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Homework 14

# Homework 14 - Su Yung Homework 14 Due Dec 4 2007 3:00 am...

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Su, Yung – Homework 14 – Due: Dec 4 2007, 3:00 am – Inst: Shinko Harper 1 This print-out should have 17 questions. Multiple-choice questions may continue on the next column or page – find 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 z n with the radius of convergence, R 2 , of the series X n = 1 n c n z n - 1 when lim n → ∞ fl fl fl c n +1 c n fl fl fl = 7 . 1. R 1 = R 2 = 1 7 correct 2. 2 R 1 = R 2 = 1 7 3. R 1 = R 2 = 7 4. 2 R 1 = R 2 = 7 5. R 1 = 2 R 2 = 1 7 6. R 1 = 2 R 2 = 7 Explanation: When lim n → ∞ fl fl fl c n +1 c n fl fl fl = 7 , the Ratio Test ensures that the series X n = 0 c n z n is (i) convergent when | z | < 1 7 , and (ii) divergent when | z | > 1 7 . On the other hand, since lim n → ∞ fl fl fl ( n + 1) c n +1 nc n fl fl fl = lim n → ∞ fl fl fl c n +1 c n fl fl fl , the Ratio Test ensures also that the series X n = 1 n c n z n - 1 is (i) convergent when | z | < 1 7 , and (ii) divergent when | z | > 1 7 . Consequently, R 1 = R 2 = 1 7 . keywords: 002 (part 1 of 1) 10 points Find a power series representation for the function f ( x ) = 1 x - 4 . 1. f ( x ) = - X n = 0 1 4 n +1 x n correct 2. f ( x ) = X n = 0 ( - 1) n - 1 4 n +1 x n 3. f ( x ) = - X n = 0 4 n x n 4. f ( x ) = X n = 0 ( - 1) n 4 n x n 5. f ( x ) = X n = 0 1 4 n +1 x n Explanation: We know that 1 1 - x = 1 + x + x 2 + . . . = X n = 0 x n .

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Su, Yung – Homework 14 – Due: Dec 4 2007, 3:00 am – Inst: Shinko Harper 2 On the other hand, 1 x - 4 = - 1 4 1 1 - ( x/ 4) · . Thus f ( x ) = - 1 4 X n = 0 x 4 · n = - 1 4 X n = 0 1 4 n x n . Consequently, f ( x ) = - X n = 0 1 4 n +1 x n with | x | < 4. keywords: 003 (part 1 of 1) 10 points Find a power series representation for the function f ( x ) = ln(4 - x ) . 1. f ( x ) = - X n = 1 x n n 4 n 2. f ( x ) = ln 4 + X n = 1 x n 4 n 3. f ( x ) = ln 4 - X n = 1 x n n 4 n correct 4. f ( x ) = X n = 0 x n n 4 n 5. f ( x ) = ln 4 - X n = 0 x n 4 n 6. f ( x ) = ln 4 + X n = 0 x n n 4 n Explanation: We can either use the known power series representation ln(1 - x ) = - X n = 1 x n n , or the fact that ln(1 - x ) = - Z x 0 1 1 - s ds = - Z x 0 n X n = 0 s n o ds = - X n = 0 Z x 0 s n ds = - X n = 1 x n n . For then by properties of logs, f ( x ) = ln 4 1 - 1 4 x · = ln 4 + ln 1 - 1 4 x · , so that f ( x ) = ln 4 - X n = 1 x n n 4 n . keywords: 004 (part 1 of 1) 10 points Find a power series representation centered at the origin for the function f ( z ) = z 3 (6 - z ) 2 . 1. f ( z ) = X n = 2 n - 1 6 n z n 2. f ( z ) = X n = 3 1 6 n - 3 z n 3. f ( z ) = X n = 2 1 6 n - 1 z n 4. f ( z ) = X n = 3 n 6 n z n 5. f ( z ) = X n = 3 n - 2 6 n - 1 z n correct Explanation:
Su, Yung – Homework 14 – Due: Dec 4 2007, 3:00 am – Inst: Shinko Harper 3 By the known result for geometric series, 1 6 - z = 1 6 1 - z 6 · = 1 6 X n = 0 z 6 · n = X n = 0 1 6 n +1 z n . This series converges on ( - 6 , 6).

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Homework 14 - Su Yung Homework 14 Due Dec 4 2007 3:00 am...

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