HWK14 - Valenica Daniel – Homework 14 – Due Dec 4 2007 3:00 am – Inst Cheng 1 This print-out should have 17 questions Multiple-choice

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Unformatted text preview: Valenica, Daniel – Homework 14 – Due: Dec 4 2007, 3:00 am – Inst: Cheng 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 y n with the radius of convergence, R 2 , of the series ∞ X n = 1 n c n y n- 1 when lim n →∞ fl fl fl c n +1 c n fl fl fl = 8 . 1. R 1 = 2 R 2 = 8 2. R 1 = R 2 = 8 3. R 1 = R 2 = 1 8 correct 4. R 1 = 2 R 2 = 1 8 5. 2 R 1 = R 2 = 1 8 6. 2 R 1 = R 2 = 8 Explanation: When lim n →∞ fl fl fl c n +1 c n fl fl fl = 8 , the Ratio Test ensures that the series ∞ X n = 0 c n y n is (i) convergent when | y | < 1 8 , and (ii) divergent when | y | > 1 8 . 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 y n- 1 is (i) convergent when | y | < 1 8 , and (ii) divergent when | y | > 1 8 . Consequently, R 1 = R 2 = 1 8 . keywords: 002 (part 1 of 1) 10 points Find a power series representation for the function f ( t ) = 1 t- 2 . 1. f ( t ) = ∞ X n = 0 (- 1) n 2 n t n 2. f ( t ) =- ∞ X n = 0 2 n t n 3. f ( t ) = ∞ X n = 0 (- 1) n- 1 2 n +1 t n 4. f ( t ) =- ∞ X n = 0 1 2 n +1 t n correct 5. f ( t ) = ∞ X n = 0 1 2 n +1 t n Explanation: We know that 1 1- x = 1 + x + x 2 + . . . = ∞ X n = 0 x n . Valenica, Daniel – Homework 14 – Due: Dec 4 2007, 3:00 am – Inst: Cheng 2 On the other hand, 1 t- 2 =- 1 2 ‡ 1 1- ( t/ 2) · . Thus f ( t ) =- 1 2 ∞ X n = 0 µ t 2 ¶ n =- 1 2 ∞ X n = 0 1 2 n t n . Consequently, f ( t ) =- ∞ X n = 0 1 2 n +1 t n with | t | < 2. keywords: 003 (part 1 of 1) 10 points Find a power series representation for the function f ( t ) = ln(2- t ) . 1. f ( t ) = ∞ X n = 0 t n n 2 n 2. f ( t ) =- ∞ X n = 1 t n n 2 n 3. f ( t ) = ln2- ∞ X n = 0 t n 2 n 4. f ( t ) = ln2 + ∞ X n = 0 t n n 2 n 5. f ( t ) = ln2 + ∞ X n = 1 t n 2 n 6. f ( t ) = ln2- ∞ X n = 1 t n n 2 n correct 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 ( t ) = ln2 ‡ 1- 1 2 t · = ln2 + ln ‡ 1- 1 2 t · , so that f ( t ) = ln2- ∞ X n = 1 t n n 2 n . keywords: 004 (part 1 of 1) 10 points Find a power series representation centered at the origin for the function f ( y ) = y 3 (7- y ) 2 . 1. f ( y ) = ∞ X n = 2 1 7 n- 1 y n 2. f ( y ) = ∞ X n = 3 n 7 n y n 3. f ( y ) = ∞ X n = 2 n- 1 7 n y n 4. f ( y ) = ∞ X n = 3 1 7 n- 3 y n 5. f ( y ) = ∞ X n = 3 n- 2 7 n- 1 y n correct Explanation: Valenica, Daniel – Homework 14 – Due: Dec 4 2007, 3:00 am – Inst: Cheng 3 By the known result for geometric series, 1 7- y = 1 7 ‡ 1- y 7 · = 1 7 ∞ X...
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This note was uploaded on 04/16/2008 for the course CALC 303L taught by Professor Cheng during the Fall '07 term at University of Texas at Austin.

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HWK14 - Valenica Daniel – Homework 14 – Due Dec 4 2007 3:00 am – Inst Cheng 1 This print-out should have 17 questions Multiple-choice

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