On the other hand we know that 1 1 t n 0 t n

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On the other hand, we know that 1 1 t = n = 0 t n . Replacing t with x 3 / 4, we thus obtain f ( x ) = 1 4 n =0 x 3 n 4 n = n =0 x 3 n 4 n +1 . keywords: 017 10.0 points Determine the value of f ( 2) when f ( x ) = x 4 2 + 2 x 3 4 4 + 3 x 5 4 6 + . . . . ( Hint : di ff erentiate the power series represen- tation of ( x 2 4 2 ) 1 .) 1. f ( 2) = 2 25 2. f ( 2) = 1 16 3. f ( 2) = 2 9 correct 4. f ( 2) = 2 9 5. f ( 2) = 1 16 Explanation: The geometric series 1 4 2 x = 1 4 2 1 1 x/ 4 2 = 1 4 2 1 + x 4 2 + x 2 4 4 + x 3 4 6 + . . . has interval of convergence ( 16 , 16), and so 1 x 4 2 = 1 4 2 x = 1 4 2 1 + x 4 2 + x 2 4 4 + x 3 4 6 + . . . on the interval ( 16 , 16). Thus, if we restrict x to the interval ( 4 , 4), we can replace x by x 2 in this series. Consequently, 1 x 2 4 2 = 1 4 2 1 + x 2 4 2 + x 4 4 4 + x 6 4 6 + . . . on the interval ( 4 , 4), On this interval the derivative of the left hand side is the term- by-term derivative of the series on the right hand. Hence 2 x ( x 2 4 2 ) 2 = 1 4 2 2 x 4 2 + 4 x 3 4 4 + 6 x 5 4 6 + . . . , and so f ( x ) = 4 2 x ( x 2 4 2 ) 2 . As x = 2 lies in ( 4 , 4), f ( 2) = 2 9 . 018 10.0 points
garcia (pgg378) – HW14 – meth – (54160) 11 Find a power series representation for the function f ( z ) = z 16 z + 1 . 019 10.0 points keywords:
garcia (pgg378) – HW14 – meth – (54160) 12 so that f ( t ) = ln(4) n = 1 t n n 4 n . 020 10.0 points Determine the interval of convergence for the power series representation of f ( x ) = tan 1 x 5 centered at the origin obtained by integrating the power series expansion for 1 / (1 + x 2 ). 1. interval of cgce. = 1 5 , 1 5 2. interval of cgce. = [ 5 , 5] correct 3. interval of cgce. = 1 5 , 1 5 4. interval of cgce. = [ 5 , 5) 5. interval of cgce. = 1 5 , 1 5 6. interval of cgce. = ( 5 , 5] Explanation: Since 1 1 x = 1 + x + x 2 + x 3 + . . . , we see that 1 1 + x 2 = 1 1 ( ( x ) 2 ) = 1 x 2 + ( x 2 ) 2 ( x 2 ) 3 + . . . = n = 0 ( 1) n x 2 n . Now x 0 1 1 + t 2 dt = tan 1 ( x ) , while x 0 n = 0 ( 1) n x 2 n dt = n = 0 ( 1) n 2 n + 1 x 2 n +1 .
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