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Unformatted text preview: suleimenov (bs26835) HW 04 rusin (55565) 1 This printout should have 17 questions. Multiplechoice questions may continue on the next column or page find all choices before answering. OMG! I just learned that I forgot to pub lish this HW. Im so sorry! Please give this material a shot I will reduce the amount of powerseries stuff on the test, but there will still be some questions about this material. Im really sorry... The due date of this homework has been set to next Friday at the classs request, so that you can complete it at your leisure after EXAM 1, and get the TAs help on Wednes day. But I urge you to work on this BEFORE the exam: questions similar to those shown will be on the exam. (Unlike the Test 1 Re view, this is a regular, graded homework.) 001 10.0 points Determine the radius of convergence, R , of the power series summationdisplay n = 1 8 x n n . 1. R = 0 2. R = 3. R = 1 correct 4. R = 8 5. R = 1 8 Explanation: The given series has the form summationdisplay n =1 a n x n with a n = 8 n . Now for this series, (i) R = 0 if it converges only at x = 0, (ii) R = if it converges for all x , while 0 < R < (iii) if 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 n n + 1 = 1 . By the Ratio Test, therefore, the given series (a) converges for all  x  < 1, and (b) diverges for all  x  > 1. Consequently, the given series has radius of convergence R = 1 . 002 10.0 points Determine the radius of convergence, R , of the series summationdisplay n =1 ( 2) n n + 3 x n . 1. R = 1 3 2. R = 0 3. R = 2 4. R = 3 5. R = 6. R = 1 2 correct Explanation: The given series has the form summationdisplay n = 1 a n x n suleimenov (bs26835) HW 04 rusin (55565) 2 with a n = ( 2) 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 2( n + 3) n + 4 = 2 . By the Ratio Test, therefore, the given series converges when  x  < 1 / 2 and diverges when  x  > 1 / 2. Consequently, R = 1 2 . 003 10.0 points Find the interval of convergence of the se ries summationdisplay n =1 2 n x n . 1. converges only at x = 0 2. interval of cgce = [ 1 , 1) 3. interval of cgce = ( 1 , 1) correct 4. interval of cgce = ( 2 , 2] 5. interval of cgce = [ 2 , 2] 6. interval of cgce = ( 1 , 1] Explanation: When a n = 2 n x n , then vextendsingle vextendsingle vextendsingle vextendsingle a n +1 a n vextendsingle vextendsingle vextendsingle vextendsingle = vextendsingle vextendsingle vextendsingle vextendsingle 2 n + 2 x n +1 2 n x n vextendsingle...
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 Spring '07
 Treisman
 Calculus

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