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# cal15 - choi(kc24547 HW15 BERG(56525 This print-out should...

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choi (kc24547) – HW15 – BERG – (56525) 1 This print-out should have 14 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. 001 10.0 points Compare the radius of convergence, R 1 , of the series summationdisplay n =0 c n x n with the radius of convergence, R 2 , of the series summationdisplay n =1 n c n x n 1 when lim n → ∞ vextendsingle vextendsingle vextendsingle c n +1 c n vextendsingle vextendsingle vextendsingle = 3 . 1. R 1 = 2 R 2 = 3 2. R 1 = 2 R 2 = 1 3 3. 2 R 1 = 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 → ∞ vextendsingle vextendsingle vextendsingle c n +1 c n vextendsingle vextendsingle vextendsingle = 3 , the Ratio Test ensures that the series summationdisplay n =0 c n x n is (i) convergent when | x | < 1 3 , and (ii) divergent when | x | > 1 3 . 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 x n 1 is (i) convergent when | x | < 1 3 , and (ii) divergent when | x | > 1 3 . Consequently, R 1 = R 2 = 1 3 . 002 10.0 points Find a power series representation for the function f ( t ) = 1 2 + t . 1. f ( t ) = summationdisplay n =0 1 2 n +1 t n 2. f ( t ) = summationdisplay n =0 ( - 1) n 2 n +1 t n 3. f ( t ) = summationdisplay n =0 ( - 1) n 2 n +1 t n correct 4. f ( t ) = summationdisplay n =0 ( - 1) n 2 t n 5. f ( t ) = summationdisplay n =0 2 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 2 + t = 1 2 parenleftBig 1 1 - ( - t/ 2) parenrightBig .

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choi (kc24547) – HW15 – BERG – (56525) 2 Thus f ( t ) = 1 2 summationdisplay n =0 parenleftbigg - t 2 parenrightbigg n = 1 2 summationdisplay n =0 ( - 1) n 2 n t n . Consequently, f ( t ) = summationdisplay n =0 ( - 1) n 2 n +1 t n with | t | < 2. 003 10.0 points Find a power series representation centered at the origin for the function f ( y ) = y 3 (6 - y ) 2 . 1. f ( y ) = summationdisplay n =3 n 6 n y n 2. f ( y ) = summationdisplay n =2 1 6 n 1 y n 3. f ( y ) = summationdisplay n =3 n - 2 6 n 1 y n correct 4. f ( y ) = summationdisplay n =3 1 6 n 3 y n 5. f ( y ) = summationdisplay n =2 n - 1 6 n y n Explanation: By the known result for geometric series, 1 6 - y = 1 6 parenleftBig 1 - y 6 parenrightBig = 1 6 summationdisplay n =0 parenleftBig y 6 parenrightBig n = summationdisplay n =0 1 6 n +1 y n . This series converges on ( - 6 , 6). On the other hand, 1 (6 - y ) 2 = d dy parenleftBig 1 6 - y parenrightBig , and so on ( - 6 , 6), 1 (6 - y ) 2 = d dy parenleftBigg summationdisplay n =0 y n 6 n +1 parenrightBigg = summationdisplay n =1 n 6 n +1 y n 1 = summationdisplay n =0 n + 1 6 n +2 y n . Thus f ( y ) = y 3 summationdisplay n =0 n + 1 6 n +2 y n = summationdisplay n =0 n + 1 6 n +2 y n +3 .
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