homework 13 - fh2825 Homework 13 Radin (58415) This...

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Unformatted text preview: fh2825 Homework 13 Radin (58415) 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 (-1)k-1 sin k=1 1 k k=1 1 k is absolutely convergent, conditionally convergent or divergent. 1. conditionally convergent correct 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 k > sin 1 k+1 3. series is divergent Explanation: Th given series can be written as an alternating series (-1)k ak k=1 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 (-1)n-1 sin k=1 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 (-4)m+1 54m For then ak and positive terms and bk are series with is absolutely convergent, conditionally convergent or divergent. 1. conditionally convergent 2. divergent 3. absolutely convergent correct Explanation: 1 sin ak k = lim lim 1 k k bk k = lim 0 sin = 1 > 0. fh2825 Homework 13 Radin (58415) The given series can be written in the form 2 m=1 (-4)m+1 = 54m is absolutely convergent, conditionally convergent, or divergent. 1. divergent correct 2. absolutely convergent an m=1 with am = (-1)m+1 But then 4m+1 4 = (-1)m+14 4 4m 5 5 = 4 ; 54 m . 3. conditionally convergent Explanation: The given series has the form am+1 am am+1 am in particular, m ak k=1 lim 4 = 4 < 1; 5 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 3(k!) . ek k+1 an+1 (k + 1)! ek = = . k+1 an k! e e Thus k lim m=1 (-4)m+1 = 54m ak+1 ak = . an m=1 is to note that |am | = so that 4 4m+1 = 4 4 4m 5 5 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 = 4/(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=2 4 m(ln m)5 converges or diverges. 1. diverges 2. converges correct Explanation: The given series can be written in the form Determine whether the series 3e k=1 -k k! m=1 f (n) fh2825 Homework 13 Radin (58415) where f is the function defined on (1, ) by f (x) = 4 . x(ln x)5 Consequently, k 3 lim ak+1 ak = 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 = 2 1 - (ln x)4 = t 2 1 1 - , 4 (ln 2) (ln t)4 keywords: 006 10.0 points in which case, t f (x) dx = lim 2 t f (x) dx 2 = lim t 1 1 - 4 (ln 2) (ln t)4 = 1 . (ln 2)4 Find the interval of convergence of the power series 5xn . n n=1 1. interval of cgce = [-1, 1) correct 2. interval of cgce = (-1, 1) 3. interval of cgce = (-5, 5) 4. interval of cgce = (-5, 5] 5. interval of cgce = [-5, 5) 6. interval of cgce = [-5, , 5] 7. interval of cgce = [-1, 1] 8. interval of cgce = (-1, 1] Explanation: When Consequently, by the Integral Test, the given series converges . 005 10.0 points Determine whether the series k=1 2k k 2 k! converges or diverges. 1. diverges 2. converges correct Explanation: The given series can be written as ak , k=1 ak = 2k k 2 . k! 5xn an = , n But then ak+1 ak 2k+1 (k + 1)2 k! = (k + 1)! 2k k 2 = 2 1 1+ k+1 k 2 then xn+1 n = n n+1 x n |x| n = |x| = n+1 n+1 an+1 an . . fh2825 Homework 13 Radin (58415) Thus n 4 xn . n+2 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 5 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 5 (-1)n n lim an+1 an = lim |x| n n+2 = |x| . n+3 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 = [-2, 2) 2. interval convergence = (-2, 2) 3. interval convergence = (-2, 2] 4. converges only at x = 0 5. interval convergence = [-1, 1) 6. interval convergence = (-1, 1) 7. interval convergence = (-1, 1] correct 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+2 (i) determine its radius of convergence, R. 1. R = 1 correct 2. R = 1 2 3. R = 2 4. R = (-, ) 5. R = 0 Explanation: The given series has the form an n=1 fh2825 Homework 13 Radin (58415) On the other hand, at the point x = 1 and x = -1, the series reduces to 5 n=1 (-1)n , n+2 n=1 1 n+2 We first apply the root test to the infinite series 8n + 5 n n |x| . 6n 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+2 then n an = 8n + 5 |x| 6n - 4 |x| 3 1 bn = , n as n . Thus the given power series will converge on the interval ( - 6/8, 6/8). 3 For convergence at the endpoints x = 4 we have to check if lim n an = lim = 1. n n + 2 bn 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] . n=0 8n + 5 6n n 3 4 n = n=0 an converges, or if n=0 8n + 5 6n n 3 - 4 n = n=0 bn converges. In the first case an = 8n + 5 6n n keywords: 009 10.0 points 3 4 n = 8n + 5 8n n , in which case Find the interval of convergence of the power series n lim an = e5/8 = 0; by the Divergence test, therefore the series n=1 8n + 5 6n n xn . an n=1 1. interval of convergence = - 6, 6 2. interval of convergence = 4 4 - , 3 3 diverges. On the other hand, bn = 8n + 5 6n = (-1)n n 3 3 3. interval of convergence = - , 4 4 4. interval of convergence = correct 3 3 - , 4 4 3 4 8n + 5 8n - n n = (-1)n an . But, as we have seen, n lim an = e5/8 . 4 4 5. interval of convergence = - , 3 3 Explanation: Thus bn oscillates between e5/8 and -e5/8 as n ; in particular, n lim bn fh2825 Homework 13 Radin (58415) 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 = 6/8 and so its interval of convergence = 3 3 - , 4 4 . 011 10.0 points Find a power series representation for the function 1 f (t) = . t-2 keywords: 010 10.0 points 1. f (t) = n=0 (-1)n 2n tn Find the interval of convergence of the series (-n)n xn . 2n n=1 2. f (t) = - 3. f (t) = - 4. f (t) = 1 2n+1 2 n tn tn correct n=0 n=0 1. interval of cgce = (-2, 2] 2. converges only at x = 0 correct 3. interval of cgce = [-2, 2) 4. interval of cgce = (-1, 1) 5. interval of cgce = (-1, 1] 6. interval of cgce = [-1, 1) Explanation: When (-1)n-1 2n+1 tn n=0 5. f (t) = n=0 1 2n+1 tn Explanation: We know that 1 = 1 + x + x2 + . . . = 1-x On the other hand, 1 1 1 . = - t-2 2 1 - (t/2) Thus 1 f (t) = - 2 n xn . n=0 (-n)n xn , an = 2n 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 t 2 1 = - 2 n=0 1 n t . 2n 2n = n |x| . 2 Consequently, n |x| = lim n 2 f (t) = - 1 2n+1 tn n=0 fh2825 Homework 13 Radin (58415) with |t| < 2. 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 42 x f (x) = . (x2 + 42 )2 As x = 2 lies in (-4, 4), we thus see that f (2) = 2 . 25 7 (Hint: differentiate the power series expansion of (x2 + 42 )-1 .) 1. f (2) = 2. f (2) = 4 5 4 25 keywords: 013 10.0 points 1 3. f (2) = 10 4. f (2) = 2 correct 25 Find a power series representation for the function z f (z) = . 4z + 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 (z) = n=0 22n z n 2. f (z) = n=0 (-1)n 2n z n 3. f (z) = n=0 2n z n 4. f (z) = n=0 22n z n+1 5. f (z) = n=0 (-1)n 22n z n+1 correct 6. f (z) = n=0 (-1)n 2n z n+1 Explanation: After simplification, f (z) = z z = . 4z + 1 1 - (-4z) On the other hand, 1 = 1-x xn . n=0 fh2825 Homework 13 Radin (58415) Thus 8 Now x f (z) = z n=0 (-4z) n 0 1 dt = tan-1 x , 1 + t2 while (-1) 2 z n 2n n = z n=0 . 0 x Consequently, (-1) x n=0 n 2n dt = n=0 (-1)n 2n+1 x . 2n + 1 f (z) = n=0 (-1)n 22n z n+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. = [-6, 6) 3. interval of cgce. = 1 1 - , 6 6 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 = 2n + 1 x 2n + 3 6 x 6 2 (-1)n x (2n + 1) 6 2n+1 . , 4. interval of cgce. = [-6, 6] correct 5. interval of cgce. = 6. interval of cgce. = 1 1 - , 6 6 1 1 - , 6 6 n lim an+1 an 2 = . 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 fh2825 Homework 13 Radin (58415) 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] . as n , so Test. 016 9 an is convergent by Ratio 10.0 points Find a power series representation centered at the origin for the function f (z) = z3 (6 - z)2 zn . keywords: 015 10.0 points 1. f (z) = Which, if any, of the following statements are true? A. The Root Test can be used to determine whether the series 1 6n-1 n=2 2. f (z) = n=2 n-1 n z 6n n n z 6n n-2 n z correct 6n-1 1 6n-3 zn k=1 k 3 + k2 3. f (z) = n=3 converges or diverges. B. The Ratio Test can be used to determine whether the series 1/n! n= 4. f (z) = n=3 5. f (z) = n=3 converges or diverges. 1. neither of them 2. A only 3. B only correct 4. both of them Explanation: A. False: when ak = then (|ak |) 1/k Explanation: By the known result for geometric series, 1 1 = z 6-z 6 1- 6 = 1 6 n=0 z 6 n = n=0 1 6n+1 zn . This series converges on (-6, 6). On the other hand, k , 3 + k2 /(1 + k ) 2 1/k 1 d 1 , = 2 (6 - z) dz 6 - z and so on (-6, 6), - 1 = k 1/k so the Root Test is inconclusive. B. True: when an = 1/n!, then an+1 an 1 = - 0 n+1 d 1 = (6 - z)2 dz n=0 zn 6n+1 = n=1 n 6n+1 z n-1 = n=0 n+1 n z . 6n+2 fh2825 Homework 13 Radin (58415) Thus 10 Thus n+1 n z = 6n+2 f (z) = z 3 n=0 n=0 n + 1 n+3 z . 6n+2 ln(1 + x) - ln(1 - x) = 2 x+ x3 x5 + +... 3 5 Consequently, f (z) = n=3 n-2 n z . 6n-1 Consequently, = 2 n=1 1 x2n-1 . 2n - 1 017 10.0 points Find a power series representation for the function 1 + 5y f (y) = ln . 1 - 5y (Hint: remember properties of logs.) f (y) = 2 n=1 52n-1 2n-1 . y 2n - 1 018 10.0 points 1. f (y) = n=1 (-1)n 52n 2n - 1 y 2n-1 Determine the interval of convergence of the series n=1 2. f (y) = n=1 2 y 2n-1 2n - 1 52n-1 2n-1 y correct 2n - 1 y 2n-1 n3 (x - 3)n . 3. f (y) = 2 1. converges only at x = 3 2. interval convergence = (2, 4) correct 3. interval convergence = [2, 4) 4. interval convergence = (-4, -2] 5. interval convergence = (-4, -2) 6. interval convergence = (-, ) Explanation: The given series has the form an (x - 3)n 4. f (y) = n=1 52n-1 n=1 2n - 1 1 2n y n 5. f (y) = n=1 6. f (y) = 2 n=1 (-1)n 52n 2n-1 y 2n - 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 = n3 . Now lim n+1 an+1 = lim n an n 3 n=1 n = 1. fh2825 Homework 13 Radin (58415) By the Ratio Test, therefore, the given series (i) converges when |x - 3| < 1, and (ii) diverges when |x - 3| > 1. On the other hand, at the points x - 3 = -1 and x - 3 = 1 the series reduces to 11 Consequently, I = C+ n=0 (-1)n y 4n+3 . (2n + 1)(4n + 3) (-1)n n3 , n=1 n=1 n3 respectively. But by the Divergence Test, both of these diverge. Consequently, interval convergence = (2, 4) . 019 10.0 points Express the indefinite integral I = as a power series. tan-1 y 2 dy 1. I = C + n=0 (-1)n y 2n+2 (2n + 1) (-1)n y 4n+3 correct (2n + 1)(4n + 3) (-1)n y 4n+3 (4n + 3) 2. I = C + n=0 3. I = C + n=0 4. I = n=0 (-1)n y 4n+3 (2n + 1)(4n + 3) 5. I = C + n=0 (-1)n y 4n+2 (2n + 1)(4n + 3) Explanation: We know that tan -1 y= n=0 y 2n+1 . (-1) 2n + 1 n Replacing y with y 2 , we get I = = n=0 tan-1 y 2 dy (-1)n (y 2 )2n+1 dy . 2n + 1 ...
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This note was uploaded on 04/30/2008 for the course M 408L taught by Professor Radin during the Spring '08 term at University of Texas at Austin.

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