Bretting nmb2278 hw14 meth 54160 10 1 f z

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bretting (nmb2278) – HW14 – meth – (54160) 10 1. f ( z ) = summationdisplay n =0 6 n z 3 n 2. f ( z ) = summationdisplay n =0 6 n z 3 n 3. f ( z ) = summationdisplay n =0 z 3 n 6 n +1 correct 4. f ( z ) = summationdisplay n =0 z 3 n 6 3 n 5. f ( z ) = summationdisplay n =0 z n 6 n +1 6. f ( z ) = summationdisplay n =0 z 3 n 6 3 n Explanation: After simplification, f ( z ) = 1 6 z 3 = 1 6 parenleftBig 1 1 ( z 3 / 6) parenrightBig . On the other hand, we know that 1 1 t = summationdisplay n =0 t n . Replacing t with z 3 / 6, we thus obtain f ( z ) = 1 6 summationdisplay n =0 z 3 n 6 n = summationdisplay n =0 z 3 n 6 n +1 . keywords: 017 10.0points Determine the value of f ( 2) when f ( x ) = x 4 2 + 2 x 3 4 4 + 3 x 5 4 6 + ... . ( Hint : differentiate the power series represen- tation of ( x 2 4 2 ) - 1 .) 1. f ( 2) = 1 16 2. f ( 2) = 2 25 3. f ( 2) = 2 9 4. f ( 2) = 1 16 5. f ( 2) = 2 9 correct Explanation: The geometric series 1 4 2 x = 1 4 2 parenleftBig 1 1 x/ 4 2 parenrightBig = 1 4 2 parenleftBig 1 + x 4 2 + x 2 4 4 + x 3 4 6 + ... parenrightBig has interval of convergence ( 16 , 16), and so 1 x 4 2 = 1 4 2 x = 1 4 2 parenleftBig 1 + x 4 2 + x 2 4 4 + x 3 4 6 + ... parenrightBig 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 parenleftBig 1 + x 2 4 2 + x 4 4 4 + x 6 4 6 + ... parenrightBig 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 parenleftBig 2 x 4 2 + 4 x 3 4 4 + 6 x 5 4 6 + ... parenrightBig , and so f ( x ) = 4 2 x ( x 2 4 2 ) 2 . As x = 2 lies in ( 4 , 4), f ( 2) = 2 9 . 018 10.0points
bretting (nmb2278) – HW14 – meth – (54160) 11 Find a power series representation for the function f ( x ) = x 16 x + 1 . 019 10.0points Find a power series representation for the function f ( z ) = ln(2 z ) .