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Unformatted text preview: fh2825 Homework 13 Radin (58415) This printout should have 19 questions. Multiplechoice 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)k1 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 pseries 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)n1 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 pseries 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 pseries 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) = . t2 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)n1 2n+1 tn
n=0 5. f (t) =
n=0 1 2n+1 tn Explanation: We know that 1 = 1 + x + x2 + . . . = 1x On the other hand, 1 1 1 . =  t2 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 termbyterm 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 = 1x xn .
n=0 fh2825 Homework 13 Radin (58415) Thus 8 Now
x f (z) = z
n=0 (4z) n 0 1 dt = tan1 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) = tan1 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 + . . . , 1x 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 6n1 n=2 2. f (z) =
n=2 n1 n z 6n n n z 6n n2 n z correct 6n1 1 6n3 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 6z 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 n1 =
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 n2 n z . 6n1 Consequently, = 2
n=1 1 x2n1 . 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 52n1 2n1 . y 2n  1 018 10.0 points 1. f (y) =
n=1 (1)n 52n 2n  1 y 2n1 Determine the interval of convergence of the series n=1 2. f (y) =
n=1 2 y 2n1 2n  1 52n1 2n1 y correct 2n  1 y 2n1 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 52n1 n=1 2n  1 1 2n y n 5. f (y) =
n=1 6. f (y) = 2
n=1 (1)n 52n 2n1 y 2n  1 Explanation: We know that x2 x3 ln(1 + x) = x  + ... 2 3 =
n=1 (1)n1 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. tan1 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 tan1 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.
 Spring '08
 RAdin

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