HW15-solutions

HW15-solutions - kim(tk5895 – HW15 – Henry –(54974 1...

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Unformatted text preview: kim (tk5895) – HW15 – Henry – (54974) 1 This print-out should have 15 questions. Multiple-choice 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 ....
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HW15-solutions - kim(tk5895 – HW15 – Henry –(54974 1...

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