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11.9_ Representations as Series-solutions

# 11.9_ Representations as Series-solutions - stiurca(mas7745...

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stiurca (mas7745) – 11.9: Representations as Series – meth – (91825) 1 This print-out should have 9 questions. Multiple-choice questions may continue on the next column or page – fnd all choices beFore answering. 001 10.0 points ±ind a power series representation For the Function f ( x ) = 1 3 + x . 1. f ( x ) = s n =0 ( - 1) n 3 n +1 x n correct 2. f ( x ) = s n 3 n +1 x n 3. f ( x ) = s n ( - 1) n 3 n +1 x n 4. f ( x ) = s n ( - 1) n 3 x n 5. f ( x ) = s n 1 3 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 3 + x = 1 3 p 1 1 - ( - x/ 3) P . Thus f ( x ) = 1 3 s n = 0 p - x 3 P n = 1 3 s n ( - 1) n 3 n x n . Consequently, f ( x ) = s n = 0 ( - 1) n 3 n +1 x n with | x | < 3. 002 10.0 points Evaluate the integral f ( y ) = i y 0 s 1 - s 3 ds . as a power series. 1. f ( y ) = s n = 0 ( - 1) n y 3 n +2 3 n + 2 2. f ( y ) = s n = 3 y 3 n 3 n + 2 3. f ( y ) = s n = 0 y 3 n +2 3 n + 2 correct 4. f ( y ) = s n = 0 y 3 n 3 n 5. f ( y ) = s n = 0 ( - 1) n y 3 n 3 n Explanation: By the geometric series representation, 1 1 - s = s n s n , and so s 1 - s 3 = s n s 3 n +1 . But then f ( y ) = i y 0 p s n = 0 s 3 n +1 P ds = s n = 0 p i y 0 s 3 n +1 ds P . Consequently, f ( y ) = s n y 3 n +2 3 n + 2 .

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stiurca (mas7745) – 11.9: Representations as Series – meth – (91825) 2 003 10.0 points Find a power series representation for 4 + x 1 + x . Hint: separate then use the series for 1 1 + x . 1. 4 + x 1 + x = 5 s k =1 x k 2. 4 + x 1 + x = 4 + 3 s k =0 ( - 1) k x k 3. 4 + x 1 + x = 4 + 5 s k x k 4. 4 + x 1 + x = 4 + 5 s k x k 5. 4 + x 1 + x = 4 + 3 s k ( - 1) k x k correct 6. 4 + x 1 + x = 3 s k ( - 1) k x k Explanation: Using the hint we get 4 + x 1 + x = 4 1 + x + x 1 + x , and 1 1 + x = 1 - x + x 2 + . . . = s k = 0 ( - 1) k x k .
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