# HW14-solutions - truong(at28889 HW14 zupan(56530 This...

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Unformatted text preview: truong (at28889) – HW14 – zupan – (56530) This print-out should have 25 questions. Multiple-choice questions may continue on the next column or page – ﬁnd all choices before answering. 001 1 We have still to check for convergence at x = ±1. But when x = 1, the series reduces to ∞ 6 √ n n=1 10.0 points Find the interval of convergence of the power series ∞ 6xn √ . n which diverges by the p-series test with p = 1 ≤ 1. On the other hand, when x = −1, the 2 series reduces to ∞ n=1 1. (−6, 6] n=1 6 (−1)n √ n which converges by the Alternating Series Test. Thus the 2. [−1, 1) correct interval of convergence = [−1, 1) . 3. (−1, 1] 002 4. [−6, , 6] 10.0 points Determine the radius of convergence, R, of the series ∞ (−4)n n x . n+2 5. [−6, 6) 6. [−1, 1] n=1 1 correct 4 7. (−6, 6) 1. R = 8. (−1, 1) 2. R = 0 Explanation: When 3. R = ∞ 6xn an = √ , n 4. R = 2 then √ n xn+1 = √ · n n+1 x √ |x| n n = |x| = √ n+1 n+1 5. R = 4 an+1 an Thus lim n→∞ an+1 an 6. R = . 1 2 Explanation: The given series has the form ∞ a n xn = |x| . By the Ratio Test, therefore, the given series converges when |x| < 1, and diverges when |x| > 1. n=1 with an = (−4)n . n+2 truong (at28889) – HW14 – zupan – (56530) 2 then Now for this series, (i) R = 0 if it converges only at x = 0, (ii) R = ∞ if it converges for all x, an+1 xn+1 5n + 1 5n + 1 = − = |x| an 5n + 6 xn 5n + 6 But while if R > 0, lim n→∞ (iii) it converges when |x| < R, and 5n + 1 = 1, 5n + 6 in which case (iv) diverges when |x| > R. But . lim n→∞ an+1 an = |x| . By the Ratio Test, therefore, the given series lim n→∞ an+1 an 4(n + 2) = lim = 4. n→∞ n+3 By the Ratio Test, therefore, the given series converges when |x| < 1/4 and diverges when |x| > 1/4. Consequently, R = 003 1 . 4 10.0 points (i) converges when |x| < 1, (ii) diverges when |x| > 1. We have still to check what happens at the endpoints x = ±1. At x = −1 the series becomes ∞ 1 5n + 1 n=1 which diverges by the Inegral Test. On the other hand, at x = 1, the series becomes ∞ Find the interval of convergence of the series ∞ xn . (−1)n 5n + 1 n=1 1. interval of cgce = [−1, 1) 2. interval of cgce = (−1, 1] correct n=1 which converges by the Alternating Series Test. Consequently, the interval of convergence = (−1, 1] . 3. converges only at x = 0 4. interval of cgce = (−5, 1] 5. interval of cgce = [−1, 5] 004 10.0 points Determine the radius of convergence, R, of the series ∞ xn . (n + 2)! n=1 6. interval of cgce = (−1, 1) 1. R = 7. interval of cgce = [−1, 1] (−1)n 5n + 1 1 2 2. R = 1 8. interval of cgce = (−∞, ∞) Explanation: When an xn = (−1)n , 5n + 1 3. R = 0 4. R = ∞ correct 5. R = 2 truong (at28889) – HW14 – zupan – (56530) Explanation: The given series has the form ∞ a n xn Explanation: We ﬁrst apply the root test to the inﬁnite series ∞ n=1 n=1 with an 1 = . (n + 2)! (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. n→∞ = lim n→∞ 1 = 0. n+3 n=0 7 7 1. − , 5 5 n 5n + 2 7n n 7 5 ∞ = an n=0 converges, or if ∞ n 5n + 2 7n 7 − 5 n ∞ bn = n=0 converges. In the ﬁrst case Find the interval of convergence of the power series n=1 5 |x| 7 −→ we have to check if 10.0 points 5n + 2 7n 5n + 2 |x| 7n For convergence at the endpoints x = ± n=0 R = ∞ . ∞ |x|n . as n → ∞. Thus the given power series will converge on the interval (−7/5, 7/5). ∞ By the Ratio Test, therefore, the given series converges for all x. Consequently, 005 n 5n + 2 7n |an| = |an |1/n = But an+1 an |an| , For this series Now for this series, lim 3 an = 5n + 2 7n n 7 5 n = 5n + 2 5n n in which case n n x . lim an = e2/5 = 0; n→∞ by the Divergence Test, therefore the series ∞ an 2. [−7, 7) n=1 3. 5 5 − , 7 7 diverges. On the other hand, 5 5 4. − , 7 7 5. 7 7 − , 5 5 bn = correct 5n + 2 7n = (−1) n n 7 − 5 n 5n + 2 5n n = (−1)n an . , 7 5 truong (at28889) – HW14 – zupan – (56530) But, as we have seen, it’s more convenient to use the Root Test to determine the interval of convergence. For then lim an = e2/5 . n→∞ 2/5 Thus bn oscillates between e n → ∞; in particular, 4 2/5 and −e |an | as 1/n n 1/n nx (−n) n 6 = = n |x| . 6 But lim bn n→∞ n |x| = ∞ n→∞ 6 for all x = 0. By the Root Test, therefore, the given series lim does not exist. Again by the Divergence Test, therefore, the series ∞ converges only at x = 0 . bn n=1 diverges. Consequently, the given power series does not converge at x = ±7/5 and so its interval of convergence = 006 7 7 − , 5 5 . 10.0 points Find the interval of convergence of the series ∞ (−n)n xn . 6n n=1 1. [−6, 6) 007 10.0 points Find the radius of convergence and interval of ∞ (−4)n xn √ convergence of the series 5 n+2 n=1 1 1 1 1. R = , I = − , 4 4 4 1 1 1 2. R = , I = − , 4 4 4 1 3. R = , I = 4 1 1 correct − , 4 4 4. diverges everywhere 5. R = 4, I = (−4, 4] Explanation: (−4)n xn an = √ , so 5 n+2 √ an+1 4n+1 |x|n+1 5 n + 2 · n n lim = lim √ 5 n→∞ an n→∞ 4 |x| n+3 2. (−1, 1] 3. [−1, 1) 4. (−6, 6] = lim 4|x| 5 n→∞ 5. converges only at x = 0 correct 6. (−1, 1) Explanation: When an (−n)n xn = , 6n n+2 n+3 = 4|x|, so by the Ratio Test, the series converges 1 1 when 4|x| < 1. Then |x| < , so R = . 4 4 1 When x = − , we get the divergent p-series 4 ∞ 1 1 √ . When x = , we get the series 5 4 n+2 n=1 truong (at28889) – HW14 – zupan – (56530) ∞ (−1)n √ , which converges by the Alternat5 n+2 n=1 1 1 ing Series Test. Thus, I = − , . 4 4 008 in particular, the series converges on the interval (−7, −5). On the other hand, at the points x + 6 = −1 and x + 6 = 1 the series reduces to ∞ ∞ √ ∞ n n(−1) , n=1 √ n n=1 respectively, both of which diverge by the Divergence test. Consequently, n(x + 6)n . interval of cgce = (−7, −5) . n=1 1. [5, 7) 009 2. (5, 7] 10.0 points Determine the radius of convergence, R, of the power series 3. (−7, −5] ∞ 4. (−7, −5) correct n=1 5. series converges only at x = −6 2. R = 7. [−7, −5) 8. series converges only at x = 6 Explanation: The given series has the form 3. R = 3 5. R = ∞ an (x + 6)n n=1 6. R = an = √ n. = so lim n→∞ 1 correct 4 Explanation: The given series has the form Now for this series, √ 1 3 4. R = 4 ∞ where (−4)n √ (x + 3)n . 4 n 1. R = 0 6. (5, 7) an+1 an √ 10.0 points Determine the interval of convergence of the power series 5 ∞ n+1 √ = n an+1 an (−1)n an (x + 3)n n+1 , n n=1 where = 1. By the Ratio test, therefore, the given series (a) converges for all |x + 6| < 1, and (b) diverges for all |x + 6| > 1; 4n an = √ . 4 n Now for this series, (i) R = 0 if it converges only at x = −3, (ii) R = ∞ if it converges for all x, truong (at28889) – HW14 – zupan – (56530) 6 with while 0 < R < ∞ (iii) if it converges for |x + 3| < R, and (iv) diverges for |x + 3| > R. 3 2 k 4 n , n+1 lim k→∞ 3 k+1 ck+1 = lim ck k k→∞ 2 an+1 an lim n→∞ = 4. (a) converges for |x + 3| < 1/4, and (b) diverges for |x + 3| > 1/4. Consequently, 3 . 2 2 , and 3 2 (ii) diverges when |x − a| > . 3 2 2 Now at the points x − a = and x − a = − 3 3 the series reduces to ∞ 1 R = . 4 010 ∞ k=0 (−1)k k 3 k , k=0 10.0 points k=0 respectively. But in both cases these series diverge by the Divergence Test. Consequently, the interval of convergence of the given series is k3 (3x + 5)k . k 2 2 2 a− , a+ 3 3 1. (−1, 1 ) 011 = 7 − , −1 3 . 10.0 points Determine the interval of convergence of the inﬁnite series 7 3 7 − , −1 3 ∞ 3 Determine the interval of convergence of the series 3. = (i) converges when |x − a| < By the Ratio test, therefore, the given series 1, 3 By the Ratio Test, the series thus in which case 2. 5 a = − . 3 , But then, But √ 4 an+1 n = 4 = 4 √ 4 an n+1 ck = k 3 ∞ (−1)k correct k=0 4. (−∞, ∞) 5. series converges only at x = − Explanation: The given series has the form ∞ k=0 ck (x − a)k 5 3 (k + 1)! (4x + 8)k . k 6 1. series converges only at x = −2 correct 2. 1 7 , 2 2 3. 1 1 − , 2 2 4. (−∞, ∞) truong (at28889) – HW14 – zupan – (56530) Thus 7 1 − , − 2 2 5. 7 4(k + 2) = ∞. 6 L = lim k→∞ Explanation: There are three possibilities for the interval of convergence of the inﬁnite series Consequently, the series converges only at x = −2 . ∞ k=0 ak (x − a)k . keywords: First set 012 L = lim k→∞ ak+1 . ak 10.0 points If the series ∞ cn 4n Then n=0 (i) when L > 0, the interval of convergence is given by a− 1 1 , a+ L L is convergent, which of the following statements must be true without further restrictions on cn ? ; ∞ cn (−3)n is divergent 1. (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, ∞ (−1) k (k k=0 = (−1) k (k k=0 4k (k + 1)! . 6k = − n=0 ∞ cn (−4)n is divergent 3. n=0 ∞ cn (−3)n is convergent correct n=0 Explanation: The series ∞ cn xn n=0 converges when x = 4, its radius of convergence, R, must have the property R ≥ 4 since the series (i) converges when |x| < R, and (ii) diverges when |x| > R. In this case ak+1 ak cn (−4)n is convergent + 1)!4k (x + 2)k , 6k so the corresponding value of ak is ak = (−1)k ∞ 2. 4. + 1)! (4x + 8)k k 6 ∞ n=0 (k + 2)!4k+1 6k · 6k+1 (k + 1)!4k 4(k + 2) . = 6 Consequently, it must be true that the series ∞ cn (−3)n is convergent n=0 truong (at28889) – HW14 – zupan – (56530) but we cannot determine whether either of the statements is (i) convergent when |t| < 2, and ∞ n cn (−4) is convergent, (i) n=0 ∞ cn (−4)n is divergent (ii) 8 (ii) divergent when |t| > 2. On the other hand, since lim |n cn|1/n = lim |cn |1/n , n=0 n→∞ must be true without further conditions on cn . n→∞ the Root Test ensures also that the series ∞ 013 10.0 points Compare the radius of convergence, R1 , of the series ∞ cn t n cn tn−1 n=1 is n (i) convergent when |t| < 2, and n=0 with the radius of convergence, R2 , of the series ∞ (ii) divergent when |t| > 2. Consequently, n cn tn−1 n=1 R = R2 = 2 . when lim |cn |1/n = n→∞ 1 . 2 1. R1 = R2 = 2 correct 2. R1 = 2R2 = 2 3. 2R1 = R2 = 2 Find a power series representation for the function 1 f (x) = . x−5 ∞ (−1)n 5n xn n=0 ∞ 2. f (x) = − 1 5. 2R1 = R2 = 2 5 n xn n=0 ∞ 1 6. R1 = R2 = 2 1 3. f (x) = 5n+1 n=0 Explanation: When 1 lim |cn |1/n = , n→∞ 2 the Root Test ensures that the series ∞ cn t n=0 10.0 points 1. f (x) = 1 2 4. R1 = 2R2 = 014 xn ∞ (−1)n−1 5n+1 xn 4. f (x) = n=0 ∞ 5. f (x) = − n Explanation: n=0 1 5n+1 xn correct truong (at28889) – HW14 – zupan – (56530) We know that 1 = 1 + x + x2 + . . . = 1−x ∞ xn . n=0 After simpliﬁcation, 1 1 1 f (x) = . = 3 3−x 3 1 − (x3 /3) On the other hand, we know that On the other hand, 1 = 1−t 1 1 1 . = − x−5 5 1 − (x/5) ∞ n x 5 n=0 ∞ 1 = − 5 n=0 ∞ tn . n=0 Replacing t with x3 /3, we thus obtain Thus 1 f (x) = − 5 9 1 n x . 5n 1 f (x) = 3 ∞ n=0 x3n = 3n ∞ n=0 x3n . 3n+1 keywords: Consequently, ∞ f (x) = − n=0 016 1 5n+1 xn 10.0 points Determine the value of f (1) when 2x3 3x5 x − 4 + 6 + ... . 22 2 2 (Hint: diﬀerentiate the power series expansion of (x2 + 22 )−1 .) f (x) = with |x| < 5. 015 10.0 points ∞ 1. f (x) = − n=0 x3n 33n ∞ 2. f (x) = − 3n x3n n=0 ∞ 3n x3n 3. f (x) = n=0 ∞ 4. f (x) = n=0 x3n correct 3n+1 ∞ 5. f (x) = − n=0 ∞ 6. f (x) = n=0 Explanation: xn 3n+1 x3n 33n 1. f (1) = 8 25 2. f (1) = 1 5 3. f (1) = 1 25 4. f (1) = 4 correct 25 5. f (1) = Find a power series representation for the function 1 f (x) = . 3 − x3 8 5 Explanation: The geometric series 1 1 1 = 2 22 + x 2 1 + x/22 1 x x2 x3 1− 2 + 4 − 6 +... 22 2 2 2 has interval of convergence (−4, 4). But if we now restrict x to the interval (−2, 2) and replace x by x2 we see that = x2 x4 x6 1 1 = 2 1− 2 + 4 − 6 + ... 2 2 + x2 2 2 2 2 truong (at28889) – HW14 – zupan – (56530) on the interval (−2, 2). 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 + 4 − 6 + ... . (x2 + 22 )2 2 2 2 2 Explanation: After simpliﬁcation, f (y) = x 2x3 3x5 − 4 + 6 + ... 22 2 2 y y = . 9y + 1 1 − (−9y) On the other hand, xn . n=0 Thus ∞ (−9y)n f (y) = y can be identiﬁed with ∞ 1 = 1−x Consequently, on the interval (−2, 2) the function f deﬁned by f (x) = 10 n=0 22 x . f (x) = (x2 + 22 )2 ∞ (−1)n 32n y n . = y n=0 As x = 1 lies in (−2, 2), we thus see that Consequently, 4 f (1) = . 25 ∞ (−1)n 32n y n+1 . f (y) = n=0 keywords: 017 10.0 points keywords: Find a power series representation for the function y f (y) = . 9y + 1 ∞ 32n y n 1. f (y) = n=0 018 10.0 points Find a power series representation for the function f (t) = ln(5 − t) . ∞ ∞ 2n n+1 2. f (y) = 1. f (t) = ln(5) + 3 y n=0 n=0 ∞ ∞ n 2n n+1 3. f (y) = (−1) 3 y correct 2. f (t) = ln(5) + n=1 n=0 ∞ ∞ n n n 4. f (y) = (−1) 3 y 3. f (t) = n=0 n=0 5. f (y) = ∞ (−1) 3 y n=0 4. f (t) = ln(5) − 6. f (y) = 3 y n=0 n=1 ∞ ∞ n n tn 5n tn n 5n ∞ n n n+1 tn n 5n 5. f (t) = ln(5) − n=0 tn correct n 5n tn 5n truong (at28889) – HW14 – zupan – (56530) ∞ 6. f (t) = − ∞ tn n5n n=1 11 n−1 n x 4n 4. f (x) = n=2 ∞ Explanation: We can either use the known power series representation ∞ ln(1 − x) = − xn n n=1 x x = − , 1 1 = x 4−x 4 1− 4 n ds x n=0 ∞ n s ds = − 0 ∞ n=1 xn . n n=0 1 1 f (t) = ln(5) 1 − t = ln(5) + ln 1 − t , 5 5 1 d = 2 (4 − x) dx ∞ n=1 tn . n 5n n = n=1 4n+1 10.0 points f (x) = x 3 n=0 Find a power series representation centered at the origin for the function f (x) = ∞ 1. f (x) = n=3 ∞ 2. f (x) = n=3 ∞ 3. f (x) = n=3 x3 (4 − x)2 1 n=0 ∞ n=0 xn . . xn 4n+1 ∞ n=0 n+1 n x = 4n+2 n+1 n x . 4n+2 ∞ n=0 n + 1 n+3 x . 4n+2 Consequently, f (x) = n=3 020 n−2 n x . 4n−1 10.0 points Determine the interval of convergence for the power series representation of xn n n x 4n 4n+1 xn−1 = ∞ n−2 n x correct 4n−1 4n−3 1 = Thus ∞ 019 ∞ 1 1 d , = 2 (4 − x) dx 4 − x so that f (t) = ln(5) − n and so on (−4, 4), For then by properties of logs, ∞ x 4 This series converges on (−4, 4). On the other hand, n=0 ∞ = − 1 = 4 ∞ s 0 n=2 1 ds 1−s 0 xn 4n−1 Explanation: By the known result for geometric series, or the fact that ln(1 − x) = − 1 5. f (x) = f (x) = tan−1 x 5 centered at the origin obtained by integrating the power series expansion for 1/(1 + x2 ). truong (at28889) – HW14 – zupan – (56530) To determine the interval of convergence of the power series, set 1 1 − , 5 5 1. interval of cgce. = Then 1 1 − , 5 5 4. interval of cgce. = 1 1 − , 5 5 an+1 an 2 , an+1 x 2 . = n→∞ an 5 By the Ratio Test, therefore, the power series converges when |x| < 5 and diverges when |x| > 5. On the other hand, at x = 5 the series reduces to ∞ (−1)n , 2n + 1 Explanation: Since n=0 1 = 1 + x + x2 + x3 + . . . , 1−x which converges by the Alternating series Test, while at x = −5 the series reduces to we see that ∞ 1 1 = 2 1+x 1 − (−(x)2 ) n=0 (−1)n+1 , 2n + 1 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 = 1 − x2 + (−x2 )2 − (x2 )3 + . . . ∞ (−1)n x2n . n=0 interval of convergence = [−5, 5] . Now x 1 dt = tan−1 (x) , 2 1+t 0 keywords: while 0 2n + 1 x 2n + 3 5 = . lim 6. interval of cgce. = [−5, 5] correct = 2n+1 and so 5. interval of cgce. = (−5, 5] x (−1)n x (2n + 1) 5 an = 2. interval of cgce. = [−5, 5) 3. interval of cgce. = 12 021 ∞ ∞ (−1)n x2n dt = n=0 n=0 (−1)n 2n+1 x . 2n + 1 10.0 points Find a power series representation for the function f (y) = ln Thus ∞ −1 tan (x) = n=0 (−1)n 2n+1 x , 2n + 1 −1 f (x) = tan x = 5 ∞ 1. f (y) = n=0 n=1 (−1)n x 2n + 1 5 . (Hint: remember properties of logs.) from which it follows that ∞ 1 + 5y 1 − 5y ∞ 2n+1 . 2. f (y) = n=1 (−1)n 52n 2n−1 y 2n − 1 52n y 2n 2n − 1 truong (at28889) – HW14 – zupan – (56530) ∞ 3. f (y) = n=1 ∞ 4. f (y) = n=1 ∞ 5. f (y) = n=1 52n−1 2n−1 y correct 2n − 1 022 10.0 points Evaluate the integral 1 y 2n−1 2n − 1 t 0 as a power series. ∞ 1. f (t) = Explanation: We know that n=0 x2 x3 + −... 2 3 ∞ (−1)n−1 n x , n = n=1 ∞ 2. f (t) = n=0 ∞ 3. f (t) = n=3 while ∞ ln(1 − x) = −x − x2 2 − ∞ = − n=1 x3 3 −... 1 n x . n Thus 4. f (t) = n=0 ∞ 5. f (t) = n=0 x3 x5 + +... 3 5 ∞ = 2 n=1 ln t3n+2 correct 3n + 2 t3n 3n + 2 t3n 3n (−1)n t3n 3n 1 x2n−1 . 2n − 1 1 = 1−s 1 = ln 2 1 + 5y 1 − 5y 1 ln(1 + 5y) − ln(1 − 5y) . = 2 ∞ sn , n=0 and so s = 1 − s3 Now by properties of logs, 1 + 5y 1 − 5y (−1)n t3n+2 3n + 2 Explanation: By the geometric series representation, ln(1 + x) − ln(1 − x) = 2 x+ s ds . 1 − s3 f (t) = 1 2n y 2n ln(1 + x) = x − 13 ∞ s3n+1 . n=0 But then t ∞ s3n+1 ds f (t) = 0 n=0 Thus t ∞ 2 f (y) = 2 ∞ n=1 1 (5y)2n−1 , 2n − 1 and so ∞ f (y) = n=1 52n−1 2n−1 . y 2n − 1 = n=0 s3n+1 ds . 0 Consequently, ∞ f (t) = n=0 t3n+2 . 3n + 2 truong (at28889) – HW14 – zupan – (56530) 023 Hint: separate then use the series for 10.0 points Express the integral 1. ∞ (−1)n x8n+5 correct (2n + 1)(8n + 5) 1. I = C + n=0 ∞ ∞ (−1)n x4n+4 (2n + 1) 3. I = C + n=0 ∞ 4−x 5. = 3 1+x (−1)n x8n+5 (8n + 5) 4. I = C + n=0 ∞ 4−x 4. = 5 1+x (2n + 1)(8n + 5) n=0 6. (−1)n x8n+5 ∞ (−1)k xk correct k=1 ∞ xk k=0 ∞ (−1)k xk k=1 ∞ xk k=1 4−x = 4+5 1+x ∞ (−1)k xk k=0 (2n + 1)(8n + 5) Explanation: Using the hint we get Explanation: We know that ∞ (−1)n tan−1 (x) = n=0 4 x 4−x = − , 1+x 1+x 1+x x2n+1 . 2n + 1 and 1 = 1 − x + x2 + . . . = 1+x 4 Replacing x with x , we get tan−1 x4 dx I = k=1 4−x = 4+3 3. 1+x (−1)n x8n+4 2. I = C + n=0 xk 4−x 2. = 4+5 1+x as a power series. 1 . 1+x ∞ 4−x = 4+3 1+x tan−1 x4 dx I = 5. I = 14 ∞ (−1)n = n=0 ∞ (−1)k xk . k=0 Thus (x4 )2n+1 2n + 1 4−x = 4 1+x dx . ∞ (−1)k xk k=0 ∞ Consequently, −x ∞ (−1)n x8n+5 I = C+ n=0 (2n + 1)(8n + 5) . ∞ ∞ (−1)k xk = k=0 10.0 points Find a power series representation for (−1)k 4 xk , k=0 while ∞ 4−x . 1+x k=0 But 4 024 (−1)k xk . ∞ k k x (−1) x k=0 (−1)k xk+1 . = k=0 truong (at28889) – HW14 – zupan – (56530) To combine the inﬁnite sums we need to express the last one as a sum of powers of xk : 5. f (y) = 2y 8 1−y 6. f (y) = 2y 8 correct (1 − y)3 ∞ k=0 (−1)k xk+1 = x − x2 + x3 − . . . ∞ (−1)k xk . = − k=1 Since the last sum now goes from k = 1 to k = ∞, we next write: ∞ ∞ k (−1) 4x k k = 4+ k=0 k (−1) 4 x , k=1 ∞ k k=0 k (−1) 4 x − − ∞ an y n n=0 n=2 n=2 (−1)k xk . n(n − 1)y n+6 ∞ k=1 =y 8 n=2 Consequently, 4−x = 4+5 1+x 025 k k (−1) x . 1. f (y) = y6 (6 − y)3 y6 2. f (y) = 6−y 3. f (y) = y8 (1 − y)3 4. f (y) = 2y 6 (6 − y)3 . n(n − 1) y ∞ yn , n=0 in which case, ∞ on (−1, 1). 1 = 1−y k=1 Determine the function f having power series representation n=2 n(n − 1)y n−2 On the other hand ∞ 10.0 points f (y) = n(n − 1)an y n−2 = ∞ ∞ k k=1 ∞ k=0 ∞ =4+ d2 dy 2 (−1)k xk+1 k (−1) 4 x − Explanation: The presence of the coeﬃcients n(n − 1) suggests that f should be related to the second derivative of some function since on the interval of convergence of the power series. Now for then we can add the two series: ∞ 15 n+6 ∞ 1 1 d = 2 (1 − y) dy 1 − y ny n−1 , = n=1 and 2 d2 1 = 3 2 1−y (1 − y) dy ∞ = n=2 n(n − 1)y n−2 , Consequently, f (y) = 2y 8 . (1 − y)3 ...
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