Exam 3, Spring '05, Solutions

Exam 3, Spring '05, Solutions - Solutions to Exam III...

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Solutions to Exam III (April 4, 2005) Math 12 Spring 2005 Tufts University (1a) . One computes: 1 x - 2 = - 1 2 - x = - 1 2 · 1 1 - x/ 2 = - 1 2 X n =0 ± x 2 ² n = X n =0 - 1 2 n +1 x n . This geometric series converges to the indicated value provided that ³ ³ ³ x 2 ³ ³ ³ < 1; the radius of convergence is R = 2. (1b) . Since d dx ´ 1 1 + x µ = - 1 (1 + x ) 2 , and since 1 1 + x = X n =0 ( - 1) n x n whenever | x | < 1, one Fnds that 1 (1 + x ) 2 = ( - 1) d dx ´ 1 1 + x µ = d dx X n =0 ( - 1) n +1 x n ! = X n =1 ( - 1) n +1 nx n - 1 whenever | x | < 1. (2a) . Recognize that X n =1 nx n - 1 = d dx X n =0 x n ! = d dx ´ 1 1 - x µ = 1 (1 - x ) 2 ; the radius of convergence is R = 1 since the equality 1 1 - x = X n =0 x n holds whenever | x | < 1, and since the radius of convergence of a power series is unchanged upon di±erentiation, (2b) . We have X n =1 n 3 n = 1 3 X n =1 n ´ 1 3 µ n - 1 ; since | 1 / 3 | < R = 1, the series X n =1 n ´ 1 3 µ n - 1 is the value at x = 1 / 3 of the series found in (2a) to represent 1 (1 - x ) 2 ; thus X n =1 n 3 n = 1 3 X n =1 n ´ 1 3 µ n - 1 = 1 3 ´ 1 (1 - 1 / 3) 2 µ = 1 3 · 1 4 / 9 = 1 3 · 9 4 = 3 4 . (3) . Since d dx ln(1 - x ) = - 1 1 - x , one knows that ln(1 - x ) = Z - 1 1 - x dx = - Z X n =0 x n ! dx = C - X n =0 x n +1 n + 1 = C - X n =1 x n n for some constant C . Since 0 = ln(1) = ln(1 - 0), and since the value of the Right-Hand-Side at x = 0 is C , we have C = 0 and so we Fnd ln(1 -
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This homework help was uploaded on 03/26/2008 for the course MATH 12 taught by Professor Garant during the Spring '08 term at Tufts.

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Exam 3, Spring '05, Solutions - Solutions to Exam III...

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