M408L hwk14 solutions

# M408L hwk14 solutions - Pham Quoc Homework 14 Due May 1...

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Pham, Quoc – Homework 14 – Due: May 1 2007, 3:00 am – Inst: Eric Katerman 1 This print-out should have 18 questions. Multiple-choice questions may continue on the next column or page – fnd all choices beFore answering. The due time is Central time. 001 (part 1 oF 1) 10 points Compare the radius oF convergence, R 1 , oF the series X n = 0 c n y n with the radius oF convergence, R 2 , oF the series X n = 1 n c n y n - 1 when lim n →∞ f f f c n +1 c n f f f = 3 . 1. R 1 = 2 R 2 = 1 3 2. 2 R 1 = R 2 = 3 3. R 1 = 2 R 2 = 3 4. R 1 = R 2 = 3 5. 2 R 1 = R 2 = 1 3 6. R 1 = R 2 = 1 3 correct Explanation: When lim n →∞ f f f c n +1 c n f f f = 3 , the Ratio Test ensures that the series X n = 0 c n y n is (i) convergent when | y | < 1 3 , and (ii) divergent when | y | > 1 3 . On the other hand, since lim n →∞ f f f ( n + 1) c n +1 nc n f f f = lim n →∞ f f f c n +1 c n f f f , the Ratio Test ensures also that the series X n = 1 n c n y n - 1 is (i) convergent when | y | < 1 3 , and (ii) divergent when | y | > 1 3 . Consequently, R 1 = R 2 = 1 3 . keywords: 002 (part 1 oF 1) 10 points ±ind a power series representation For the Function f ( t ) = 1 6 + t . 1. f ( t ) = X n = 0 6 n +1 t n 2. f ( t ) = X n = 0 1 6 n +1 t n 3. f ( t ) = X n = 0 ( - 1) n 6 t n 4. f ( t ) = X n = 0 ( - 1) n 6 n +1 t n correct 5. f ( t ) = X n = 0 ( - 1) n 6 n +1 t n Explanation: We know that 1 1 - x = 1 + x + x 2 + . . . = X n = 0 x n .

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Pham, Quoc – Homework 14 – Due: May 1 2007, 3:00 am – Inst: Eric Katerman 2 On the other hand, 1 6 + t = 1 6 1 1 - ( - t/ 6) · . Thus f ( t ) = 1 6 X n = 0 µ - t 6 n = 1 6 X n = 0 ( - 1) n 6 n t n . Consequently, f ( t ) = X n = 0 ( - 1) n 6 n +1 t n with | t | < 6. keywords: 003 (part 1 of 1) 10 points Find a power series representation for the function f ( t ) = t 16 t + 1 . 1. f ( t ) = X n = 0 4 n t n 2. f ( t ) = X n = 0 4 2 n t n +1 3. f ( t ) = X n = 0 ( - 1) n 4 n t n 4. f ( t ) = X n = 0 4 2 n t n 5. f ( t ) = X n = 0 ( - 1) n 4 n t n +1 6. f ( t ) = X n = 0 ( - 1) n 4 2 n t n +1 correct Explanation: After simpli±cation, f ( t ) = t 16 t + 1 = t 1 - ( - 16 t ) . On the other hand, 1 1 - x = X n = 0 x n . Thus f ( t ) = t n X n = 0 ( - 16 t ) n o = t n X n = 0 ( - 1) n 4 2 n t n . Consequently, f ( t ) = X n = 0 ( - 1) n 4 2 n t n +1 . keywords: 004 (part 1 of 1) 10 points Find a power series representation centered at the origin for the function f ( y ) = y 3 (3 - y ) 2 . 1. f ( y ) = X n = 3 n - 2 3 n - 1 y n correct 2. f ( y ) = X n = 2 1 3 n - 1 y n 3. f ( y ) = X n = 3 1 3 n - 3 y n 4. f ( y ) = X n = 2 n - 1 3 n y n 5. f ( y ) = X n = 3 n 3 n y n Explanation:
Pham, Quoc – Homework 14 – Due: May 1 2007, 3:00 am – Inst: Eric Katerman 3 By the known result for geometric series, 1 3 - y = 1 3 1 - y 3 · = 1 3 X n = 0 y 3 · n = X n = 0 1 3 n +1 y n . This series converges on (

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M408L hwk14 solutions - Pham Quoc Homework 14 Due May 1...

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