11.8_ Power Series-solutions

# x n an by the ratio test therefore the given series

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Unformatted text preview: when |x| &gt; 1. x→∞ 005 n=1 √ 1 . n+5 Applying the Integral Test with ∞ cn 7n n=0 is convergent, which of the following statements must be true without further restrictions on cn ? ∞ 1 f ( x) = √ x+5 1. we see that f is continuous, positive, and decreasing on [1, ∞), but the improper integral 2. n=0 n=0 f (x) dx ∞ (‡) n=1 (−1)n √ . n+5 cn (−3)n is convergent correct ∞ 1 diverges, so the inﬁnite series (∗) diverges also. On the other hand, at x = −1, the series becomes cn (−3)n is divergent ∞ ∞ I= 10.0 points If the series We have still to check what happens at the endpoints x = ±1. At x = 1 the series becomes ( ∗) 1 = 0, x+5 however, the Alternating Series Test ensures that (‡) converges. Consequently, the so ∞ 1 x+5 the same continuous, positive and decreasing function as before. Since an+1 an lim a n = f ( n) with xn =√ , n+5 √ xn+1 n+5 =√ n + 6 xn √ n+5 = | x| = | x| √ n+6 (−1)n an , 3. n=0 cn (−7)n is convergent ∞ 4. n=0 cn (−7)n is divergent Explanation: stiurca (mas7745) – 11.8: Power Series – meth – (91825) The series with ∞ cn xn an = n=0 converges when x = 7, its radius of convergence, R, must have the property R ≥ 7 since the series (i) converges when |x| &lt; R, and (ii) diverges when |x| &gt; R. Consequently, it must be true that the series ∞ n=0 cn (−3) n is convergent but we cannot determine whether either of the statements ∞ (i) n=0 ∞ cn (−7)n is convergent, n (ii) n=0 cn (−7) is divergent For the series ∞ xn n=1 (n + 6)! , (i) determine its radius of convergence, R. 1 . (n + 6)! Now for this series, (i) R = 0 if it converges only at x = 0, (ii) R = ∞ if it converges for all x, while 0 &lt; R &lt; ∞, (iii) if it converges when |x| &lt; R, and (iv) diverges when |x| &gt; R. But lim n→∞ an+1 an = lim n→∞ 1 = 0. n+7 By the Ratio Test, therefore, the given series converges for all x. Consequently, R=∞ . must be true without further conditions on cn . 006 (part 1 of 2) 10.0 points 4 007 (part 2 of 2) 10.0 points (ii) Determine the interval of convergence of the series. 1. (−6, 6) 1. R = 0 2. [−1, 1] 2. R = 6 3. (−∞, ∞) correct 4. converges only at x = 0 3. R = 1 5. [−6, 6] 1 4. R = 6 6. (−1, 1] 5. R = ∞ correct Explanation: The given series has the form Explanation: Since R = ∞, the given series converges for all x, and so has ∞ a n xn n=1 interval convergence = (−∞, ∞) . stiurca (mas7745) – 11.8: Power Series – meth – (91825) 008 10.0 points Find the interval of convergence of the power series ∞ 4 xn √. n which diverges by the p-series test with p = 1 ≤ 1. On the other hand, when x = −1, the 2 series reduces to ∞ n=1 1. [−1, 1) correct 5 n=1 4 (−1)n √ n which converges by the Alternating Series Test. Thus the 2. (−1, 1) interval of convergence = [−1, 1) . 3. (−4, 4) 009 4. [−4, 4) 10.0 points Determine the interval of convergence of the series of the series 5. (−4, 4] ∞ 6. [−1, 1] n=1...
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