quiz07_s107_solns

# quiz07_s107_solns - ∑ ∞ n =0 c n-1 2 n is convergent(a...

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Math 1B Quiz 7 Solutions Section 107 October 25, 2009 1. (5 points) (a) Showing your work very clearly , ﬁnd a power series representation for the function f ( x ) = x 4 (4 x - 1) 2 I spy a derivative of 1 1 - 4 x . Calculate d dx 1 1 - 4 x = 4 (4 x - 1) 2 so f ( x ) = x 4 4 d dx 1 1 - 4 x = x 4 4 d dx X n =0 (4 x ) n = x 4 4 X n =0 4 n nx n - 1 = X n =0 4 n - 1 nx n +3 = X n =4 4 n - 4 ( n - 3) x n (b) Find the radius and interval of convergence for your power series. Using the Ratio Test, the power series converges if lim n →∞ ± ± 4 n - 3 ( n - 2) x n +1 4 n - 4 ( n - 3) x n ± ± < 1 lim n →∞ ± ± 4( n - 2) x ( n - 3) ± ± < 1 | 4 x | < 1 1

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so the radius of convergence is 1 / 4. To ﬁnd the interval of convergence we must check the endpoints. If | x | = 1 / 4, then the terms are ± 4 - 4 ( n - 3), so the series diverges by the test for divergence. Therefore, the interval of convergence is ( - 1 / 4 , 1 / 4). 2. (3 points) Suppose
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Unformatted text preview: ∑ ∞ n =0 c n (-1 2 ) n is convergent. (a) Does it follow that ∑ ∞ n =0 c n ( 1 3 ) n is convergent? Answer: Yes (it is within the radius of convergence) (b) Does it follow that ∑ ∞ n =1 c n ( 1 2 ) n is convergent? Answer: No (if the radius of convergence is 1/2, it could diverge) (c) Does it follow that ∑ ∞ n =1 | c n (1) n | is divergent? Answer: No (maybe the c n were already positive and the ROC is 5) 3. (2 points) Give an example of a non-polynomial function that is equal to its MacLaurin series on its entire domain. Possible answers: e x , cos x, sin x 2...
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quiz07_s107_solns - ∑ ∞ n =0 c n-1 2 n is convergent(a...

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