then Consequently, by the Ratio Test, the given series is absolutely convergent . An alternative method for determining the behavior of the series ∞ n! . e3n (n + 1)! e3n n+1 an+1 = = . 3n+3 an n! e e3 Thus n→∞ lim m=1 (−6)m+1 = 54m ∞ an+1 an = ∞. an m=1 is to note that |am | = so that 6 6m+1 =6 4 4m 5 5 ∞ m Consequently, by the Ratio Test, the given series is divergent . 004 10.0 points , Determine whether the series |am | ∞ m=1 is a geometric series with r = 6/(54 ) < 1. Since r < 1, the geometric series converges. Hence we see again that the given series is absolutely convergent. keywords: 003 10.0 points m=5 3 m(ln m)4 converges or diverges. 1. converges correct 2. diverges Explanation: The given series can be written in the form ∞ Determine whether the series ∞ e n=1 −3n n! m=1 f (n) tgo72 – Homework 13 – Gompf – (58370) where f is the function defined on (1, ∞) by f ( x) = 3 . x(ln x)4 Consequently, n→∞ 3 lim an+1 an = 0, Now f is positive and decreasing on (1, ∞); in addition, t and so by the Ratio Test, the given series converges . f (x) dx = 5 1 − (ln x)3 = t 5 1 1 − , 3 (ln 5) (ln t)3 keywords: 006 10.0 points in which case, ∞ t f (x) dx = lim 5 t→∞ f (x) dx 5 = lim t→∞ 1 1 − 3 (ln 5) (ln t)3 = 1 . (ln 5)3 Find the interval of convergence of the power series ∞ 4 xn √. n n=1 Consequently, by the Integral Test, the given series converges . 005 10.0 points 1. interval of cgce = (−1, 1) 2. interval of cgce = (−1, 1] 3. interval of cgce = (−4, 4] 4. interval of cgce = [−1, 1) correct 5. interval of cgce = [−4, , 4] 6. interval of cgce = (−4, 4) 7. interval of cgce = [−4, 4) 8. interval of cgce = [−1, 1] Explanation: When 4 xn an = √ , n Determine whether the series ∞ n=1 4n n2 n! converges or diverges. 1. diverges 2. converges correct Explanation: The given series can be written as ∞ an , n=1 an = 4n n2 . n! But then an+1 an 4n+1 (n + 1)2 n! = 4n n2 (n + 1)! = 4 1 1+ n+1 n 2 then √ xn+1 n =√ ·n n+1 x √ n | x| n = | x| =√ n+1 n+1 an+1 an . . tgo72 – Homework 13 – Gompf – (58370) Thus n→∞ 4 xn . n+3 lim an+1 an with = | x| . an = (−1)n Now for this series, By the Ratio Test, therefore, the given series converges when |x| < 1, and diverges when |x| > 1. We have still to check for convergence at x = ±1. But when x = 1, the series reduces to ∞ 4 √ n n=1 (i) R = 0 if it converges only at x = 0, (ii) R = ∞ if it converges for all x, while 0 < R < ∞, (iii) if it converges when |x| < R, and (iv) diverges when |x| > R. But n→∞ which diverges by the p-series test with p = 1 2 ≤ 1. On the other hand, when x = −1, the series reduces to ∞ n=1 4 (−1)n √ n lim an+1 an = lim |x| n→∞ n+3 = | x| . n+4 which converges by the Alternating Series Test. Thus the interval of convergence = [−1, 1) . keywords: 007 (part 1 of 2) 10.0 points For the series ∞ By the Ratio Test, therefore, the given series converges when |x| < 1 and diverges when |x| > 1. Consequently, R=1 . 008 (part 2 of 2) 10.0 points (ii) Determine the interval of convergence of the series. 1. interval convergence = (−3, 3) 2. interval convergence = (−1, 1] correct 3. interval convergence = (−1, 1) 4. interval convergence = [−1, 1) 5. interval convergence = [−3, 3) 6. converges only at x = 0 7. interval convergence = (−3, 3] Explanation: Since R = 1, the given series (i) converges when |x| < 1, and (ii) diverges when |x| > 1. n=1 (−1)n n x, n+3 (i) determine its radius of convergence, R. 1. R = 0 2. R = 1 correct 3. R = 1 3 4. R = (−∞, ∞) 5. R = 3 Explanation: The given series has the form ∞ an n=1 tgo72 – Homework 13 – Gompf – (58370) On the other hand, at the point x = 1 and x = −1, the series reduces to ∞ 5 n=1 (−1)n , n+3 ∞ n=1 1 n+3 We first apply the root test to the infinite series ∞ 3n + 2 n n | x| . 6n n=1 For this series an 1/n respectively. But by the Alternating Series Test, the first of these series converges. On the other hand, if we set 1 an = , n+3 then n→∞ = 3n + 2 | x| 6n −→ 1 | x| 2 1 bn = , n as n → ∞. Thus the given power series will converge on the interval ( − 6/3, 6/3). For convergence at the endpoints x = ±2 we have to check if ∞ lim an n = lim = 1. n→∞ n + 3 bn n=0 3n + 2 6n n 2 n ∞ = n=0 an By the p-series Test with p = 1, however, the series n bn diverges. Thus by the Limit Comparison test, the series n an also diverges. Consequently, the given series has interval convergence = (−1, 1] . keywords: 009 10.0 points converges, or if ∞ n=0 3n + 2 6n n −2 n ∞ = n=0 bn converges. In the first case an = 3n + 2 6n n 2 n = 3n + 2 3n n , in which case Find the interval of convergence of the power series ∞ n→∞ lim an = e2/3 = 0; by the Divergence test, therefore the series ∞ n=1 3n + 2 6n n x. n=1 n an − 2, 2 11 −, 22 diverges. On the other hand, bn = 3n + 2 6n = (−1)n n 1. interval of convergence correct 2. interval of convergence = = −2 n 3n + 2 3n n = (−1)n an . 3. interval of convergence = − 6, 6 11 4. interval of convergence = − , 22 5. interval of convergence = − 2, 2 Explanation: But, as we have seen, n→∞ lim an = e2/3 . Thus bn oscillates between e2/3 and −e2/3 as n → ∞; in particular, n→∞ lim bn tgo72 – Homework 13 – Gompf – (58370) does not exist. Again by the Divergence Test, therefore, the series ∞ 6 for all x = 0. By the Root Test, therefore, the given series converges only at x = 0 . 011 10.0 points bn n=1 diverges. Consequently, the given power series does not converge at x = ±6/3 and so its interval of convergence = (−2, 2) . keywords: 010 10.0 points Find a power series representation for the function 1 f ( t) = . t−2 ∞ 1. f (t) = n=0 ∞ 1 2n+1 1 tn tn correct 2. f (t) = − 3. f (t) = Find the interval of convergence of the series ∞ (−n)n xn . 2n n=1 n=0 2n+1 ∞ n=0 (−1)n 2n tn 2 n tn ∞ 1. interval of cgce = (−1, 1] 2. interval of cgce = [−1, 1) 3. interval of cgce = (−1, 1) 4. converges only at x = 0 correct 5. interval of cgce = [−2, 2) 6. interval of cgce = (−2, 2] (−n)n xn , 2n it’s more convenient to use the Root Test to determine the interval of convergence. For then an = |an |1/n = (−n)n But 1/n xn 4. f (t) = − 5. f (t) = n=0 ∞ n=0 (−1)n−1 2n+1 tn Explanation: We know that 1 = 1 + x + x2 + . . . = 1−x On the other hand, ∞ xn . n=0 Explanation: When 1 1 1 . =− t−2 2 1 − (t/2) Thus 1 f ( t) = − 2 ∞ n=0 t 2 n 1 =− 2 ∞ n=0 1n t. 2n Consequently, ∞ 2n = n | x| . 2 f ( t) = − with |t| < 2. 1 2n+1 tn n=0 n | x| =∞ lim n→∞ 2 tgo72 – Homework 13 – Gompf – (58370) can be identified with 012 10.0 points Determine the value of f (1) when f ( x) = 2 x3 3 x5 x − 4 + 6 + ... . 22 2 2 22 x f ( x) = . ( x2 + 2 2 ) 2 As x = 1 lies in (−2, 2), we thus see that f (1) = 4 . 25 7 (Hint: differentiate the power series expansion of (x2 + 22 )−1 .) 1. f (1) = 8 25 keywords: 1 2. f (1) = 25 3. f (1) = 4. f (1) = 1 5 8 5 013 10.0 points Find a power series representation for the function x f ( x) = . 16x + 1 ∞ 4 correct 5. f (1) = 25 Explanation: The geometric series 1 1 1 =2 22 + x 2 1 + x/22 = x x2 x3 1 1− 2 + 4 − 6 +... 22 2 2 2 1. f (x) = n=0 ∞ (−1)n 4n xn+1 (−1)n 4n xn 42n xn+1 2. f (x) = n=0 ∞ 3. f (x) = n=0 ∞ has interval of convergence (−4, 4). But if we now restrict x to the interval (−2, 2) and replace x by x2 we see that 1 x2 x4 x6 1 = 2 1− 2 + 4 − 6 + ... 2 2 + x2 2 2 2 2 on the interval (−2, 2). In addition, in this interval the series expansion of the derivative of the left hand side is the term-by-term derivative of the series on the right: − 2x 2 x 4 x3 6 x5 1 = 2 − 2 + 4 − 6 + ... . ( x2 + 2 2 ) 2 2 2 2 2 4. f (x) = n=0 ∞ 4 n xn 5. f (x) = n=0 ∞ (−1)n 42n xn+1 correct 42n xn 6. f (x) = n=0 Explanation: After simplification, f ( x) = x x = . 16x + 1 1 − (−16x) Consequently, on the interval (−2, 2) the function f defined by f ( x) = x − 4 + 6 + ... 22 2 2 2 x3 3 x5 On the other hand, 1 = 1−x ∞ xn . n=0 tgo72 – Homework 13 – Gompf – (58370) Thus ∞ 8 Now x f ( x) = x n=0 (−16x) ∞ n 0 1 dt = tan−1 x , 1 + t2 while (−1) 4 x n 2n n =x n=0 . 0 x ∞ ∞ Consequently, ∞ n=0 (−1) x n 2n dt = n=0 (−1)n 2n+1 x . 2n + 1 f ( x) = n=0 (−1)n 42n xn+1 . Thus ∞ keywords: 014 10.0 points tan −1 x= n=0 (−1)n 2n+1 x , 2n + 1 Determine the interval of convergence for the power series representation of x f (x) = tan−1 6 centered at the origin obtained by integrating the power series expansion for 1/(1 − x). 1. interval of cgce. = [−6, 6) 2. interval of cgce. = 3. interval of cgce. = 11 −, 66 11 −, 66 from which it follows that f (x) = tan −1 x = 6 ∞ n=0 (−1)n x 2n + 1 6 2n+1 . To determine the interval of convergence of the power series, set an = Then an+1 an and so n→∞ (−1)n x (2n + 1) 6 2n+1 . = 2n + 1 x 2n + 3 6 x 6 2 , 4. interval of cgce. = [−6, 6] correct 5. interval of cgce. = 11 −, 66 lim an+1 an 2 = . 6. interval of cgce. = (−6, 6] Explanation: Since 1 = 1 + x + x 2 + x3 + . . . , 1−x we see that 1 1 = 2 1+x 1 − (−(x)2 ) = 1 − x2 + (−x2 )2 − (x2 )3 + . . . ∞ By the Ratio Test, therefore, the power series converges when |x| < 6 and diverges when |x| > 6. On the other hand, at x = 6 the series reduces to ∞ n=0 (−1)n , 2n + 1 which converges by the Alternating series Test, while at x = −6 the series reduces to ∞ = n=0 (−1) x n 2n . n=0 (−1)n+1 , 2n + 1 tgo72 – Homework 13 – Gompf – (58370) which converges again by the Alternating Series Test. Consequently, the power series representation for f (x) obtained from the series expansion for 1/(1 − x) has interval of convergence = [−6, 6] . B. False: when an = 1/n2 , then an+1 an = n n+1 2 9 −→ 1 as n → ∞, so the Ratio Test is inconclusive. 016 10.0 points keywords: 015 10.0 points Find a power series representation centered at the origin for the function f ( t) = ∞ t3 (2 − t)2 . Which, if any, of the following statements are true? A. The Root Test can be used to determine whether the series ∞ 1. f (t) = n=2 ∞ n−1 n t 2n n−2 n t correct 2n−1 1 2n−1 1 2n−3 nn t 2n tn tn 2. f (t) = n=3 ∞ k=1 k 3 + k2 3. f (t) = n=2 ∞ converges or diverges. B. The Ratio Test can be used to determine whether the series ∞ 4. f (t) = n=3 ∞ n=1 1 n2 5. f (t) = n=3 converges or diverges. 1. A only 2. neither of them correct 3. B only 4. both of them Explanation: A. False: when ak = then (|ak |)1/k = k 1/k /(1 + k 2 )1/k −→ 1 so the Root Test is inconclusive. k , 3 + k2 Explanation: By the known result for geometric series, 1 1 = t 2−t 2 1− 2 1 = 2 ∞ n=0 t 2 n ∞ = n=0 1 2n+1 tn . and so on (−2, 2), d 1 = (2 − t)2 dt ∞ This series converges on (−2, 2). On the other hand, 1 1 d , = 2 (2 − t) dt 2 − t ∞ tn 2n+1 ∞ n=0 = n=1 n 2n+1 t n−1 = n=0 n+1 n t. 2n+2 tgo72 – Homework 13 – Gompf – (58370) Thus ∞ 10 Thus n+1 n t= 2n+2 ∞ f ( t) = t 3 n=0 n=0 n + 1 n+3 t . 2n+2 ln(1 + x) − ln(1 − x) = 2 x+ x3 x5 + +... 3 5 ∞ Consequently, ∞ f ( t) = n=3 n−2 n t. 2n−1 Consequently, =2 n=1 1 x2n−1 . 2n − 1 017 10.0 points Find a power series representation for the function 1 + 2z f (z ) = ln . 1 − 2z (Hint: remember properties of logs.) ∞ ∞ f (z ) = 2 n=1 22n−1 2n−1 . z 2n − 1 018 10.0 points 1. f (z ) = n=1 ∞ (−1)n 22n 2n − 1 z 2n−1 Determine the interval of convergence of the series ∞ n=1 2. f (z ) = 2 3. f (z ) = n=1 ∞ 22n−1 n=1 ∞ 22n−1 2n−1 z correct 2n − 1 z 2n−1 n4 (x − 7)n . 2n − 1 1. interval convergence = (−∞, ∞) 2. interval convergence = (−8, −6) 3. interval convergence = [6, 8) 4. converges only at x = 7 5. interval convergence = (−8, −6] 6. interval convergence = (6, 8) correct Explanation: The given series has the form an (x − 7)n 4. f (z ) = n=1 ∞ 2 z 2n−1 2n − 1 1 2n z n (−1)n 22n 2n−1 z 2n − 1 5. f (z ) = n=1 ∞ 6. f (z ) = 2 n=1 Explanation: We know that x2 x3 ln(1 + x) = x − + −... 2 3 = n=1 (−1)n−1 n xn , while ln(1 − x) = −x − =− x2 2 − −... 3 1n x. n x3 n=1 with an = n4 . Now lim n+1 an+1 = lim n→∞ an n 4 n=1 n→∞ = 1. tgo72 – Homework 13 – Gompf – (58370) By the Ratio Test, therefore, the given series (i) converges when |x − 7| < 1, and (ii) diverges when |x − 7| > 1. On the other hand, at the points x − 7 = −1 and x − 7 = 1 the series reduces to ∞ ∞ n=1 11 Consequently, ∞ I = C+ n=0 (−1)n y 12n+7 . (2n + 1)(12n + 7) (−1)n n4 , n4 n=1 respectively. But by the Divergence Test, both of these diverge. Consequently, interval convergence = (6, 8) . 019 10.0 points Express the indefinite integral I= as a power series. ∞ tan−1 y 6 dy 1. I = C + n=0 ∞ (−1)n y 12n+6 (2n + 1)(12n + 7) 2. I = n=0 (−1)n y 12n+7 (2n + 1)(12n + 7) ∞ 3. I = C + n=0 ∞ (−1)n y 6n+6 (2n + 1) 4. I = C + n=0 ∞ (−1)n y 12n+7 correct (2n + 1)(12n + 7) (−1)n y 12n+7 (12n + 7) 5. I = C + n=0 Explanation: We know that ∞ tan −1 y= n=0 y 2n+1 . (−1) 2n + 1 n Replacing y with y 6 , we get I= = n=0 tan−1 y 6 dy ∞ (−1)n (y 6 )2n+1 dy . 2n + 1 ...
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