Homework #13 with Solutions - First-Year Interest Group Seminar | N 1, Assignments of Health sciences

Download Homework #13 with Solutions - First-Year Interest Group Seminar | N 1 and more Assignments Health sciences in PDF only on Docsity! Version PREVIEW – Homework13 – Berg – (58460) 1 This print-out should have 19 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. CalC12h01a 001 10.0 points Find the interval of convergence of the power series ∞ ∑ n = 1 4xn√ n . 1. interval of cgce = [−4, , 4] 2. interval of cgce = [−1, 1) correct 3. interval of cgce = (−4, 4] 4. interval of cgce = (−1, 1) 5. interval of cgce = (−1, 1] 6. interval of cgce = (−4, 4) 7. interval of cgce = [−1, 1] 8. interval of cgce = [−4, 4) Explanation: When an = 4xn√ n , then ∣ ∣ ∣ ∣ an+1 an ∣ ∣ ∣ ∣ = ∣ ∣ ∣ ∣ xn+1√ n + 1 · √ n xn ∣ ∣ ∣ ∣ = |x|√n√ n + 1 = |x| (√ n n + 1 ) . Thus lim n→∞ ∣ ∣ ∣ ∣ an+1 an ∣ ∣ ∣ ∣ = |x| . By the Ratio Test, therefore, the given series converges when |x| < 1, and diverges when |x| > 1. We have still to check for convergence at x = ±1. But when x = 1, the series reduces to ∞ ∑ n =1 4√ n which diverges by the p-series test with p = 1 2 ≤ 1. On the other hand, when x = −1, the series reduces to ∞ ∑ n =1 (−1)n 4√ n which converges by the Alternating Series Test. Thus the interval of convergence = [−1, 1) . keywords: CalC12h04a 002 (part 1 of 2) 10.0 points For the series ∞ ∑ n =1 (−1)n n + 8 xn , (i) determine its radius of convergence, R. 1. R = 1 correct 2. R = 0 3. R = 1 8 4. R = (−∞, ∞) 5. R = 8 Explanation: The given series has the form ∞ ∑ n =1 an with an = (−1)n xn n + 8 . Now for this series, Version PREVIEW – Homework13 – Berg – (58460) 2 (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→∞ ∣ ∣ ∣ an+1 an ∣ ∣ ∣ = lim n→∞ |x|n + 8 n + 9 = |x| . By the Ratio Test, therefore, the given series converges when |x| < 1 and diverges when |x| > 1. Consequently, R = 1 . 003 (part 2 of 2) 10.0 points (ii) Determine the interval of convergence of the series. 1. interval convergence = (−8, 8] 2. converges only at x = 0 3. interval convergence = [−8, 8) 4. interval convergence = [−1, 1) 5. interval convergence = (−8, 8) 6. interval convergence = (−1, 1] correct 7. interval convergence = (−1, 1) Explanation: Since R = 1, the given series (i) converges when |x| < 1, and (ii) diverges when |x| > 1. On the other hand, at the point x = 1 and x = −1, the series reduces to ∞ ∑ n =1 (−1)n n + 8 , ∞ ∑ n =1 1 n + 8 respectively. But by the Alternating Series Test, the first of these series converges. On the other hand, if we set an = 1 n + 8 , bn = 1 n , then lim n→∞ an bn = lim n→∞ n n + 8 = 1 . By the p-series Test with p = 1, however, the series ∑ n bn diverges. Thus by the Limit Comparison test, the series ∑ n an also di- verges. Consequently, the given series has interval convergence = (−1, 1] . keywords: CalC12h07a 004 (part 1 of 2) 10.0 points For the series ∞ ∑ n = xn (n + 3)! , (i) determine its radius of convergence, R. 1. R = 1 3 2. R = 3 3. R = 0 4. R = ∞ correct 5. R = 1 Explanation: The given series has the form ∞ ∑ n = 1 anx n with an = 1 (n + 3)! . Now for this series, Version PREVIEW – Homework13 – Berg – (58460) 5 an = (−8)n xn 3 √ n + 5 , so lim n→∞ ∣ ∣ ∣ ∣ an+1 an ∣ ∣ ∣ ∣ = lim n→∞ 8n+1|x|n+1 3 √ n + 6 · 3 √ n + 5 8n|x|n = lim n→∞ 8|x| 3 √ n + 5 n + 6 = 8|x|, so by the Ratio Test, the series converges when 8|x| < 1. Then |x| < 1 8 , so R = 1 8 . When x = −1 8 , we get the divergent p-series ∞ ∑ n=1 1 3 √ n + 5 . When x = 1 8 , we get the series ∞ ∑ n=1 (−1)n 3 √ n + 5 , which converges by the Alternat- ing Series Test. Thus, I = ( −1 8 , 1 8 ] . CalC12h15a 009 10.0 points Find the radius of convergence, R, of the power series ∞ ∑ n =1 4 √ n(x + 6)n . 1. R = ∞ 2. R = 1 correct 3. R = 0 4. R = 1 6 5. R = 6 Explanation: The given series has the form ∞ ∑ n = 1 an(x + 6) n where an = 4 √ n . Now for this series, (i) R = 0 if it converges only at x = −6, (ii) R = ∞ if it converges for all x, while 0 < R < ∞ (iii) if it converges for |x + 6| < R, and (iv) diverges for |x + 6| > R. But ∣ ∣ ∣ an+1 an ∣ ∣ ∣ = 4 √ n + 1 4 √ n = 4 √ n + 1 n , so lim n→∞ ∣ ∣ ∣ an+1 an ∣ ∣ ∣ = 1 . By the Ratio test, therefore, the given series (a) converges for all |x + 6| < 1, and (b) diverges for all |x + 6| > 1. Consequently, R = 1 . CalC12h18a 010 10.0 points Determine the radius of convergence, R, of the power series ∞ ∑ n = 1 (−5)n√ n (x − 2)n . 1. R = 2 2. R = 0 3. R = 1 2 4. R = ∞ 5. R = 1 5 correct 6. R = 5 Version PREVIEW – Homework13 – Berg – (58460) 6 Explanation: The given series has the form ∞ ∑ n =1 (−1)nan(x − 2)n where an = 5n√ n . Now for this series, (i) R = 0 if it converges only at x = 2, (ii) R = ∞ if it converges for all x, while 0 < R < ∞ (iii) if it converges for |x − 2| < R, and (iv) diverges for |x − 2| > R. But ∣ ∣ ∣ an+1 an ∣ ∣ ∣ = 5 ( √ n√ n + 1 ) = 5 ( √ n n + 1 ) , in which case lim n→∞ ∣ ∣ ∣ an+1 an ∣ ∣ ∣ = 5 . By the Ratio test, therefore, the given series (a) converges for |x − 2| < 1/5, and (b) diverges for |x − 2| > 1/5. Consequently, R = 1 5 . CalC12h23d 011 10.0 points Determine the interval of convergence of the series of the series ∞ ∑ n =1 n! (5x − 1)n . 1. converges only at x = 1 5 correct 2. interval convergence = [ − 1 5 , 1 5 ] 3. interval convergence = (−5, 5) 4. interval convergence = (−∞, ∞) 5. converges only at x = 5 6. interval convergence = [0, 10) Explanation: The given series has the form ∑ n = 1 an(x − a)n with an = n! 5 n, a = 1 5 . But if x 6= a, lim n→∞ n!5n (x − a)n = ∞ , and so by the Divergence Test, this series ∞ ∑ n = 1 n!5n (x − a)n diverges whenever x 6= a. Consequently, the given series converges only at x = 1 5 . CalC12i01a 012 10.0 points Compare the radius of convergence, R1, of the series ∞ ∑ n = 0 cnt n with the radius of convergence, R2, of the series ∞ ∑ n = 1 n cn t n−1 when lim n→∞ |cn|1/n = 1 5 . Version PREVIEW – Homework13 – Berg – (58460) 7 1. 2R1 = R2 = 5 2. 2R1 = R2 = 1 5 3. R1 = 2R2 = 5 4. R1 = R2 = 1 5 5. R1 = 2R2 = 1 5 6. R1 = R2 = 5 correct Explanation: When lim n→∞ |cn|1/n = 1 5 , the Root Test ensures that the series ∞ ∑ n = 0 cnt n is (i) convergent when |t| < 5, and (ii) divergent when |t| > 5. On the other hand, since lim n→∞ |n cn|1/n = lim n→∞ |cn|1/n , the Root Test ensures also that the series ∞ ∑ =1 n cn t n−1 is (i) convergent when |t| < 5, and (ii) divergent when |t| > 5. Consequently, R = R2 = 5 . CalC12i03c 013 10.0 points Find a power series representation for the function f(t) = 1 t − 6 . 1. f(t) = ∞ ∑ n =0 1 6n+1 tn 2. f(t) = ∞ ∑ n =0 (−1)n−16n+1 tn 3. f(t) = − ∞ ∑ n =0 6n tn 4. f(t) = − ∞ ∑ n =0 1 6n+1 tn correct 5. f(t) = ∞ ∑ n =0 (−1)n6n tn Explanation: We know that 1 1 − x = 1 + x + x 2 + . . . = ∞ ∑ n =0 xn . On the other hand, 1 t − 6 = − 1 6 ( 1 1 − (t/6) ) . Thus f(t) = −1 6 ∞ ∑ n = 0 ( t 6 )n = −1 6 ∞ ∑ n =0 1 6n tn . Consequently, f(t) = − ∞ ∑ n =0 1 6n+1 tn with |t| < 6. CalC12i05s 014 10.0 points Version PREVIEW – Homework13 – Berg – (58460) 10 1. f(z) = ∞ ∑ n = 0 z4n 4n 2. f(z) = ∞ ∑ n = 0 z4n+2 4n + 2 correct 3. f(z) = ∞ ∑ n = 0 (−1)nz4n+2 4n + 2 4. f(z) = ∞ ∑ n = 4 z4n 4n + 2 5. f(z) = ∞ ∑ n = 0 (−1)nz4n 4n Explanation: By the geometric series representation, 1 1 − s = ∞ ∑ n = 0 sn , and so s 1 − s4 = ∞ ∑ n = 0 s4n+1 . But then f(z) = ∫ z 0 { ∞ ∑ n = 0 s4n+1 } ds = ∞ ∑ n =0 { ∫ z 0 s4n+1 ds } . Consequently, f(z) = ∞ ∑ n =0 z4n+2 4n + 2 . CalC12i26a 018 10.0 points Use the Taylor series representation cen- tered at the origin for e−x 2 to evaluate the definite integral I = ∫ 2 0 6 e−x 2 dx. 1. I = ∞ ∑ k = 0 (−1)k 2k + 1 6 · 22k+1 2. I = ∞ ∑ k = 0 (−1)k k!(2k + 1) 6 · 22k+1 correct 3. I = ∞ ∑ k = 0 1 k! 6 · 22k 4. I = n ∑ k = 0 1 k!(2k + 1) 6 · 22k+1 5. I = ∞ ∑ k = 0 (−1)k k! 6 · 22k Explanation: The Taylor series for ex is given by ex = 1 + x + 1 2! x2 + . . . + 1 n! xn + . . . and its interval of convergence is (−∞, ∞). Thus we can substitute x → −x2 for all val- ues of x, showing that e−x 2 = ∞ ∑ k =0 (−1)k k! x2k everywhere on (−∞, ∞). Thus I = ∫ 2 0 6 ( ∞ ∑ k =0 (−1)k k! x2k ) dx. But we can change the order of summation and integration on the interval of convergence, so I = 6 ∞ ∑ k =0 ( ∫ 2 0 (−1)k k! x2k ) dx = 6 ∞ ∑ k =0 [ (−1)k k!(2k + 1) x2k+1 ]2 0 . Consequently, I = ∞ ∑ k =0 (−1)k k!(2k + 1) 6 · 22k+1 . Version PREVIEW – Homework13 – Berg – (58460) 11 CalC12i38s 019 10.0 points Determine the function f having power se- ries representation f(x) = ∞ ∑ n =2 n(n − 1) xn+5 on (−1, 1). 1. f(x) = 2x7 (1 − x)3 correct 2. f(x) = 2x7 1 − x 3. f(x) = 2x5 (5 − x)3 4. f(x) = x7 (1 − x)3 5. f(x) = x5 5 − x 6. f(x) = x5 (5 − x)3 Explanation: The presence of the coefficients n(n − 1) suggests that f should be related to the sec- ond derivative of some function since d2 dx2 ( ∞ ∑ n = 0 anx n ) = ∞ ∑ n = 2 n(n − 1)anxn−2 on the interval of convergence of the power series. Now ∞ ∑ n =2 n(n − 1)xn+5 = x7 { ∞ ∑ n = 2 n(n − 1)xn−2 } . On the other hand 1 1 − x = ∞ ∑ n = 0 xn , in which case, 1 (1 − x)2 = d dx ( 1 1 − x ) = ∞ ∑ n =1 nxn−1 , and 2 (1 − x)3 = d2 dx2 ( 1 1 − x ) = ∞ ∑ n = 2 n(n − 1)xn−2 , Consequently, f(x) = 2x7 (1 − x)3 .