HW 5-solutions - morris (gmm643) – HW 5 – mann –...

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Unformatted text preview: morris (gmm643) – HW 5 – mann – (54675) 1 This print-out should have 17 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. Sections 12.9, 12.10, 12.11 001 10.0 points Compare the radius of convergence, R 1 , of the series ∞ summationdisplay n = 0 c n t n with the radius of convergence, R 2 , of the series ∞ summationdisplay n =1 n c n t n − 1 when lim n →∞ vextendsingle vextendsingle vextendsingle c n +1 c n vextendsingle vextendsingle vextendsingle = 4 . 1. 2 R 1 = R 2 = 1 4 2. R 1 = 2 R 2 = 4 3. 2 R 1 = R 2 = 4 4. R 1 = 2 R 2 = 1 4 5. R 1 = R 2 = 1 4 correct 6. R 1 = R 2 = 4 Explanation: When lim n →∞ vextendsingle vextendsingle vextendsingle c n +1 c n vextendsingle vextendsingle vextendsingle = 4 , the Ratio Test ensures that the series ∞ summationdisplay n = 0 c n t n is (i) convergent when | t | < 1 4 , and (ii) divergent when | t | > 1 4 . On the other hand, since lim n →∞ vextendsingle vextendsingle vextendsingle ( n + 1) c n +1 nc n vextendsingle vextendsingle vextendsingle = lim n →∞ vextendsingle vextendsingle vextendsingle c n +1 c n vextendsingle vextendsingle vextendsingle , the Ratio Test ensures also that the series ∞ summationdisplay n = 1 n c n t n − 1 is (i) convergent when | t | < 1 4 , and (ii) divergent when | t | > 1 4 . Consequently, R 1 = R 2 = 1 4 . 002 10.0 points Find a power series representation for the function f ( t ) = 1 5 + t . 1. f ( t ) = ∞ summationdisplay n =0 (- 1) n 5 n +1 t n correct 2. f ( t ) = ∞ summationdisplay n =0 5 n +1 t n 3. f ( t ) = ∞ summationdisplay n =0 (- 1) n 5 t n 4. f ( t ) = ∞ summationdisplay n =0 1 5 n +1 t n 5. f ( t ) = ∞ summationdisplay n =0 (- 1) n 5 n +1 t n Explanation: We know that 1 1- x = 1 + x + x 2 + . . . = ∞ summationdisplay n =0 x n . On the other hand, 1 5 + t = 1 5 parenleftBig 1 1- (- t/ 5) parenrightBig . morris (gmm643) – HW 5 – mann – (54675) 2 Thus f ( t ) = 1 5 ∞ summationdisplay n =0 parenleftbigg- t 5 parenrightbigg n = 1 5 ∞ summationdisplay n = 0 (- 1) n 5 n t n . Consequently, f ( t ) = ∞ summationdisplay n = 0 (- 1) n 5 n +1 t n with | t | < 5. 003 10.0 points Find a power series representation for the function f ( y ) = ln(2- y ) . 1. f ( y ) = ln 2 + ∞ summationdisplay n = 0 y n n 2 n 2. f ( y ) = ln 2- ∞ summationdisplay n = 0 y n 2 n 3. f ( y ) =- ∞ summationdisplay n = 1 y n n 2 n 4. f ( y ) = ∞ summationdisplay n = 0 y n n 2 n 5. f ( y ) = ln 2 + ∞ summationdisplay n = 1 y n 2 n 6. f ( y ) = ln 2- ∞ summationdisplay n = 1 y n n 2 n correct Explanation: We can either use the known power series representation ln(1- x ) =- ∞ summationdisplay n =1 x n n , or the fact that ln(1- x ) =- integraldisplay x 1 1- s ds =- integraldisplay x braceleftBig ∞ summationdisplay n = 0 s n bracerightBig ds =- ∞ summationdisplay n = 0 integraldisplay x s n ds =- ∞ summationdisplay n =1 x n n ....
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This note was uploaded on 11/17/2010 for the course PHY 56630 taught by Professor Coker during the Spring '10 term at University of Texas.

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HW 5-solutions - morris (gmm643) – HW 5 – mann –...

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