11.9 Representing Functions as Power Series-solutions

11.9 Representing Functions as Power Series-solutions -...

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im (gi768) – 11.9 Representing Functions as Power Series – sadun – (53088) 1 This print-out should have 7 questions. Multiple-choice questions may continue on the next column or page – fnd all choices be±ore answering. 001 10.0 points Find a power series representation ±or the ±unction f ( x ) = 1 x - 6 . 1. f ( x ) = s n =0 ( - 1) n - 1 6 n +1 x n 2. f ( x ) = s n =0 ( - 1) n 6 n x n 3. f ( x ) = - s n =0 6 n x n 4. f ( x ) = - s n =0 1 6 n +1 x n correct 5. f ( x ) = s n =0 1 6 n +1 x n Explanation: We know that 1 1 - x = 1 + x + x 2 + . . . = s n = 0 x n . On the other hand, 1 x - 6 = - 1 6 p 1 1 - ( x/ 6) P . Thus f ( x ) = - 1 6 s n = 0 p x 6 P n = - 1 6 s n = 0 1 6 n x n . Consequently, f ( x ) = - s n = 0 1 6 n +1 x n with | x | < 6. 002 10.0 points Find a power series representation ±or 2 + x 1 + x . Hint: separate then use the series ±or 1 1 + x . 1. 2 + x 1 + x = 2 + 3 s k =1 x k 2. 2 + x 1 + x = s k =1 ( - 1) k x k 3. 2 + x 1 + x = 3 s k = 1 x k 4. 2 + x 1 + x = 2 + s k = 1 ( - 1) k x k correct 5. 2 + x 1 + x = 2 + 3 s k =0 x k 6. 2 + x 1 + x = 2 + s k = 0 ( - 1) k x k Explanation: Using the hint we get 2 + x 1 + x = 2 1 + x + x 1 + x , and 1 1 + x = 1 - x + x 2 + . . . = s k = 0 ( - 1) k x k .
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