# Ii find an explicit formula for f x 1 f x 4 7 x 1 x 2

• 12
• 100% (1) 1 out of 1 people found this document helpful

This preview shows page 6 - 9 out of 12 pages.

(ii) Find an explicit formula for f ( x ). But summationdisplay n ( a n + b n ) = parenleftBig summationdisplay n a n parenrightBig + parenleftBig summationdisplay n b n parenrightBig when both of summationdisplay n a n summationdisplay n b n converge. Consequently, on ( 1 , 1) f ( x ) = 7 + 4 x 1 x 2 . 010 10.0points If the series summationdisplay n =0 c n x n converges when x = 4 and diverges when x = 6, which of the following series converge without further restrictions on { c n } ?
vidal (mav3385) – HW14 – schultz – (53130) 7 8. B only correct Explanation: A. The series summationdisplay n =0 c n x n diverges for all | x | > 6, hence for x = 8. B. Since summationdisplay n =0 c n ( 4) n +1 = 4 parenleftBigg summationdisplay n =0 c n ( 4) n parenrightBigg the series summationdisplay n =0 c n ( 4) n +1 converges. C. The series summationdisplay n =0 c n x n diverges for all | x | > 6, hence for x = 7. 011 10.0points Determine the interval of convergence of the series summationdisplay k =1 ( 1) k 1 k 2 k (2 x + 1) k . 1. interval convergence = parenleftBig 3 2 , 1 2 parenrightBig 2. interval convergence = bracketleftBig 3 2 , 1 2 bracketrightBig 3. interval convergence = bracketleftBig 3 2 , 1 2 parenrightBig 4. interval convergence = parenleftBig 3 2 , 1 2 bracketrightBig cor- rect 5. interval convergence = ( −∞ , ) 6. converges only at x = 1 2 Explanation: summationdisplay k =1 c k parenleftBig x + 1 2 parenrightBig k , c k = ( 1) k k . But then lim k → ∞ vextendsingle vextendsingle vextendsingle c k +1 c k vextendsingle vextendsingle vextendsingle = lim k → ∞ k k + 1 = 1 . By the Ratio Test, therefore, the series (i) convergeswhen | x + 1 2 | < 1, (ii) divergeswhen | x + 1 2 | > 1. Now at the point x + 1 2 = 1, the series reduces to summationdisplay k =1 ( 1) k k , while at x + 1 2 = 1, it reduces to summationdisplay k =1 1 k . In the first case the series converges by the Alternating Series Test, while in the second it diverges by the p -Series Test with p = 1. Consequently, the series has interval of convergence = parenleftBig 3 2 , 1 2 bracketrightBig . 012 10.0points Find a power series representation centered at the origin for the function f ( x ) = 1 (7 x ) 2 . 1. f ( x ) = summationdisplay n =0 n + 1 7 n x n 2. f ( x ) = summationdisplay n =0 ( n + 1) x n 3. f ( x ) = summationdisplay n =1 n 7 n +1 x n - 1 correct
vidal (mav3385) – HW14 – schultz – (53130) 8 4. f ( x ) = summationdisplay n =1 1 7 n +1 x n 5. f ( x ) = summationdisplay n =0 1 7 n +1 x n 6. f ( x ) = summationdisplay n =1 n 7 n x n - 1 Explanation: By the known result for geometric series, 1 7 x = 1 7 parenleftBig 1 x 7 parenrightBig = 1 7 summationdisplay n =0 parenleftBig x 7 parenrightBig n = summationdisplay n =0 1 7 n +1 x n . This series converges on ( 7 , 7). On the other hand, 1 (7 x ) 2 = d dx parenleftBig 1 7 x parenrightBig , and so on ( 7 , 7), 1 (7 x ) 2 = d dx parenleftBigg summationdisplay n =0 x n 7 n +1 parenrightBigg = summationdisplay n =1 n 7 n +1 x n - 1 .