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int cal 15

# int cal 15 - pokharel(yp624 HW15 Radin(57410 This print-out...

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pokharel (yp624) – HW15 – Radin – (57410) 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 - 6) n . 1. b 9 = f (6) (6) 9! 2. b 9 = f (6) (9) 6! 3. b 9 = f (9) (6) 9! correct 4. b 9 = f (9) (6) 9 5. b 9 = f (9) (9) 6! Explanation: We know that if f ( x ) = summationdisplay n =0 b n ( x - 6) n , then the series representation is the Taylor series representation of f centered at x = 6, in which case b n = f ( n ) (6) n ! . Consequently, b 9 = f (9) (6) 9! . 002 10.0 points Find the Taylor series centered at x = 0 for f ( x ) = cos 3 x. 1. summationdisplay n =0 ( - 1) n 3 2 n (2 n )! x n 2. summationdisplay n =0 ( - 1) n 3 2 n n ! x 2 n 3. summationdisplay n =0 ( - 1) n 3 2 n (2 n )! x 2 n correct 4. summationdisplay n =0 ( - 1) n 3 2 n n ! x n 5. summationdisplay n =1 3 n (2 n )! x n 6. summationdisplay n =1 3 n (2 n )! x 2 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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pokharel (yp624) – HW15 – Radin – (57410) 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 = 1 when f ( x ) = 4 + 2 x - 3 x 2 . 1. f ( x ) = 4 + 2( x - 1) - 3( x - 1) 2 2. f ( x ) = 3 - 4( x - 1) - 6( x - 1) 2 3. f ( x ) = 4 - 4( x - 1) + 6( x - 1) 2 4. f ( x ) = 3 - 4( x - 1) - 3( x - 1) 2 correct 5. f ( x ) = 3 + 2( x - 1) + 3( x - 1) 2 6. f ( x ) = 4 + 2( x - 1) - 6( x - 1) 2 Explanation: For a function f the Taylor series represen- tation centered at x = 1 is given by f ( x ) = summationdisplay n =0 1 n ! f ( n ) (1)( x - 1) 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 ) = 2 - 6 x, f ′′ ( x ) = - 6 . Thus f (1) = 3 , f (1) = - 4 , f ′′ (1) = - 6 . Consequently, f ( x ) = 3 - 4( x - 1) - 3( x - 1) 2 . 004 (part 1 of 5) 10.0 points When f is the function f ( x ) = e (4 x +5) , (i) find the third derivative, f ′′′ , of f . 1. f ′′′ ( x ) = 6 · 4 3 e (4 x +5) 2. f ′′′ ( x ) = 3! e (4 x +5) 3. f ′′′ ( x ) = 4 3 e (4 x +5) correct 4. f ′′′ ( x ) = e (4 x +5) 5. f ′′′ ( x ) = 1 3!
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int cal 15 - pokharel(yp624 HW15 Radin(57410 This print-out...

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