M408D Quest Homework 4-solutions - hernandez(ah29758 M408D Quest Homework 4 pascale(54550 This print-out should have 15 questions Multiple-choice

# M408D Quest Homework 4-solutions - hernandez(ah29758 M408D...

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hernandez (ah29758) – M408D Quest Homework 4 – pascaleff – (54550) 1 This print-out should have 15 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. 001 10.0points Find a power series representation for the function f ( x ) = 1 4 + x . 1. f ( x ) = summationdisplay n =0 ( - 1) n 4 n +1 x n correct 2. f ( x ) = summationdisplay n =0 4 n +1 x n 3. f ( x ) = summationdisplay n =0 1 4 n +1 x n 4. f ( x ) = summationdisplay n =0 ( - 1) n 4 x n 5. f ( x ) = summationdisplay n =0 ( - 1) n 4 n +1 x n Explanation: We know that 1 1 - x = 1 + x + x 2 + ... = summationdisplay n =0 x n . On the other hand, 1 4 + x = 1 4 parenleftBig 1 1 - ( - x/ 4) parenrightBig . Thus f ( x ) = 1 4 summationdisplay n =0 parenleftBig - x 4 parenrightBig n = 1 4 summationdisplay n =0 ( - 1) n 4 n x n . Consequently, f ( x ) = summationdisplay n =0 ( - 1) n 4 n +1 x n with | x | < 4. 002 10.0points Find a power series representation for the function f ( y ) = ln parenleftbigg 1 + 5 y 1 - 5 y parenrightbigg . ( Hint : remember properties of logs.) 1. f ( y ) = summationdisplay n =1 2 2 n - 1 y 2 n 1 2. f ( y ) = summationdisplay n =1 5 2 n 1 2 n - 1 y 2 n 1 3. f ( y ) = 2 summationdisplay n =1 5 2 n 1 2 n - 1 y 2 n 1 correct 4. f ( y ) = 2 summationdisplay n =1 ( - 1) n 5 2 n 2 n - 1 y 2 n 1 5. f ( y ) = summationdisplay n =1 1 n y 2 n 6. f ( y ) = summationdisplay n =1 ( - 1) n 5 2 n 2 n - 1 y 2 n 1 Explanation: We know that ln(1 + x ) = x - x 2 2 + x 3 3 - ... = summationdisplay n =1 ( - 1) n 1 n x n , while ln(1 - x ) = - x - x 2 2 - x 3 3 - ... = - summationdisplay n =1 1 n x n .
hernandez (ah29758) – M408D Quest Homework 4 – pascaleff – (54550) 2 Thus ln(1 + x ) - ln(1 - x ) = 2 braceleftBig x + x 3 3 + x 5 5 + ... bracerightBig = 2 braceleftBig summationdisplay n =1 1 2 n - 1 x 2 n 1 bracerightBig . Consequently, f ( y ) = 2 summationdisplay n =1 5 2 n 1 2 n - 1 y 2 n 1 . 003 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 expan- sion of ( x 2 + 4 2 ) 1 .) 1. f (2) = 4 5 2. f (2) = 1 50 3. f (2) = 1 10 4. f (2) = 2 25 correct 5. f (2) = 4 25 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). But if we now restrict x to the interval ( - 4 , 4) and replace x by x 2 we see that 1 4 2 + x 2 = 1 4 2 parenleftBig 1 - x 2 4 2 + x 4 4 4 - x 6 4 6 + ... parenrightBig on the interval ( - 4 , 4). In addition, in this interval the series expansion of the deriva- tive of the left hand side is the term-by-term derivative of the series on the right: - 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 . Consequently, on the interval ( - 4 , 4) the function f defined by f ( x ) = x 4 2 - 2 x 3 4 4 + 3 x 5 4 6 + ... can be identified with f ( x ) = 4 2 x ( x 2 + 4 2 ) 2 . As x = 2 lies in ( - 4 , 4), we thus see that f (2) = 2 25 . keywords: 004 10.0points Express the integral I = integraldisplay tan 1 ( z 4 ) dz as a power series.