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Unformatted text preview: pokharel (yp624) – HW14 – Radin – (57410) 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 = (−8, 8) 2. interval convergence = [−8, 8) 3. interval convergence = (−8, 8] 4. interval convergence = (−1, 1] correct 5. interval convergence = [−1, 1) 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. On the other hand, at the point x = 1 and x = −1, the series reduces to ∞ n=1 (−1)n n x, n+8 (i) determine its radius of convergence, R. 1. R = 0 2. R = 1 correct 3. R = 8 4. R = (−∞, ∞) 1 8 Explanation: The given series has the form 5. R = ∞ an n=1 xn an = (−1) . n+8 Now for this series, n with n=1 (−1)n , n+8 ∞ n=1 1 n+8 (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+8 bn = 1 , n lim lim an+1 an n+8 = lim |x| = | x| . n→∞ n+9 an n = lim = 1. n→∞ n + 8 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] . pokharel (yp624) – HW14 – Radin – (57410) 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. converges only at x = 0 2. interval of cgce = (−4, 4] 3. interval of cgce = (−1, 1) correct 4. interval of cgce = [−4, 4] 5. interval of cgce = (−1, 1] 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 + 3)! n=1 = | x| But 4n + 4 xn+1 √ 4 n xn √ 4n + 4 √ = | x| 4n 4n + 4 = 1, 4n 1. R = 3 2. R = 4n + 4 . 4n 1 3 3. R = 1 4. R = 0 n→∞ lim 5. R = ∞ correct 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 + 3)! (i) R = 0 if it converges only at x = 0, pokharel (yp624) – HW14 – Radin – (57410) (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+4 006 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 7n xn . 7+8 3n 1. interval = 2. interval = Find the interval of convergence of the series ∞ (−n)n xn . 2n n=1 77 −, 33 11 −, 77 correct 1. interval of cgce = (−2, 2] 2. interval of cgce = [−1, 1) 3. interval of cgce = (−1, 1] 4. interval of cgce = (−1, 1) 5. interval of cgce = [−2, 2) 6. converges only at x = 0 correct Explanation: When 3. interval = [−7, 7) 4. interval = [−7, 7] 5. interval = 6. interval = 11 −, 77 77 −, 33 Explanation: We first apply the Ratio Test to the infinite series ∞ 7n | x| n . 3n7 + 8 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 For this series 7n+1 3n7 + 8 an+1 | x| . = an 7n 3(n + 1)7 + 8 But 3n7 + 8 = 3(n + 1)7 + 8 3 8 n7 n+1 7 8 + n7 n xn = (−n)n n 2 1/n n | x| = . 2 3+ , pokharel (yp624) – HW14 – Radin – (57410) so n→∞ 4 lim an+1 an n→∞ 3. R = 1 8 n7 8 n7 4. R = = 7 | x| . = lim 7 | x| 3 + 3 n+1 7 n 1 4 + 5. R = 4 correct Explanation: The given series has the form ∞ Thus the given power series will converge on 1 the interval (− 7 , 1 ). 7 For convergence at the endpoints x = we have to check which, if any, of the series ∞ 1 ±7 a n xn n=1 n=0 1 , 7+8 3n ∞ n=0 1 (−1) 7+8 3n n with an = Now for this series, (−1)n . 4n n5 converges. Now the Integral test ensures that the series ∞ 1 , n7 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 an+1 an (iii) if it converges when |x| < R, and hence by the Comparison test, the series ∞ n=0 1 7+8 3n converges also. As ∞ n=0 1 = (−1) 7+8 3n n ∞ n=0 1 7+8 3n n→∞ lim = lim n→∞ 1 n 4 n+1 5 = 1 . 4 1 the power series thus converges at x = ± 7 . Consequently, By the Ratio Test, therefore, the given series converges when |x| < 4 and diverges when |x| > 4. Consequently, R=4 . 008 (part 2 of 2) 10.0 points (ii) Determine the interval of convergence of the series. 1. interval convergence = (−∞, ∞) 2. interval convergence = [−1, 1) 3. interval convergence = [−1, 1] 4. interval convergence = [−4, 4) 5. interval convergence = [−4, 4] correct interval = 11 −, 77 . 007 (part 1 of 2) 10.0 points For the series ∞ n=1 xn , 4n n5 (i) determine its radius of convergence, R. 1. R = (−∞, ∞) 2. R = 0 pokharel (yp624) – HW14 – Radin – (57410) 6. interval convergence = [−1/4, 1/4] 7. converges only at x = 0 n=1 5 Explanation: The given series has the form an (x + 3)n with an = Now for this series (i) R = 0 if it converges only at x = −3, Explanation: Since R = 4, the given series 8. interval convergence = [−1/4, 1/4) (−1)n . n 5n (ii) diverges when |x| > 4. (i) converges when |x| < 4, and On the other hand, at the point x = −4 and x = 4, the series reduces to ∞ (ii) R = ∞ if it converges for all x, while if R > 0, (−1)n n5 ∞ , n=1 n=1 1 n5 (iii) it coverges when |x + 3| < R, and (iv) diverges when |x + 3| > R. an+1 an But respectively. But ∞ n=1 (−1)n = n5 ∞ n=1 1 , n5 n→∞ lim = lim n→∞ 1 n =. 5(n + 1) 5 so by the p-Series Test with p = 5 both series converge. Consequently, the given series has interval convergence = [−4, 4] . 009 (part 1 of 2) 10.0 points For the series ∞ By the Ratio Test, therefore, the given series converges when |x + 3| < 5 and diverges when |x + 3| > 5. Consequently, R=5 . 010 (part 2 of 2) 10.0 points n=1 (−1)n (x + 3)n , n n5 (ii) Determine the interval of convergence of the series. 1. interval convergence = (−8, 2) 2. interval convergence = [−2, 8] 3. interval convergence = [−2, 8) 4. interval convergence = (−∞, ∞) 5. converges only at x = −3 6. interval convergence = (−8, 2] correct (i) determine its radius of convergence, R. 1. R = ∞ 2. R = 5 correct 3. R = 0 4. R = 3 5. R = 1 5 pokharel (yp624) – HW14 – Radin – (57410) Explanation: Since R = 5, the given series (i) converges when |x + 3| < 5, and (ii) diverges when |x + 3| > 5. On the other hand, at the points x + 3 = −5 and x + 3 = 5 the series reduces to ∞ 6 with ck = k 2 But then, k→∞ 3 2 k , 5 a=− . 3 lim ck+1 3 k+1 = lim ck k k→∞ 2 2 = 3 . 2 n=1 1 , n ∞ (−1)n n By the Ratio Test, the series thus (i) converges when |x − a| < (ii) diverges when |x − a| > Now at the points x − a = the series reduces to ∞ ∞ 2 3 2 3, 2 3. and n=1 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 = (−8, 2] . 011 10.0 points 2 and x − a = − 3 k, k=0 k=0 2 (−1)k k 2 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 7 2 2 (a − 3 , a + 3 ) = − , −1 3 012 10.0 points k=0 k2 (3x + 5)k . 2k 1. interval convergence = (−1, 1 ) 2. interval convergence = (−∞, ∞) 3. series converges only at x = − 4. interval convergence = rect 5. interval convergence = 7 1, 3 5 3 cor- Determine the interval of convergence of the infinite series ∞ k=0 (−1)k (k + 1)! 5x + 8 2k 6 ,2 5 k . 1. interval conv. = 7 − , −1 3 8 2. series converges only at x = − cor5 rect 3. interval conv. = 66 −, 55 Explanation: The given series has the form ∞ 4. interval conv. = (−∞, ∞) 5. interval conv. = −2, − 6 5 k=0 ck (x − a)k Explanation: pokharel (yp624) – HW14 – Radin – (57410) There are three possibilities for the interval of convergence of the infinite series ∞ 7 Consequently, series converges only at x = − keywords: 8 . 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 = −4 and diverges when x = 6, which of the following series converge without further restrictions on {cn }? ∞ A. n=0 ∞ cn 4n cn n=0 ∞ B. C. n=0 cn (−2)n 1. all of them k k=0 (−1)k = (k + 1)! 5x + 8 2k (−1)k 2. A only k ∞ k=0 8 (k + 1)!5k x+ k 5 2 , 3. B and C only correct 4. A and C only 5. C only 6. none of them so the corresponding value of ak is ak = (−1)k In this case ak+1 ak =− (k + 2)!5k+1 2k+1 = Thus 5(k + 2) = ∞. L = lim 2 k→∞ 5(k + 2) . 2 (k + 1)!5k 2k 5k (k + 1)! . 2k 7. B only 8. A and B only Explanation: A. The series ∞ cn 4n n=0 pokharel (yp624) – HW14 – Radin – (57410) need not converge even though ∞ 8 3. R1 = 2R2 = 2 (−1)n cn 4n 4. R1 = 2R2 = 5. 2R1 = R2 = 6. R1 = R2 = Explanation: When lim 1 2 1 2 1 correct 2 n=0 converges. B. The interval of convergence of series ∞ cn xn n=0 contains (−4, 4). Since x = 1 belongs to this interval, the series ∞ cn n=0 cn+1 = 2, n→∞ cn the Ratio Test ensures that the series ∞ also converges. C. The interval of convergence of series ∞ n=0 cn y n is 1 (i) convergent when |y | < 2 , and cn xn n=0 contains (−4, 4). Since x = −2 belongs to this interval, the series cn (−2)n 1 (ii) divergent when |y | > 2 . 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 y n−1 n=1 Compare the radius of convergence, R1 , of the series ∞ is (i) convergent when |y | < 1 , and 2 1 (ii) divergent when |y | > 2 . cn y n n=0 with the radius of convergence, R2 , of the series ∞ Consequently, R1 = R2 = 015 1 . 2 n cn y n−1 n=1 when n→∞ lim cn+1 cn = 2. 10.0 points 1. R1 = R2 = 2 2. 2R1 = R2 = 2 Find a power series representation for the function 1 . f ( x) = 2 − x3 pokharel (yp624) – HW14 – Radin – (57410) ∞ 9 tn n2n 1. f (x) = n=0 ∞ x3n correct 2n+1 x3n 23n xn 2n+1 ∞ 3. f (t) = − 4. f (t) = n=1 ∞ 2. f (x) = n=0 ∞ n=0 tn n 2n ∞ 3. f (x) = − 4. f (x) = n=0 5. f (t) = ln 2 − 6. f (t) = ln 2 + n=1 ∞ tn correct n 2n tn 2n ∞ 2n x3n n=0 ∞ n=1 5. f (x) = − 6. f (x) = − x3n 23n 2n x3n n=0 ∞ Explanation: We can either use the known power series representation ∞ 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 (t) = ln 2 1 − t = ln 2 + ln 1 − t , 2 2 so that n=0 keywords: 016 10.0 points ∞ Find a power series representation for the function f (t) = ln(2 − t) . ∞ f (t) = ln 2 − 017 n=1 tn . n 2n 10.0 points 1. f (t) = ln 2 + n=0 ∞ tn n 2n tn 2n Find a power series representation centered at the origin for the function f ( t) = t3 (4 − t)2 . 2. f (t) = ln 2 − n=0 pokharel (yp624) – HW14 – Radin – (57410) ∞ 10 10.0 points 1. f (t) = n=2 ∞ n−1 n t 4n 1 4n−1 tn 018 2. f (t) = n=2 ∞ Find a power series representation for the function f (y ) = ln 1 + 4y . 1 − 4y 3. f (t) = n=3 ∞ n−2 n t correct 4n−1 1 4n−3 nn t 4n t n (Hint: remember properties of logs.) ∞ 4. f (t) = n=3 ∞ 1. f (y ) = n=1 ∞ 42n y 2n 2n − 1 1 y 2n−1 2n − 1 (−1)n 42n 2n−1 y 2n − 1 42n−1 2n−1 y correct 2n − 1 1 2n y 2n 5. f (t) = n=3 2. f (y ) = n=1 ∞ Explanation: By the known result for geometric series, 1 = 4−t 1 = 4 ∞ 3. f (y ) = n=1 ∞ 1 4 1− t 4 t 4 n ∞ 4. f (y ) = n=1 ∞ = n=0 1 4n+1 tn . 5. f (y ) = n=1 n=0 This series converges on (−4, 4). On the other hand, 1 d 1 , = 2 (4 − t) dt 4 − t and so on (−4, 4), d 1 = 2 (4 − t) dt ∞ ∞ Explanation: We know that ln(1 + x) = x − = n=1 x2 x3 + −... 2 3 ∞ (−1)n−1 n x, n tn 4n+1 ∞ while ln(1 − x) = −x − =− Thus ln(1 + x) − ln(1 − x) = 2 x+ x3 x5 + +... 3 5 ∞ n=0 = n=1 n 4n+1 t n−1 = n=0 n+1 n t. 4n+2 x2 x3 − −... 2 3 ∞ Thus ∞ n=1 1n x. n f ( t) = t 3 n=0 n+1 n t= 4n+2 ∞ n=0 n + 1 n+3 t . 4n+2 Consequently, ∞ f ( t) = n=3 n−2 n t. 4n−1 =2 n=1 1 x2n−1 . 2n − 1 pokharel (yp624) – HW14 – Radin – (57410) Now by properties of logs, ln 1 + 4y 1 1 + 4y = ln 1 − 4y 2 1 − 4y But then y ∞ 11 f (y ) = 0 n=0 s4n+1 ds ∞ y 0 1 ln(1 + 4y ) − ln(1 − 4y ) . = 2 2 2 ∞ Thus f (y ) = and so ∞ = n=0 s4n+1 ds . n=1 1 (4y )2n−1 , 2n − 1 42n−1 2n−1 . y 2n − 1 Consequently, ∞ f (y ) = f (y ) = n=1 n=0 y 4n+2 . 4n + 2 020 10.0 points 019 10.0 points Evaluate the definite integral y Find a power series representation for the function f (x) = x2 tan−1 x on (−1, 1). ∞ f (y ) = 0 s ds . 1 − s4 as a power series. ∞ 1. f (y ) = n=0 ∞ (−1)n y 4n+2 4n + 2 y 4n 4n + 2 y 4n+2 4n + 2 y 4n 4n (−1)n y 4n 4n correct 1. f (x) = n=0 ∞ (−1)n x2n+3 (2n + 1)! (−1)n 2n+3 x correct 2n + 1 1 xn+3 n+1 (−1)n n+3 x (n + 1)! 1 x2n+3 2n + 1 (−1)n n+3 x n+1 2. f (x) = n=0 ∞ 2. f (y ) = n=4 ∞ 3. f (x) = n=0 ∞ 3. f (y ) = n=0 ∞ 4. f (y ) = n=0 ∞ 4. f (x) = n=0 ∞ 5. f (y ) = 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 pokharel (yp624) – HW14 – Radin – (57410) 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+3 x . 2n + 1 ...
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This note was uploaded on 12/23/2009 for the course M 408L taught by Professor Radin during the Fall '08 term at University of Texas.

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