Exam_solutions_4_

# Exam_solutions_4_ - MATH 141 4th Examination Prof Jonathan...

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Unformatted text preview: MATH 141, 4th Examination Prof. Jonathan Rosenberg Grading and Solution Key 1. (30 points) Find the Taylor series around a = 0 (i.e., the Maclaurin series) for f ( x ) = (2+ x ) − 2 . You should find an explicit formula (in terms of n ) for the coefficient of x n . There are several acceptable methods: using the general formula in terms of f ( n ) (0) , using facts about binomial series, or starting with the series for (1- x ) − 1 and doing some manipulation and/or differentiating. You do not need to say anything about convergence. Solution. Write (2 + x ) − 1 = 1 2 + x = 1 2 parenleftBig 1 + x 2 parenrightBig − 1 = 1 2 ∞ summationdisplay n =0 parenleftbigg- x 2 parenrightbigg n = ∞ summationdisplay n =0 (- 1) n 2 − ( n +1) x n . Now differentiate term by term. We get d dx (2 + x ) − 1 =- (2 + x ) − 2 = ∞ summationdisplay n =0 (- 1) n 2 − ( n +1) nx n − 1 . So (2 + x ) − 2 =- ∞ summationdisplay n =0 (- 1) n 2 − ( n +1) nx n − 1 = ∞ summationdisplay n =0 (- 1) n 2 − ( n +2) ( n + 1) x n . Alternative Solution. If f ( x ) = (2 + x ) − 2 , then f ′ ( x ) =- 2(2 + x ) − 3 , f ′′ ( x ) = (- 3)(- 2)(2 + x ) − 4 , f ′′′ ( x ) = (- 4)(- 3)(- 2)(2 + x ) − 5 , and in general, f ( n ) ( x ) = (- 1) n ( n + 1)!(2 + x ) − n − 2 . Substituting x = 0 into this, we get f ( n ) (0) = (- 1) n ( n + 1)! 2 − n − 2 , and so the Taylor series is ∞ summationdisplay n =0 f ( n ) (0) n !...
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## This note was uploaded on 09/12/2009 for the course MATH 141 taught by Professor Hamilton during the Spring '07 term at Maryland.

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Exam_solutions_4_ - MATH 141 4th Examination Prof Jonathan...

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