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M408L_HW15 - hyun(hh7953 HW15 gogolev(57440 This print-out...

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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) 0 cos(3 x ) 1 1 - 3 sin(3 x ) 0 2 - 9 cos(3 x ) - 9 3 27 sin(3 x ) 0 4 81 cos(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 .
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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 . Consequently, f ( x ) = 4 - 5( x - 2) - 2( x - 2) 2 . 004 (part 1 of 5) 10.0 points When f is the function f ( x ) = e (2 x +3) , (i) find the third derivative, f ′′′ , of f . 1. f ′′′ ( x ) = 1 3! 2 3 e (2 x +3) 2. f ′′′ ( x ) = e (2 x +3) 3. f ′′′ ( x ) = 3! e (2 x +3) 4. f ′′′ ( x ) = 2 3 e (2 x +3) correct 5. f ′′′ ( x ) = 6 · 2 3 e (2 x +3) Explanation: Applying the Chain Rule successively to the function f ( x ) = e (2 x +3) we see that f ( x ) = 2 e (2 x +3) , f ′′ ( x ) = 2 2 e (2 x +3) , and f ′′′ ( x ) = 2 3 e (2 x +3) .
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