4 426 16 a c2 a2 b2 9 25 34 c 34 y 0

Info iconThis preview shows page 1. Sign up to view the full content.

View Full Document Right Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: nverges. Radius of convergence is 1, interval of k2 + 4 convergence is [−2, 0]. 28. ρ = lim k→+∞ ∞ x = 2, k=1 k ln(k + 1) |x − 3| = |x − 3|, converges if |x − 3| < 1, diverges if |x − 3| > 1. If (k + 1) ln k (−1)k ln k converges; if x = 4, k ∞ k=1 ln k diverges. Radius of convergence is 1, interval of k convergence is [2, 4). π |x − 1|2 = 0, radius of convergence +∞, interval of convergence (−∞, +∞). k→+∞ (2k + 3)(2k + 2) 29. ρ = lim 30. ρ = lim k→+∞ 1 1 1 |2x − 3| = |2x − 3|, converges if |2x − 3| < 1 or |x − 3/2| < 8, diverges if 16 16 16 ∞ ∞ (−1)k diverges; if x = 19/2, |x − 3/2| > 8. If x = −13/2, k=0 1 diverges. Radius of convergence k=0 is 8, interval of convergence is (−13/2, 19/2). 31. ρ = lim k k→+∞ |uk | = |x| = 0, the series converges absolutely for all x so the interval of k→+∞ ln k lim convergence is (−∞, +∞). 32. ρ = lim k→+∞ 2k + 1 |x| = 0 (2k )(2k − 1) 10 33. (a) so R = +∞. -1 1 -1 Exercise Set 11.8 392 |x|2 = 0, R = +∞ k→+∞ 4(k + 1)(k + 2) 34. Ratio Test: ρ = lim 35. By the Ratio Test for absolute convergence, (pk + p)!(k !)p (pk + p)(pk + p − 1)(pk + p − 2) · · · (pk + p − [p − 1]) |x| = lim |x| k→+∞ (pk )![(k + 1)!]p k→+∞ (k + 1)p ρ = lim = lim p p − k→+∞ 1 k+1 p− 2 k+1 ··· p − p−1 k+1 |x| = pp |x|, converges if |x| < 1/pp , diverges if |x| > 1/pp . Radius of convergence is 1/pp . 36. By the Ratio Test for absolute convergence, ρ = lim k→+∞ (k + 1 + p)!k !(k + q )! k+1+p |x| = lim |x| = 0, k→+∞ (k + 1)(k + 1 + q ) (k + p)!(k + 1)!(k + 1 + q )! radius of convergence is +∞. 37. (a) By Theorem 11.4.3(b) both series converge or diverge together, so they have the same radius of convergence. (b) By Theorem 11.4.3(a) the series (ck + dk )(x − x0 )k converges if |x − x0 | < R; if |x − x0 | > R ck (x − x0 )k would converge by then (ck + dk )(x − x0 )k cannot converge, as otherwise the same Theorem. Hence the radius of convergence of (ck + dk )(x − x0 )k is R. (c) Let r be the radius of convergence of (ck + dk )(x − x0 )k . If |x − x0 | < min(R1 , R2 ) k k then ck (x − x0 ) and dk (x − x0 ) converge, so (ck + dk )(x − x0 )k converges. Hence r ≥ min(R1 , R2 ) (to see that r > min(R1 , R2 ) is possible consider the case ck = −dk = 1). If in addition R1 = R2 , and R1 < |x − x0 | < R2 (or R2 < |x − x0 | < R1 ) then (ck + dk )(x − x0 )k cannot converge, as otherwise all three series would converge. Thus in this case r = min(R1 , R2 ). 38. By the Root Test for absolute convergence, ρ = lim |ck |1/k |x| = L|x|, L|x| < 1 if |x| < 1/L so the radius of convergence is 1/L. k→+∞ ∞ ck x converges if |x| < R so 39. By assumption |x| < √ ∞ k k=0 ∞ R. Moreover, ck x has radius of convergence √ k=0 2 ck x2k k=0 ∞ ∞ ck (−R)k is divergent. Suppose that ck Rk is convergent and k=0 ∞ ck Rk k=0 ck (−R)k is also absolutely convergent and hence convergent is absolutely convergent then k=0 ∞ because |ck Rk | = |ck (−R)k |, which contradicts the assumption that ck (−R)k is divergent so k=0 ∞ ck R must be conditionally convergent. k=0 ∞ R. k=0 k √ R. Thus k=0 ∞ 40. The assumption is that ck (x2 )k converges if |x2 | < R, = ck (x ) diverges if |x | > R, |x| > 2k = k=0 ck x k=0 ∞ 2k ∞ 2k 393 Chapter 11 EXERCISE SET 11.9 1. sin 4◦ = sin π (π/45)3 (π/45)5 π = − + − ··· 45 45 3! 5! (a) Method 1: |Rn (π/45)| ≤ sin 4◦ ≈ (π/45)n+1 < 0.000005 for n + 1 = 4, n = 3; (n + 1)! (π/45)3 π − ≈ 0.069756 45 3! (b) Method 2: The first term in the alternating series that is less than 0.000005 is the result is the same as in part (a). 2. cos 3◦ = cos (π/45)5 , so 5! (π/60)2 (π/60)4 π =1− + − ··· 60 2 4! (π/60)n+1 (π/60)2 < 0.0005 for n = 2; cos 3◦ ≈ 1 − ≈ 0.9986. (n + 1)! 2 (π/60)4 (b) Method 2: The first term in the alternating series that is less than 0.0005 is , so the 4! result is the same as in part (a). (a) Method 1: |Rn (π/60)| ≤ 3. f (k) (x) = ex , |f (k) (x)| ≤ e1/2 < 2 on [0, 1/2], let M = 2, e1/2 = 1 + |Rn (1/2)| ≤ e1/2 ≈ 1 + 1 1 11 ++ + + ···; 2 8 48 24 · 16 M 2 (1/2)n+1 ≤ (1/2)n+1 ≤ 0.00005 for n = 5; (n + 1)! (n + 1)! 1 1 11 ++ + ≈ 1.64844, calculator value 1.64872 2 8 48 24 · 16 4. f (x) = ex , f (k) (x) = ex , |f (k) (x)| ≤ 1 on [−1, 0], |Rn (x)| ≤ if n = 6, so e−1 ≈ 1 − 1 + 5. |Rn (0.1)| ≤ 1 1 (1)n+1 = < 0.5 × 10−3 (n + 1)! (n + 1)! 1 1 1 1 1 −+−+ ≈ 0.3681, calculator value 0.3679 2! 3! 4! 5! 6! (0.1)n+1 ≤ 0.000005 for n = 3; cos 0.1 ≈ 1 − (0.1)2 /2 = 0.99500, calculator value (n + 1)! 0.995004 . . . 6. (0.1)3 /3 < 0.5 × 10−3 so tan−1 (0.1) ≈ 0.100, calculator value ≈ 0.0997 7. Expand about π/2 to get sin x = 1 − |Rn (x)| ≤ 1 1 (x − π/2)2 + (x − π/2)4 − · · ·, 85◦ = 17π/36 radians, 2! 4! (π/36)n+1 |x − π/2|n+1 |17π/36 − π/2|n+1 , |Rn (17π/36)| ≤ = < 0.5 × 10−4 (n + 1)! (n + 1)! (n + 1)! 1 if n = 3, sin 85◦ ≈ 1 − (...
View Full Document

This document was uploaded on 02/23/2014 for the course MANAGMENT 2201 at University of Michigan.

Ask a homework question - tutors are online