M408L_HW14 - hyun (hh7953) – HW14 – gogolev – (57440)...

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Unformatted text preview: hyun (hh7953) – HW14 – gogolev – (57440) This print-out should have 20 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. 001 (part 1 of 2) 10.0 points For the series ∞ 1 002 (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 = (−1, 1] correct 4. interval convergence = [−1, 1) 5. interval convergence = (−2, 2) 6. interval convergence = (−1, 1) 7. converges only at x = 0 Explanation: Since R = 1, the given series (i) converges when |x| < 1, and (ii) diverges when |x| > 1. On the other hand, at the point x = 1 and x = −1, the series reduces to ∞ n=1 (−1)n n x, n+2 (i) determine its radius of convergence, R. 1. R = 1 correct 2. R = (−∞, ∞) 3. R = 1 2 4. R = 2 5. R = 0 Explanation: The given series has the form ∞ an n=1 xn an = (−1) . n+2 Now for this series, n with n=1 (−1)n , n+2 ∞ n=1 1 n+2 (i) R = 0 if it converges only at x = 0, (ii) R = ∞ if it converges for all x, while 0 < R < ∞, (iv) diverges when |x| > R. But n→∞ respectively. But by the Alternating Series Test, the first of these series converges. On the other hand, if we set an = then n→∞ (iii) if it converges when |x| < R, and 1 , n+2 bn = 1 , n lim lim an+1 an n+2 = lim |x| = | x| . n→∞ n+3 an n = lim = 1. n→∞ n + 2 bn By the Ratio Test, therefore, the given series converges when |x| < 1 and diverges when |x| > 1. Consequently, R=1 . 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] . hyun (hh7953) – HW14 – gogolev – (57440) keywords: 003 10.0 points while at x = −1 the series becomes ∞ 2 n=1 √ (−1)n 4n . Find the interval of convergence of the series ∞√ 4 n xn . n=1 But as n → ∞, neither of √ √ 4n, (−1)n 4n −→ 0 holds, so both of ∞ 1. interval of cgce = (−1, 1) correct 2. interval of cgce = (−1, 1] 3. converges only at x = 0 4. interval of cgce = (−4, 4] 5. interval of cgce = [−4, 4] 6. interval of cgce = [−1, 1) Explanation: When an = then an+1 an = √ √ 4 n xn , √ ∞ 4n, n=1 n=1 √ (−1)n 4n diverge. Consequently, the interval of convergence = (−1, 1) . 004 10.0 points Determine the radius of convergence, R, of the series ∞ xn . (n + 2)! n=1 = | x| But 4n + 4 xn+1 √ 4 n xn √ 4n + 4 √ = | x| 4n 4n + 4 = 1, 4n 1. R = 2 2. R = ∞ correct 4n + 4 . 4n 3. R = 1 4. R = 1 2 n→∞ lim 5. R = 0 Explanation: The given series has the form ∞ so an+1 = | x| . n→∞ an By the Ratio Test, therefore, the given series lim (ii) diverges when |x| > 1. (i) converges when |x| < 1, a n xn n=1 with an = Now for this series, We have still to check what happens at the endpoints x = ±1. At x = 1 the series becomes ∞√ 4n , n=1 1 . (n + 2)! (i) R = 0 if it converges only at x = 0, hyun (hh7953) – HW14 – gogolev – (57440) (ii) R = ∞ if it converges for all x, while 0 < R < ∞, (iv) diverges when |x| > R. But n→∞ 3 n | x| =∞ 2 But n→∞ lim (iii) if it converges when |x| < R, and for all x = 0. By the Root Test, therefore, the given series converges only at x = 0 . lim an+1 an = lim n→∞ 1 = 0. n+3 006 10.0 points By the Ratio Test, therefore, the given series converges for all x. Consequently, R=∞ . 005 10.0 points Find the interval of convergence of the power series ∞ n=0 3n xn . 3+3 4n 1. interval = 2. interval = Find the interval of convergence of the series ∞ (−n)n xn . 2n n=1 33 −, 44 11 −, 33 correct 1. interval of cgce = (−1, 1] 2. interval of cgce = (−2, 2] 3. converges only at x = 0 correct 4. interval of cgce = (−1, 1) 5. interval of cgce = [−2, 2) 6. interval of cgce = [−1, 1) (−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 3. interval = [−3, 3] 4. interval = 33 −, 44 5. interval = [−3, 3) 6. interval = 11 −, 33 Explanation: When Explanation: We first apply the Ratio Test to the infinite series ∞ 3n | x| n . 4n3 + 3 n=0 For this series 3n+1 4n3 + 3 an+1 | x| . = an 3n 4(n + 1)3 + 3 But 4n3 + 3 = 4(n + 1)3 + 3 4 3 n3 n+1 3 3 + n3 n xn = (−n)n n 2 1/n n | x| = . 2 4+ , hyun (hh7953) – HW14 – gogolev – (57440) so n→∞ 4 lim an+1 an n→∞ 3. R = 0 4. R = 3 correct 3 | x| 4 + 4 n+1 3 n 3 n3 3 n3 = lim + = 3 | x| . Thus the given power series will converge on 1 the interval (− 3 , 1 ). 3 1 For convergence at the endpoints x = ± 3 we have to check which, if any, of the series ∞ 1 3 Explanation: The given series has the form 5. R = ∞ a n xn n=1 n=0 4n3 1 , +3 ∞ n=0 (−1)n 4n3 1 +3 with an = Now for this series, (−1)n . 3n n5 converges. Now the Integral test ensures that the series ∞ 1 , n3 n=0 (i) R = 0 if it converges only at x = 0, (ii) R = ∞ if it converges for all x, while 0 < R < ∞, (iv) diverges when |x| > R. But (iii) if it converges when |x| < R, and hence by the Comparison test, the series ∞ n=0 4n3 1 +3 converges also. As ∞ n=0 1 = (−1) 3+3 4n n ∞ n=0 1 3+3 4n n→∞ lim an+1 an = lim n→∞ n 1 3 n+1 5 = 1 . 3 1 the power series thus converges at x = ± 3 . Consequently, By the Ratio Test, therefore, the given series converges when |x| < 3 and diverges when |x| > 3. Consequently, R=3 . 008 (part 2 of 2) 10.0 points (ii) Determine the interval of convergence of the series. 1. interval convergence = [−1, 1) 2. interval convergence = [−1/3, 1/3] 3. interval convergence = [−3, 3) 4. interval convergence = (−∞, ∞) interval = 11 −, 33 . 007 (part 1 of 2) 10.0 points For the series ∞ n=1 xn , 3n n5 (i) determine its radius of convergence, R. 1. R = 1 2. R = (−∞, ∞) hyun (hh7953) – HW14 – gogolev – (57440) 5. interval convergence = [−3, 3] correct 6. interval convergence = [−1, 1] 7. interval convergence = [−1/3, 1/3) 8. converges only at x = 0 Explanation: Since R = 3, the given series (i) converges when |x| < 3, and with an = Now for this series (i) R = 0 if it converges only at x = −5, 5. R = 1 2 5 Explanation: The given series has the form an (x + 5)n n=1 (−1)n . n 2n (ii) diverges when |x| > 3. On the other hand, at the point x = −3 and x = 3, the series reduces to ∞ (ii) R = ∞ if it converges for all x, while if R > 0, n=1 (−1)n , n5 ∞ n=1 1 n5 (iii) it coverges when |x + 5| < R, and (iv) diverges when |x + 5| > R. an+1 an But respectively. But ∞ (−1)n n5 ∞ = n=1 n=1 1 , n5 n→∞ lim = lim n→∞ 1 n =. 2(n + 1) 2 so by the p-Series Test with p = 5 both series converge. Consequently, the given series has interval convergence = [−3, 3] . 009 (part 1 of 2) 10.0 points For the series ∞ By the Ratio Test, therefore, the given series converges when |x + 5| < 2 and diverges when |x + 5| > 2. Consequently, R=2 . 010 (part 2 of 2) 10.0 points (ii) Determine the interval of convergence of the series. 1. interval convergence = (−7, −3] correct 2. interval convergence = (−7, −3) 3. converges only at x = −5 4. interval convergence = [3, 7] 5. interval convergence = [3, 7) n=1 (−1)n (x + 5)n , n 2n (i) determine its radius of convergence, R. 1. R = 5 2. R = 0 3. R = ∞ 4. R = 2 correct hyun (hh7953) – HW14 – gogolev – (57440) with 6. interval convergence = (−∞, ∞) Explanation: Since R = 2, the given series (i) converges when |x + 5| < 2, and But then, lim ck+1 1 k+1 = lim ck k k→∞ 2 3 6 5 a=− . 2 1 . 2 ck = k 3 1k , 2 k→∞ = (ii) diverges when |x + 5| > 2. By the Ratio Test, the series thus (i) converges when |x − a| < 2, and (ii) diverges when |x − a| > 2. Now at the points x − a = 2 and x − a = −2 the series reduces to ∞ ∞ On the other hand, at the points x + 5 = −2 and x + 5 = 2 the series reduces to ∞ n=1 1 , n ∞ n=1 (−1)n n respectively. But by the p-series Test with p = 1, the first of these series diverges; while by the Alternating Series Test the second is convergent. Consequently, interval convergence = (−7, −3] . 011 10.0 points k, k=0 k=0 3 (−1)k k 3 Determine the interval of convergence of the series ∞ respectively. But in both cases these series diverge by the Divergent Test. Consequently, the interval of convergence of the given series is 1 9 (a − 2, a + 2) = − , − 2 2 012 10.0 points k=0 k3 (2x + 5)k . k 4 19 , 22 = 1 9 − ,− 2 2 11 −, 22 Determine the interval of convergence of the infinite series ∞ 1. interval convergence = 2. interval convergence correct 3. interval convergence = k=0 (−1)k (k + 1)! 5x + 9 2k 7 11 , 55 77 −, 55 k . 1. interval conv. = 2. interval conv. = 4. interval convergence = (−∞, ∞) 5. series converges only at x = − Explanation: The given series has the form ∞ 5 2 9 3. series converges only at x = − cor5 rect 4. interval conv. = − 11 7 ,− 5 5 k=0 ck (x − a) k 5. interval conv. = (−∞, ∞) Explanation: hyun (hh7953) – HW14 – gogolev – (57440) There are three possibilities for the interval of convergence of the infinite series ∞ 7 Consequently, series converges only at x = − keywords: 9 . 5 k=0 ak (x − a)k . First set L = lim Then (i) when L > 0, the interval of convergence is given by 1 1 ; a− , a+ L L (ii) when L = 0, the interval of convergence is given by (−∞, ∞); (iii) when L = ∞, the interval of convergence reduces to the point {a}, i.e., the series converges only at x = a. For the given series, ∞ k→∞ ak+1 . ak If the series 013 10.0 points ∞ cn xn n=0 converges when x = −5 and diverges when x = 6, which of the following series converge without further restrictions on {cn }? ∞ A. n=0 ∞ cn cn (−5)n+1 cn (−2)n B. n=0 ∞ C. n=0 1. B only k k=0 (−1)k = (k + 1)! 5x + 9 2k (−1)k 2. A only k ∞ k=0 9 (k + 1)!5k x+ k 5 2 3. all of them correct , 4. C only 5. B and C only 6. A and B only 7. none of them 8. A and C only Explanation: A. The interval of convergence of series ∞ so the corresponding value of ak is ak In this case ak+1 ak (k + 2)!5k+1 =− 2k+1 5(k + 2) . = 2 Thus 5(k + 2) = ∞. L = lim 2 k→∞ (k + 1)!5k 2k 5k (k + 1)! = (−1)k . 2k cn xn n=0 hyun (hh7953) – HW14 – gogolev – (57440) contains (−5, 5). Since x = 1 belongs to this interval, the series ∞ 8 1 8 3. R1 = 2R2 = 4. R1 = R2 = 8 5. 2R1 = R2 = cn n=0 also converges. B. Since ∞ ∞ 1 8 1 correct 8 n=0 cn (−5)n+1 = −5 n=0 cn (−5)n 6. R1 = R2 = Explanation: When lim the series ∞ n=0 cn (−5)n+1 cn+1 = 8, n→∞ cn the Ratio Test ensures that the series ∞ converges. C. The interval of convergence of series ∞ n=0 cn xn is cn xn n=0 1 (i) convergent when |x| < 8 , and contains (−5, 5). Since x = −2 belongs to this interval, the series cn (−2)n 1 (ii) divergent when |x| > 8 . On the other hand, since n→∞ lim n=0 (n + 1)cn+1 ncn = lim n→∞ cn+1 , cn converges also. 014 10.0 points the Ratio Test ensures also that the series ∞ n cn xn−1 n=1 Compare the radius of convergence, R1 , of the series ∞ is (i) convergent when |x| < 1 , and 8 1 (ii) divergent when |x| > 8 . cn xn n=0 with the radius of convergence, R2 , of the series ∞ Consequently, R1 = R2 = 1 . 8 n cn xn−1 n=1 when n→∞ lim cn+1 cn = 8. 015 10.0 points 1. R1 = 2R2 = 8 2. 2R1 = R2 = 8 Find a power series representation for the function 1 . f ( x) = 2 − x3 hyun (hh7953) – HW14 – gogolev – (57440) ∞ ∞ 9 xn correct n 2n xn 2n 1. f (x) = − 2. f (x) = − 3. f (x) = − 4. f (x) = 2n x3n n=0 ∞ 3. f (x) = ln 2 − 4. f (x) = ln 2 − ∞ n=1 ∞ n=0 ∞ x3n 23n xn 2n+1 n=0 n=0 5. f (x) = − n=1 xn n2n ∞ ∞ n=0 ∞ x3n 23n x3n 2n+1 2n x3n correct 6. f (x) = ln 2 + n=1 xn 2n 5. f (x) = n=0 ∞ Explanation: We can either use the known power series representation ∞ 6. f (x) = n=0 Explanation: After simplification, f ( x) = 1 1 1 . = 3 2−x 2 1 − (x3 /2) 1 = 1−t 3 ∞ ln(1 − x) = − or the fact that x n=1 xn , n On the other hand, we know that t. n=0 n ln(1 − x) = − x ∞ 0 1 ds 1−s ds ∞ =− sn n=0 ∞ x 0 0 Replacing t with x /2, we thus obtain f ( x) = 1 2 ∞ =− . n=0 sn ds = − n=1 xn . n x3n 2n ∞ = n=0 x3n 2n+1 For then by properties of logs, 1 1 f (x) = ln 2 1 − x = ln 2 + ln 1 − x , 2 2 so that n=0 keywords: 016 10.0 points ∞ Find a power series representation for the function f (x) = ln(2 − x) . ∞ f (x) = ln 2 − 017 n=1 xn . n 2n 10.0 points 1. f (x) = n=0 xn n 2n ∞ Find a power series representation centered at the origin for the function xn n 2n f ( t) = t3 (2 − t)2 . 2. f (x) = ln 2 + n=0 hyun (hh7953) – HW14 – gogolev – (57440) ∞ 10 10.0 points 1. f (t) = n=2 ∞ n−1 n t 2n n−2 n t correct 2n−1 nn t 2n 1 2n−3 1 2n−1 t n 018 2. f (t) = n=3 ∞ Find a power series representation for the function f (z ) = ln 1 + 2z . 1 − 2z 3. f (t) = n=3 ∞ (Hint: remember properties of logs.) ∞ 4. f (t) = n=3 ∞ 1. f (z ) = n=1 ∞ 1 2n z 2n (−1)n 22n 2n−1 z 2n − 1 22n−1 2n−1 z correct 2n − 1 22n z 2n 2n − 1 1 z 2n−1 2n − 1 5. f (t) = n=2 t n 2. f (z ) = n=1 ∞ Explanation: By the known result for geometric series, 1 = 2−t 1 = 2 ∞ 3. f (z ) = n=1 ∞ 1 2 1− t 2 t 2 n ∞ 4. f (z ) = n=1 ∞ = n=0 1 2n+1 tn . 5. f (z ) = n=1 n=0 This series converges on (−2, 2). On the other hand, 1 d 1 , = 2 (2 − t) dt 2 − t and so on (−2, 2), d 1 = 2 (2 − t) dt ∞ ∞ Explanation: We know that ln(1 + x) = x − = n=1 x2 x3 + −... 2 3 ∞ (−1)n−1 n x, n tn 2n+1 ∞ while ln(1 − x) = −x − =− Thus ln(1 + x) − ln(1 − x) = 2 x+ x3 x5 + +... 3 5 ∞ n=0 = n=1 n 2n+1 t n−1 = n=0 n+1 n t. 2n+2 x2 x3 − −... 2 3 ∞ Thus ∞ n=1 1n x. n f ( t) = t 3 n=0 n+1 n t= 2n+2 ∞ n=0 n + 1 n+3 t . 2n+2 Consequently, ∞ f ( t) = n=3 n−2 n t. 2n−1 =2 n=1 1 x2n−1 . 2n − 1 hyun (hh7953) – HW14 – gogolev – (57440) Now by properties of logs, ln 1 + 2z 1 1 + 2z = ln 1 − 2z 2 1 − 2z But then t ∞ 11 f ( t) = 0 n=0 s4n+1 ds ∞ t 0 1 = ln(1 + 2z ) − ln(1 − 2z ) . 2 2 2 ∞ Thus f (z ) = and so ∞ = n=0 s4n+1 ds . n=1 1 (2z )2n−1 , 2n − 1 22n−1 2n−1 . z 2n − 1 Consequently, ∞ f ( t) = f (z ) = n=1 n=0 t4n+2 . 4n + 2 020 10.0 points 019 10.0 points Evaluate the definite integral t Find a power series representation for the function f (x) = x4 tan−1 x on (−1, 1). ∞ f ( t) = 0 s ds . 1 − s4 as a power series. ∞ 1. f (t) = n=4 ∞ t4n 4n + 2 t4n 4n t4n+2 4n + 2 1. f (x) = n=0 ∞ (−1)n n+5 x n+1 1 x2n+5 2n + 1 1 xn+5 n+1 (−1)n 2n+5 x correct 2n + 1 (−1)n x2n+5 (2n + 1)! (−1)n n+5 x (n + 1)! 2. f (x) = n=0 ∞ 2. f (t) = n=0 ∞ 3. f (x) = correct 4. f (x) = n=0 ∞ n=0 ∞ 3. f (t) = n=0 ∞ 4. f (t) = n=0 ∞ (−1)n t4n+2 4n + 2 (−1)n t4n 4n 5. f (t) = n=0 5. f (x) = n=0 ∞ Explanation: By the geometric series representation, 1 = 1−s and so s = 1 − s4 ∞ 6. f (x) = n=0 sn , n=0 ∞ Explanation: The interval of convergence of the geometric series 1 = 1 + x + x2 + . . . 1−x s4n+1 . n=0 hyun (hh7953) – HW14 – gogolev – (57440) is (−1, 1). Thus on (−1, 1) 1 = 1 − x 2 + x4 − . . . = 1 + x2 On the other hand, tan−1 x = 0 x ∞ 12 n=0 (−1)n x2n . 1 dt . 1 + t2 Thus on (−1, 1) tan −1 x ∞ x= 0 ∞ n=0 x 0 ∞ (−1)n t2n dt = n=0 (−1)n t2n dt (−1)n 2n+1 x . 2n + 1 = n=0 Consequently, on (−1, 1) ∞ f ( x) = n=0 (−1)n 2n+5 x . 2n + 1 ...
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This note was uploaded on 01/19/2010 for the course M 57440 taught by Professor Gogolev during the Fall '09 term at University of Texas at Austin.

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