CalcHW13Answers - nav277 Homework 13 Odell (58340) This...

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Unformatted text preview: nav277 Homework 13 Odell (58340) 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 2k converges if and only if the series Determine whether the series (-1)k-1 sin k=1 1 2k k=1 1 2k is absolutely convergent, conditionally convergent or divergent. 1. series is divergent 2. absolutely convergent 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 2k > sin 1 2(k + 1) 3. conditionally convergent correct Explanation: Th given series can be written as an alternating series while (-1)k ak k=1 1 = 0, 2k k the Alternating series test applies, showing that the series lim sin 1 > 0. 2k To test for absolute convergence, i.e., convergence of the series ak = sin with (-1)n-1 sin k=1 1 2k 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 2k , bk 1 = . 2k Determine whether the series m=1 (-3)m+1 23m For then ak and positive terms and bk are series with is absolutely convergent, conditionally convergent or divergent. 1. conditionally convergent 2. absolutely convergent correct 3. divergent Explanation: 1 sin ak 2k = lim lim 1 k k bk 2k = lim 0 sin = 1 > 0. nav277 Homework 13 Odell (58340) The given series can be written in the form 2 m=1 (-3)m+1 = 23m is absolutely convergent, conditionally convergent, or divergent. 1. absolutely convergent 2. divergent correct an m=1 with am = (-1)m+1 But then 3m+1 3 = (-1)m+13 3 3m 2 2 = 3 ; 23 m . 3. conditionally convergent Explanation: The given series has the form am+1 am am+1 am in particular, m ak k=1 lim 3 = 3 < 1; 2 where ak = But then Consequently, by the Ratio Test, the given series is absolutely convergent . An alternative method for determining the behavior of the series 4(k!) . e5k an+1 (k + 1)! e5k k+1 = . = 5k+5 an k! e e5 Thus k lim m=1 (-3)m+1 = 23m ak+1 ak = . an m=1 is to note that |am | = so that 3 3m+1 = 3 3 3m 2 2 m Consequently, by the Ratio Test, the given series is divergent . 004 10.0 points , Determine whether the series m=1 |am | is a geometric series with r = 3/(23 ) < 1. Since r < 1, the geometric series converges. Hence we see again that the given series is absolutely convergent. keywords: 003 10.0 points k=4 5 k(ln k)6 converges or diverges. 1. diverges 2. converges correct Explanation: The given series can be written in the form Determine whether the series 4e k=1 -5k k! k=1 f (n) nav277 Homework 13 Odell (58340) where f is the function defined on (1, ) by f (x) = 5 . x(ln x)6 Consequently, m 3 lim am+1 am = 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 = 4 1 - (ln x)5 = t 4 1 1 - , 5 (ln 4) (ln t)5 keywords: 006 10.0 points in which case, t f (x) dx = lim 4 t f (x) dx 4 = lim t 1 1 - 5 (ln 4) (ln t)5 = 1 . (ln 4)5 Find the interval of convergence of the power series 4xn . n n=1 1. interval of cgce = [-4, , 4] 2. interval of cgce = [-1, 1) correct 3. interval of cgce = (-4, 4] 4. interval of cgce = (-1, 1) 5. interval of cgce = (-1, 1] 6. interval of cgce = (-4, 4) 7. interval of cgce = [-1, 1] 8. interval of cgce = [-4, 4) Explanation: When Consequently, by the Integral Test, the given series converges . 005 10.0 points Determine whether the series m=1 2m m2 m! converges or diverges. 1. converges correct 2. diverges Explanation: The given series can be written as am , m=1 am = 2m m2 . m! 4xn an = , n But then am+1 am 2m+1 (m + 1)2 m! = 2m m2 (m + 1)! = 2 1 1+ m+1 m 2 then xn+1 n = n n+1 x n |x| n = |x| = n+1 n+1 an+1 an . . nav277 Homework 13 Odell (58340) Thus n 4 xn . n+6 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+6 = |x| . n+7 which converges by the Alternating Series Test. Thus the interval of convergence = [-1, 1) . By the Ratio Test, therefore, the given series converges when |x| < 1 and diverges when |x| > 1. Consequently, R = 1 . keywords: 007 (part 1 of 2) 10.0 points For the series 008 (part 2 of 2) 10.0 points (ii) Determine the interval of convergence of the series. 1. interval convergence = [-6, 6) 2. interval convergence = (-6, 6] 3. interval convergence = (-1, 1] correct 4. interval convergence = [-1, 1) 5. interval convergence = (-6, 6) 6. converges only at x = 0 7. interval convergence = (-1, 1) 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+6 (i) determine its radius of convergence, R. 1. R = (-, ) 2. R = 6 3. R = 0 4. R = 1 6 5. R = 1 correct Explanation: The given series has the form an n=1 nav277 Homework 13 Odell (58340) On the other hand, at the point x = 1 and x = -1, the series reduces to 5 n=1 (-1)n , n+6 n=1 1 n+6 We first apply the root test to the infinite series 5n + 7 n n |x| . 4n n=1 For this series 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+6 then n an = 5n + 7 |x| 4n - 5 |x| 4 1 bn = , n as n . Thus the given power series will converge on the interval ( - 4/5, 4/5). 4 For convergence at the endpoints x = 5 we have to check if lim an n = lim = 1. n n + 6 bn n=0 5n + 7 4n n 4 5 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] . converges, or if n=0 5n + 7 4n n 4 - 5 n = n=0 bn converges. In the first case an = 5n + 7 4n n keywords: 009 10.0 points 4 5 n = 5n + 7 5n n , in which case Find the interval of convergence of the power series n lim an = e7/5 = 0; by the Divergence test, therefore the series n=1 5n + 7 4n n xn . an n=1 5 5 1. interval of convergence = - , 4 4 2. interval of convergence = 5 5 - , 4 4 diverges. On the other hand, bn = 5n + 7 4n = (-1)n n 3. interval of convergence = - 4, 4 4 4 4. interval of convergence = - , 5 5 5. interval of convergence = correct Explanation: 4 4 - , 5 5 4 5 5n + 7 5n - n n = (-1)n an . But, as we have seen, n lim an = e7/5 . Thus bn oscillates between e7/5 and -e7/5 as n ; in particular, n lim bn nav277 Homework 13 Odell (58340) 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 . bn n=1 diverges. Consequently, the given power series does not converge at x = 4/5 and so its interval of convergence = 4 4 - , 5 5 . 011 10.0 points Find a power series representation for the function 1 f (y) = . y-3 keywords: 010 10.0 points 1. f (y) = n=0 (-1)n 3n y n Find the interval of convergence of the series (-n)n xn . 4n n=1 2. f (y) = - 3. f (y) = 1 3n+1 1 3n+1 y n correct n=0 yn n=0 1. interval of cgce = (-1, 1] 2. interval of cgce = (-1, 1) 3. interval of cgce = [-4, 4) 4. converges only at x = 0 correct 5. interval of cgce = (-4, 4] 6. interval of cgce = [-1, 1) Explanation: When 4. f (y) = n=0 (-1)n-1 3n+1 y n 5. f (y) = - 3n y n n=0 Explanation: We know that 1 = 1 + x + x2 + . . . = 1-x On the other hand, 1 1 1 . = - y-3 3 1 - (y/3) Thus 1 f (y) = - 3 xn . n=0 (-n)n xn , an = 4n it's more convenient to use the Root Test to determine the interval of convergence. For then |an |1/n = (-n)n But 1/n xn n=0 y 3 n 1 = - 3 n=0 1 n y . 3n 4n = n |x| . 4 Consequently, n |x| = lim n 4 f (y) = - 1 3n+1 yn n=0 nav277 Homework 13 Odell (58340) with |y| < 3. 012 10.0 points Determine the value of f (2) when f (x) = x 2x3 3x5 - 4 + 6 + ... . 42 4 4 can be identified with f (x) = 42 x . (x2 + 42 )2 7 As x = 2 lies in (-4, 4), we thus see that f (2) = 2 . 25 (Hint: differentiate the power series expansion of (x2 + 42 )-1 .) 1. f (2) = 2. f (2) = 4 25 1 10 keywords: 013 10.0 points 2 3. f (2) = correct 25 4. f (2) = 4 5 Find a power series representation for the function t f (t) = . 16t + 1 1 5. f (2) = 50 Explanation: The geometric series 1 1 1 = 2 42 + x 4 1 + x/42 1 x x2 x3 = 2 1- 2 + 4 - 6 +... 4 4 4 4 has interval of convergence (-16, 16). But if we now restrict x to the interval (-4, 4) and replace x by x2 we see that 1 x2 x4 x6 1 = 2 1- 2 + 4 - 6 + ... 4 2 + x2 4 4 4 4 on the interval (-4, 4). 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 1 2x 4x3 6x5 - 2 = 2 - 2 + 4 - 6 + ... . (x + 42 )2 4 4 4 4 Consequently, on the interval (-4, 4) the function f defined by f (x) = x - 4 + 6 + ... 42 4 4 2x3 3x5 1. f (t) = n=0 4 n tn 2. f (t) = n=0 (-1)n 4n tn 3. f (t) = n=0 (-1)n 42n tn+1 correct 4. f (t) = n=0 42n tn+1 5. f (t) = n=0 42n tn 6. f (t) = n=0 (-1)n 4n tn+1 Explanation: After simplification, f (t) = t t = . 16t + 1 1 - (-16t) On the other hand, 1 = 1-x xn . n=0 nav277 Homework 13 Odell (58340) Thus 8 Now x f (t) = t n=0 (-16t) n 0 1 dt = tan-1 x , 1 + t2 while (-1) 4 t n 2n n = t n=0 . 0 x Consequently, (-1) x n=0 n 2n dt = n=0 (-1)n 2n+1 x . 2n + 1 f (t) = n=0 (-1)n 42n tn+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 4 centered at the origin obtained by integrating the power series expansion for 1/(1 - x). 1. interval of cgce. = 2. interval of cgce. = 1 1 - , 4 4 1 1 - , 4 4 from which it follows that f (x) = tan -1 x = 4 n=0 (-1)n x 2n + 1 4 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) 4 2n+1 . 3. interval of cgce. = [-4, 4) 4. interval of cgce. = (-4, 4] 5. interval of cgce. = 1 1 - , 4 4 = 2n + 1 x 2n + 3 4 x 4 2 , lim an+1 an 2 = . 6. interval of cgce. = [-4, 4] correct 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| < 4 and diverges when |x| > 4. On the other hand, at x = 4 the series reduces to n=0 (-1)n , 2n + 1 which converges by the Alternating series Test, while at x = -4 the series reduces to = n=0 (-1) x n 2n . n=0 (-1)n+1 , 2n + 1 nav277 Homework 13 Odell (58340) 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 = [-4, 4] . 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 (y) = y3 (5 - y)2 . Which, if any, of the following statements are true? A. The Root Test can be used to determine whether the series 1. f (y) = n=3 n n y 5n n-1 n y 5n n-2 n y correct 5n-1 1 5n-3 1 5n-1 yn yn 2. f (y) = n=2 k=1 ln k 2+k k 3. f (y) = n=3 converges or diverges. B. The Ratio Test can be used to determine whether the series 4. f (y) = n=3 n=1 1 n2 5. f (y) = n=2 converges or diverges. 1. B only 2. neither of them 3. A only correct 4. both of them Explanation: A. True: when ak = then |ak |1/k = as k , so Test. ln k 2+k k Explanation: By the known result for geometric series, 1 1 = y 5-y 5 1- 5 1 = 5 n=0 y 5 n = n=0 1 5n+1 yn . This series converges on (-5, 5). On the other hand, 1 1 d , = (5 - y)2 dy 5 - y , and so on (-5, 5), d 1 = (5 - y)2 dy ln k - 0 2+k n=0 yn 5n+1 ak is convergent by the Root = n=1 n 5n+1 y n-1 = n=0 n+1 n y . 5n+2 nav277 Homework 13 Odell (58340) Thus 10 Thus n+1 n y = 5n+2 f (y) = y 3 n=0 n=0 n + 1 n+3 y . 5n+2 ln(1 + x) - ln(1 - x) = 2 x+ x3 x5 + +... 3 5 Consequently, f (y) = n=3 n-2 n y . 5n-1 Consequently, = 2 n=1 1 x2n-1 . 2n - 1 017 10.0 points Find a power series representation for the function 1 + 5z f (z) = ln . 1 - 5z (Hint: remember properties of logs.) f (z) = 2 n=1 52n-1 2n-1 . z 2n - 1 018 10.0 points 1. f (z) = 2 n=1 (-1)n 52n 2n - 1 z 2n-1 Determine the interval of convergence of the series n=1 2. f (z) = n=1 1 2n z n 52n-1 2n-1 z 2n - 1 (-1)n 52n 2n-1 z 2n - 1 2 z 2n-1 2n - 1 52n-1 2n-1 z correct 2n - 1 n4 (x - 8)n . 3. f (z) = n=1 1. interval convergence = (7, 9) correct 2. interval convergence = (-9, -7] 3. interval convergence = (-9, -7) 4. interval convergence = (-, ) 5. interval convergence = [7, 9) 6. converges only at x = 8 Explanation: The given series has the form an (x - 8)n 4. f (z) = n=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 x3 xn , while ln(1 - x) = -x - = - x2 2 - -... 3 1 n x . n n=1 with an = n4 . Now lim n+1 an+1 = lim n an n 4 n=1 n = 1. nav277 Homework 13 Odell (58340) By the Ratio Test, therefore, the given series (i) converges when |x - 8| < 1, and (ii) diverges when |x - 8| > 1. On the other hand, at the points x - 8 = -1 and x - 8 = 1 the series reduces to 11 Consequently, I = C+ n=0 (-1)n x4n+3 . (2n + 1)(4n + 3) (-1)n n4 , n=1 n=1 n4 respectively. But by the Divergence Test, both of these diverge. Consequently, interval convergence = (7, 9) . 019 10.0 points Express the indefinite integral I = as a power series. tan-1 x2 dx 1. I = C + n=0 (-1)n x4n+3 correct (2n + 1)(4n + 3) (-1)n x4n+2 (2n + 1)(4n + 3) 2. I = C + n=0 3. I = n=0 (-1)n x4n+3 (2n + 1)(4n + 3) 4. I = C + n=0 (-1)n x4n+3 (4n + 3) (-1)n x2n+2 (2n + 1) 5. I = C + n=0 Explanation: We know that tan -1 x= n=0 x2n+1 . (-1) 2n + 1 n Replacing x with x2 , we get I = = n=0 tan-1 x2 dx (-1)n (x2 )2n+1 dx . 2n + 1 ...
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