Keywords 015 100 points find a power series

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keywords: 015 10.0 points Find a power series representation for the function f ( z ) = ln ± 1 + 4 z 1 4 z ² . ( Hint : remember properties of logs.) 1. f ( z ) = s n = 1 2 2 n 1 z 2 n 1 2. f ( z ) = s n = 1 ( 1) n 4 2 n 2 n 1 z 2 n 1 3. f ( z ) = 2 s n = 1 4 2 n 1 2 n 1 z 2 n 1 correct 4. f ( z ) = 2 s n = 1 ( 1) n 4 2 n 2 n 1 z 2 n 1 5. f ( z ) = s n = 1 1 n z 2 n 6. f ( z ) = s n = 1 4 2 n 1 2 n 1 z 2 n 1 Explanation: We know that ln(1 + x ) = x x 2 2 + x 3 3 . . . = s n = 1 ( 1) n 1 n x n , while ln(1 x ) = x x 2 2 x 3 3 . . . = s n = 1 1 n x n . Thus ln(1 + x ) ln(1 x ) = 2 p x + x 3 3 + x 5 5 + . . . P = 2 p s n = 1 1 2 n 1 x 2 n 1 P . Consequently, f ( z ) = 2 s n = 1 4 2 n 1 2 n 1 z 2 n 1 .
may (am76336) – HW14 – he – (52980) 10 016 10.0 points Evaluate the integral f ( y ) = i y 0 s 1 s 3 ds . as a power series. 1. f ( y ) = s n = 0 y 3 n 3 n 2. f ( y ) = s n = 0 ( 1) n y 3 n +2 3 n + 2 3. f ( y ) = s n = 0 ( 1) n y 3 n 3 n 4. f ( y ) = s n = 3 y 3 n 3 n + 2 5. f ( y ) = s n = 0 y 3 n +2 3 n + 2 correct Explanation: By the geometric series representation, 1 1 s = s n = 0 s n , and so s 1 s 3 = s n = 0 s 3 n +1 . But then f ( y ) = i y 0 p s n = 0 s 3 n +1 P ds = s n = 0 p i y 0 s 3 n +1 ds P . Consequently, f ( y ) = s n = 0 y 3 n +2 3 n + 2 . 017 10.0 points Express the integral I = i tan 1 ( t 4 ) dt as a power series. 1. I = C + s n = 0 ( 1) n t 8 n +4 (2 n + 1)(8 n + 5) 2. I = C + s n = 0 ( 1) n t 8 n +5 (8 n + 5) 3. I = s n = 0 ( 1) n t 8 n +5 (2 n + 1)(8 n + 5) 4. I = C + s n = 0 ( 1) n t 4 n +4 (2 n + 1) 5. I = C + s n = 0 ( 1) n t 8 n +5 (2 n + 1)(8 n + 5) correct Explanation: We know that tan 1 ( t ) = s n = 0 ( 1) n t 2 n +1 2 n + 1 . Replacing t with t 4 , we get I = i tan 1 ( t 4 ) dt = i p s n = 0 ( 1) n ( t 4 ) 2 n +1 2 n + 1 P dt . Consequently, I = C + s n = 0 ( 1) n t 8 n +5 (2 n + 1)(8 n + 5) . 018 10.0 points Use the Taylor series for e x 2 to evaluate the integral I = i 3 0 2 e x 2 dx .
may (am76336) – HW14 – he – (52980) 11 1. I = n s k = 0 1 k !(2 k + 1) 2 · 3 2 k +1 2. I = s k = 0 ( 1) k k !(2 k + 1) 2 · 3 2 k +1 correct 3. I = s k = 0 ( 1) k 2 k + 1 2 · 3 2 k +1 4. I = s k = 0 1 k ! 2 · 3 2 k 5. I = s k = 0 ( 1) k k ! 2 · 3 2 k Explanation: The Taylor series for e x is given by e x = 1 + x + 1 2! x 2 + . . . + 1 n ! x n + . . . and its interval of convergence is ( −∞ , ). Thus we can substitute x → − x 2 for all val- ues of x , showing that e x 2 = s k = 0 ( 1) k k ! x 2 k everywhere on ( −∞ , ). Thus I = i 3 0 2 p s k = 0 ( 1) k k ! x 2 k P dx. But we can change the order of summation and integration on the interval of convergence, so I = 2 s k = 0 p i 3 0 ( 1) k k ! x 2 k P dx = 2 s k = 0 b ( 1) k k !(2 k + 1) x 2 k +1 B 3 0 . Consequently, I = s k = 0 ( 1) k k !(2 k + 1) 2 · 3 2 k +1 . 019 10.0 points Find the Taylor series representation for f centered at x = 2 when f ( x ) = 8 + 4 x 3 x 2 . 1. f ( x ) = 8 + 4( x 2) 3( x 2) 2 2. f ( x ) = 4 + 4( x 2) + 3( x 2) 2 3. f ( x ) = 8 8( x 2) + 6( x 2) 2 4. f ( x ) = 4 8( x 2) 3( x 2) 2 correct 5. f ( x ) = 8 + 4( x 2) 6( x 2) 2 6. f ( x ) = 4 8( x 2) 6( x 2) 2 Explanation: For a function f the Taylor series represen- tation centered at x = 2 is given by f ( x ) = s n = 0 1 n !