This preview shows pages 1–3. Sign up to view the full content.
This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
Unformatted text preview: kim (tk5895) HW15 Henry (54974) 1 This printout should have 15 questions. Multiplechoice questions may continue on the next column or page find all choices before answering. 001 10.0 points Determine the radius of convergence, R , of the series summationdisplay n = 1 ( 5) n n + 3 x n . 1. R = 3 2. R = 0 3. R = 5 4. R = 1 3 5. R = 1 5 correct 6. R = Explanation: The given series has the form summationdisplay n =1 a n x n with a n = ( 5) n n + 3 . Now for this series, (i) R = 0 if it converges only at x = 0, (ii) R = if it converges for all x , while if R > 0, (iii) it converges when  x  < R , and (iv) diverges when  x  > R . But lim n vextendsingle vextendsingle vextendsingle a n +1 a n vextendsingle vextendsingle vextendsingle = lim n 5( n + 3) n + 4 = 5 . By the Ratio Test, therefore, the given series converges when  x  < 1 / 5 and diverges when  x  > 1 / 5. Consequently, R = 1 5 . 002 10.0 points Find the interval of convergence of the power series summationdisplay n = 1 parenleftBig 8 n + 7 2 n parenrightBig n x n . 1. interval of convergence = parenleftBig 4 , 4 parenrightBig 2. interval of convergence = bracketleftBig 4 , 4 parenrightBig 3. interval of convergence = bracketleftBig 1 4 , 1 4 parenrightBig 4. interval of convergence = [ 2 , 2) 5. interval of convergence = parenleftBig 1 4 , 1 4 parenrightBig correct Explanation: We first apply the root test to the infinite series summationdisplay n = 1  a n  ,  a n  = parenleftBig 8 n + 7 2 n parenrightBig n  x  n . For this series  a n  1 /n = parenleftBig 8 n + 7 2 n parenrightBig  x  4  x  as n . Thus the given power series will converge on the interval ( 1 / 4 , 1 / 4). For convergence at the endpoints x = 1 4 we have to check if summationdisplay n =0 parenleftBig 8 n + 7 2 n parenrightBig n parenleftBig 1 4 parenrightBig n = summationdisplay n =0 a n kim (tk5895) HW15 Henry (54974) 2 converges, or if summationdisplay n =0 parenleftBig 8 n + 7 2 n parenrightBig n parenleftBig 1 4 parenrightBig n = summationdisplay n = 0 b n converges. In the first case a n = parenleftBig 8 n + 7 2 n parenrightBig n parenleftBig 1 4 parenrightBig n = parenleftBig 8 n + 7 8 n parenrightBig n , in which case lim n a n = e 7 / 8 negationslash = 0; by the Divergence test, therefore the series summationdisplay n = 1 a n diverges. On the other hand, b n = parenleftBig 8 n + 7 2 n parenrightBig n parenleftBig 1 4 parenrightBig n = ( 1) n parenleftBig 8 n + 7 8 n parenrightBig n = ( 1) n a n . But, as we have seen, lim n a n = e 7 / 8 ....
View
Full
Document
 Fall '08
 Cepparo
 Calculus

Click to edit the document details