M408L_HW15 - hyun (hh7953) – HW15 – gogolev – (57440)...

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Unformatted text preview: hyun (hh7953) – HW15 – gogolev – (57440) 1 This print-out should have 16 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. 001 10.0 points Determine an expression for b 9 if f ( x ) = ∞ summationdisplay n = 0 b n ( x- 3) n . 1. b 9 = f (3) (3) 9! 2. b 9 = f (9) (3) 9! correct 3. b 9 = f (9) (3) 9 4. b 9 = f (9) (9) 3! 5. b 9 = f (3) (9) 3! Explanation: We know that if f ( x ) = ∞ summationdisplay n = 0 b n ( x- 3) n , then the series representation is the Taylor series representation of f centered at x = 3, in which case b n = f ( n ) (3) n ! . Consequently, b 9 = f (9) (3) 9! . 002 10.0 points Find the Taylor series centered at x = 0 for f ( x ) = cos 3 x . 1. ∞ summationdisplay n = 1 3 n (2 n )! x n 2. ∞ summationdisplay n = 1 3 n (2 n )! x 2 n 3. ∞ summationdisplay n = 0 (- 1) n 3 2 n n ! x 2 n 4. ∞ summationdisplay n = 0 (- 1) n 3 2 n (2 n )! x 2 n correct 5. ∞ summationdisplay n = 0 (- 1) n 3 2 n (2 n )! x n 6. ∞ summationdisplay n = 0 (- 1) n 3 2 n n ! x n Explanation: The Taylor series centered at x = 0 for any f is f (0) + f ′ (0) x + 1 2! f ′′ (0) x 2 + . . . = ∞ summationdisplay n = 0 1 n ! f ( n ) (0) x n . But when f ( x ) = cos(3 x ), n f ( n ) ( x ) f ( n ) (0) cos(3 x ) 1 1- 3 sin(3 x ) 2- 9 cos(3 x )- 9 3 27 sin(3 x ) 4 81cos(3 x ) 81 . . . . . . in other words, f ′ (0) = f ′′′ (0) = f (5) (0) = . . . = 0 , while f (0) = 1 , f ′′ (0) =- 3 2 , f (4) (0) = 3 4 . Thus in general, f (2 n +1) (0) = 0 , f (2 n ) (0) = (- 1) n 3 2 n . hyun (hh7953) – HW15 – gogolev – (57440) 2 Consequently, the Taylor series of cos 3 x cen- tered at x = 0 is ∞ summationdisplay n = 0 (- 1) n 3 2 n (2 n )! x 2 n . 003 10.0 points Find the Taylor series representation for f centered at x = 2 when f ( x ) = 6 + 3 x- 2 x 2 . 1. f ( x ) = 6 + 3( x- 2)- 2( x- 2) 2 2. f ( x ) = 4 + 3( x- 2) + 2( x- 2) 2 3. f ( x ) = 6- 5( x- 2) + 4( x- 2) 2 4. f ( x ) = 6 + 3( x- 2)- 4( x- 2) 2 5. f ( x ) = 4- 5( x- 2)- 4( x- 2) 2 6. f ( x ) = 4- 5( x- 2)- 2( x- 2) 2 correct Explanation: For a function f the Taylor series represen- tation centered at x = 2 is given by f ( x ) = summationdisplay n = 0 1 n ! f ( n ) (2)( x- 2) n . Since f is a polynomial of degree 2, however, f ( n ) = 0 for all n ≥ 3, so we have only to calculate derivatives of f up to order 2: f ′ ( x ) = 3- 4 x, f ′′ ( x ) =- 4 . Thus f (2) = 4 , f ′ (2) =- 5 , f ′′ (2) =- 4 ....
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This note was uploaded on 01/19/2010 for the course CH 53435 taught by Professor Sparks during the Fall '09 term at University of Texas.

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M408L_HW15 - hyun (hh7953) – HW15 – gogolev – (57440)...

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