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Unformatted text preview: Su, Yung – Homework 14 – Due: Dec 4 2007, 3:00 am – Inst: Shinko Harper 1 This printout should have 17 questions. Multiplechoice 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 . 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 ) = ln4 + ∞ X n = 1 x n 4 n 3. f ( x ) = ln4 ∞ X n = 1 x n n 4 n correct 4. f ( x ) = ∞ X n = 0 x n n 4 n 5. f ( x ) = ln4 ∞ X n = 0 x n 4 n 6. f ( x ) = ln4 + ∞ 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 1 1 s ds = Z x n ∞ X n = 0 s n o ds = ∞ X n = 0 Z x s n ds = ∞ X n = 1 x n n . For then by properties of logs, f ( x ) = ln4 ‡ 1 1 4 x · = ln4 + ln ‡ 1 1 4 x · , so that f ( x ) = ln4 ∞ 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 ∞...
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 Spring '09
 Harper
 Calculus, Power Series, Taylor Series, Shinko Harper

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