11.9_ Representations as Series-solutions

The ratio test therefore the power series converges

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Unformatted text preview: 3, 3] correct 11 −, 33 6. interval of cgce. = 3 Explanation: Since n=0 which converges by the Alternating series Test, while at x = −3 the series reduces to 1 = 1 + x + x2 + x3 + . . . , 1−x we see that ∞ 1 1 = 2 1+x 1 − (−(x)2 ) n=0 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 . = (−1)n+1 , 2n + 1 n=0 Now interval of convergence = [−3, 3] . x 0 1 dt = tan−1 (x) , 1 + t2 while x keywords: ∞ ∞ (−1)n (−1)n x2n dt = 0 n=0 n=0 2n + 1 x2n+1 . Thus ∞ tan−1 (x) = n=0 005 Find a power series representation for the function 1 . f (z ) = 5 + 16z 2 (−1)n 2n+1 x , 2n + 1 from which it follows that f (x) = tan−1 x = 3 ∞ n=0 ∞ (−1)n x 2n + 1 3 2n+1 1. f (z ) = . To determine the interval of convergence of the power series, set n=0 ∞ 2. f (z ) = n=0 ∞ an (−1)n x = (2n + 1) 3 2n+1 . 10.0 points 3. f (z ) = n=0 42n z 2n 5n+1 4n z n 5n+1 (−1)n 42n z 2n correct 5n+1 stiurca (mas7745) – 11.9: Representations as Series – meth – (91825) ∞ Explanation: We can either use the known power series representation (−1)n 42n z 2n 5n 4. f (z ) = n=0 ∞ ∞ (−1)n 4n z n 5n+1 5. f (z ) = n=0 n=1 or the fact that x 1 1 f (z ) = = 5 + 16z 2 5 1 1 − (− 16 z 2 ) 5 1 5 ln(1 − x) = − 1 16 1 + z2 5 0 ∞ − n=0 16 2 z 5 s n ds =− 0 n=0 . ∞ ∞ = n=0 (−1)n 42n z 2n 5n+1 ∞ x sn ds = − =− n=0 n 1 ds 1−s ∞ x On the other hand, 1 5 xn , n ln(1 − x) = − Explanation: After simplification, = 4 0 n=1 xn . n For then by properties of logs, 1 1 f (y ) = ln(8) 1 − y = ln(8) + ln 1 − y , 8 8 so that keywords: 006 10.0 points ∞ f (y ) = ln(8) − n=1 Find a power series representation for the function f (y ) = ln(8 − y ) . ∞ 1. f (y ) = ln(8) + n=1 ∞ 2. f (y ) = ln(8) − n=1 ∞ 3. f (y ) = − n=1 4. f (y ) = ln(8) − n=0 ∞ 5. f (y ) = ln(8) + n=0 6. f (y ) = n=0 yn n 8n f (x) = x4 tan−1 (x) correct on (−1, 1). ∞ yn n8n ∞ ∞ n 8n 1. f (x) = n=0 yn 8n yn n 8n 10.0 points Find a power series representation for the function yn 8n yn 007 yn . n 8n ∞ 2. f (x) = n=0 ∞ 3. f (x) = n=0 ∞ 4. f (x) = n=0 (−1)n n+5 x (n + 1)! (−1)n x2n+5 (2n + 1)! (−1)n n+5 x n+1 (−1)n 2n+5 x correct 2n + 1 stiurca (mas7745) – 11.9: Representations as Series – meth – (91825) when ∞ 1 xn+5 n+1 5. f (x) = n=0 ∞ n→∞ n=0 = 2. 1. R1 = R2 = 2 1 x2n+5 2n + 1 6. f (x) = cn+1 cn lim 5 Explanation: The interval of convergence of the geometric series 1 = 1 + x + x2 + . . . 1−x 2. R1 = 2R2 = 2 1 2 3. 2R1 = R2 = 4. 2R1 = R2 = 2 is (−1, 1). Thus on (−1, 1) 1 = 1 − x2 + x4 − . . . = 1 + x2 ∞ (−1)n x2n . n=0 1 correct 2 6. R1 = R2 = On the other hand, x tan−1 (x) = 0 1 dt . 1 + t2 Thus on (−1, 1) 1 2 5. R1 = 2R2 = Explanation: When lim cn+1 cn n→∞ ∞ x (−1)n t2n dt tan−1 (x) = 0 the Ratio Test ensures that the series n=0 ∞ ∞ cn z n x (−1)n t2n dt = = 2, n=0 0 n=0 is ∞ = n=0 (−1)n 2n + 1 x2n+1 . 1 (i) convergent when |z | < , and 2 1 (ii) divergent when |z | > . 2 Consequently, on (−1, 1) ∞ f ( x) = n=0 008 (−1)n 2n+5 x . 2n + 1 On the other hand, since lim 10.0 points Compare the radius of convergence, R1 , of the series n→∞ (n + 1)cn+1 ncn = lim n→∞ cn+1 , cn the Ratio Test ensures also that the series ∞ ∞ cn z n n cn z n−1 n=0 with the radius of convergence, R2 , of the series n=1 is ∞ n cn z n−1 n=1 1 (i) convergent when |z | < , and 2 stiurca (mas7745) – 11.9: Representations as Series – meth – (91825) 1 (ii) divergent when |z | > . 2 Thus ∞ f ( x) = x3 Consequently, 009 n=0 1 . 2 R1 = R2 = x3 ∞ n=2 ∞ 2. f (x) = n=3 ∞ 3. f (x) = n=3 ∞ 4. f (x) = n=3 ∞ 5. f (x) = n=2 (6 − x)2 ∞ . n−1 n x 6n 1 6n−3 xn nn x 6n n−2 n x correct 6n−1 1 6n−1 xn Explanation: By the known result for geometric series, 1 1 = x 6−x 6 1− 6 = 1 6 ∞ x 6 n=0 n ∞ 1 = n=0 6n+1 xn . This series converges on (−6, 6). On the other hand, 1 1 d , = 2 (6 − x) dx 6 − x and so on (−6, 6), d 1 = (6 − x)2 dx ∞ = n=1 n 6n+1 ∞ n=0 xn 6n+1 ∞ x n−1 = n=0 ∞ n=0 n + 1 n+3 x . 6n+2 Consequently, Find a power series representation centered at the origin for the function 1. f (x) = n+1 n x= 6n+2 10.0 points f ( x) = 6 n+1 n x. 6n+2 f ( x) = n=3 n−2 n x. 6n−1...
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This document was uploaded on 09/17/2013.

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