Representation Of Functions By Power Series Homework Online

Unformatted text preview: 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 10.0 points 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 10.0 points 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 10.0 points Determine the value of f (2) when f ( x ) = x 4 2- 2 x 3 4 4 + 3 x 5 4 6 + . . . . ( Hint : 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 + . . ....
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