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# h5 - moore(jwm2685 HW05 gilbert(55485 This print-out should...

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moore (jwm2685) – HW05 – gilbert – (55485) 1 This print-out should have 12 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. 001 10.0points Determine the value of f ( - 1) when f ( x ) = x 2 2 + 2 x 3 2 4 + 3 x 5 2 6 + ... . ( Hint : differentiate the power series represen- tation of ( x 2 - 2 2 ) 1 .) 1. f ( - 1) = 1 8 2. f ( - 1) = - 1 8 3. f ( - 1) = 4 9 4. f ( - 1) = - 4 9 correct 5. f ( - 1) = - 4 25 Explanation: The geometric series 1 2 2 - x = 1 2 2 parenleftBig 1 1 - x/ 2 2 parenrightBig = 1 2 2 parenleftBig 1 + x 2 2 + x 2 2 4 + x 3 2 6 + ... parenrightBig has interval of convergence ( - 4 , 4), and so 1 x - 2 2 = - 1 2 2 - x = - 1 2 2 parenleftBig 1 + x 2 2 + x 2 2 4 + x 3 2 6 + ... parenrightBig on the interval ( - 4 , 4). Thus, if we restrict x to the interval ( - 2 , 2), we can replace x by x 2 in this series. Consequently, 1 x 2 - 2 2 = - 1 2 2 parenleftBig 1 + x 2 2 2 + x 4 2 4 + x 6 2 6 + ... parenrightBig on the interval ( - 2 , 2), 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 - 2 2 ) 2 = 1 2 2 parenleftBig 2 x 2 2 + 4 x 3 2 4 + 6 x 5 2 6 + ... parenrightBig , and so f ( x ) = 2 2 x ( x 2 - 2 2 ) 2 . As x = - 1 lies in ( - 2 , 2), f ( - 1) = - 4 9 . 002 10.0points Find a power series representation for the function f ( t ) = ln(8 - t ) . 1. f ( t ) = summationdisplay n =0 t n n 8 n 2. f ( t ) = ln 8 - summationdisplay n =1 t n n 8 n correct 3. f ( t ) = ln 8 + summationdisplay n =0 t n n 8 n 4. f ( t ) = ln 8 - summationdisplay n =0 t n 8 n 5. f ( t ) = - summationdisplay n =1 t n n 8 n 6. f ( t ) = ln 8 + summationdisplay n =1 t n 8 n Explanation: We can either use the known power series representation ln(1 - x ) = - summationdisplay n =1 x n n ,

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moore (jwm2685) – HW05 – gilbert – (55485) 2 or the fact that ln(1 - x ) = - integraldisplay x 0 1 1 - s ds = - integraldisplay x 0 braceleftBig summationdisplay n =0 s n bracerightBig ds = - summationdisplay n =0 integraldisplay x 0 s n ds = - summationdisplay n =1 x n n .
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