samplefinalunkownyear

# samplefinalunkownyear - MATH 262: Sample Final with...

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Unformatted text preview: MATH 262: Sample Final with Solutions Problem 1. Determine the values of x for which the series X n =1 (4 x + 1) n n converges absolutely, converges conditionally, or diverges. Solution. Apply the Ratio Test. lim n (4 x + 1) n +1 n + 1 / (4 x + 1) n n = lim n (4 x + 1) n +1 n (4 x + 1) n ( n + 1) = | 4 x + 1 | lim n n n + 1 = | 4 x + 1 | . Hence, the series converges absolutely if | 4 x + 1 | < 1, and diverges if | 4 x + 1 | > 1. Check the case when | 4 x + 1 | = 1. If 4 x + 1 = 1 then X n =1 (4 x + 1) n n = X n =1 1 n diverges. If 4 x + 1 =- 1 then X n =1 (4 x + 1) n n = X n =1 (- 1) n n converges conditionally. Problem 2. Given the function F ( x ) = Z x sin t t dt. (a) find the Maclaurin series for F ( x ), (b) evaluate F (1) with error less than 10- 5 . Solution. (a) We have sin t = X k =0 (- 1) k (2 k + 1)! t 2 k +1 . Hence, sin t t = X k =0 (- 1) k (2 k + 1)! t 2 k and F ( x ) = Z x X k =0 (- 1) k (2 k + 1)! t 2 k ! dt = X k =0 (- 1) k (2 k + 1)! t 2 k +1 2 k + 1 . 1 (b) F (1) = X k =0 (- 1) k (2 k + 1) (2 k + 1)! is an alternating series and we can use the alternating series error estimation. Hence, | E (1)- s n | (- 1) n (2 n + 1) (2 n + 1)! = 1 (2 n + 1) (2 n + 1)! , where s n is the sum of the first n terms of the series. Observe that the solution of the inequality 1 (2 n + 1) (2 n + 1)! < 10- 5 provides the required number of terms. The solution is n 4, so we need at least 4 first terms of the series. Problem 3. (a) Without using lHopitals Rule, compute lim x ( e 2 x- 1) 2 ln(1 + x )- x . (b) Determine power series representation for the function f ( x ) = 1- x 1 + x in powers if x- 2. Solution. (a) We have e 2 x = 1 + 2 x 1!...
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## This note was uploaded on 01/15/2010 for the course MATH MATH 262 taught by Professor Gregrix during the Spring '09 term at McGill.

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samplefinalunkownyear - MATH 262: Sample Final with...

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