A e ca b e 4d 4 if 0 r0 d r1 4d 5

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Unformatted text preview: 1)k =1+ k=1 sinh−1 x = x + ∞ (−1)k k=1 1 · 3 · 5 · · · (2k − 1) 2k x, 2k k ! 1 · 3 · 5 · · · (2k − 1) 2k+1 x 2k k !(2k + 1) (c) R = 1 36. (a) sin−1 x = (1 − x2 )−1/2 dx − C = 3 57 1 x + · · · − C, sin−1 0 = 0 so C = 0 x + x3 + x5 + 6 40 112 = (b) 1 − x2 1 3 5 1 + x2 + x4 + x6 + · · · dx − C 2 8 16 −1/2 ∞ =1+ k=1 ∞ =1+ k=1 ∞ =1+ sin−1 x = x + k=1 ∞ k=1 (−1/2)(−3/2)(−5/2) · · · (−1/2 − k + 1) −x2 k! k (−1)k (1/2)k (1)(3)(5) · · · (2k − 1) (−1)k x2k k! 1 · 3 · 5 · · · (2k − 1) 2k x 2k k ! 1 · 3 · 5 · · · (2k − 1) 2k+1 x 2k k !(2k + 1) (c) R = 1 ∞ 37. (a) y (t) = y0 k=0 (−1)k (0.000121)k tk k! (b) y (1) ≈ y0 (1 − 0.000121t) t=1 = 0.999879y0 (c) y0 e−0.000121 ≈ 0.9998790073y0 38. (a) If ct ct ≈ 0 then e−ct/m ≈ 1 − , and v (t) ≈ m m 1− ct m v0 + mg mg cv0 − = v0 − + g t. c c m (b) The quadratic approximation is v0 ≈ 1− (ct)2 ct + m 2m2 v0 + mg c2 mg mg 2 cv0 − = v0 − +g t+ t. v0 + c c m 2m2 c 403 Chapter 11 39. θ0 = 5◦ = π/36 rad, k = sin(π/72) L = 2π g (a) T ≈ 2π 1/9.8 ≈ 2.00709 L g (b) T ≈ 2π 1+ k2 4 ≈ 2.008044621 (c) 2.008045644 40. The third order model gives the same result as the second, because there is no term of degree three π /2 1·3π in (5). By the Wallis sine formula, , and sin4 φ dφ = 2·4 2 0 T ≈4 π /2 L g 0 1·3 1 1 + k 2 sin2 φ + 2 k 4 sin4 φ 2 2 2! π k 2 π 3k 4 3π + + 2 24 8 16 L g dφ = 4 L g = 2π 1+ k2 9k 4 + 4 64 mg mgR2 = = mg 1 − 2h/R + 3h2 /R2 − 4h3 /R3 + · · · (R + h)2 (1 + h/R)2 41. (a) F = (b) If h = 0, then the binomial series converges to 1 and F = mg . (c) Sum the series to the linear term, F ≈ mg − 2mgh/R. (d) 2h 2 · 29,028 mg − 2mgh/R =1− =1− ≈ 0.9973, so about 0.27% less. mg R 4000 · 5280 42. (a) We can diﬀerentiate term-by-term: ∞ y= k=1 (−1)k x2k−1 = 22k−1 k !(k − 1)! ∞ xy + y + xy = k=0 ∞ xy + y + xy = k=0 ∞ (b) y = k=0 ∞ k=0 (−1)k+1 x2k+1 ,y = 22k+1 (k + 1)!k ! (−1)k+1 (2k + 1)x2k+1 + 22k+1 (k + 1)!k ! ∞ k=0 k=0 k=0 (−1)k+1 (2k + 1)x2k , and 22k+1 (k + 1)!k ! (−1)k+1 x2k+1 + 22k+1 (k + 1)!k ! ∞ k=0 (−1)k x2k+1 , 22k (k !)2 1 (−1)k+1 x2k+1 2k + 1 + −1 =0 22k (k !)2 2(k + 1) 2(k + 1) (−1)k (2k + 1)x2k ,y = 22k+1 k !(k + 1)! Since J1 (x) = ∞ ∞ ∞ k=1 (−1)k (2k + 1)x2k−1 . 22k (k − 1)!(k + 1)! (−1)k x2k+1 and x2 J1 (x) = 2k+1 k !(k + 1)! 2 ∞ k=1 (−1)k−1 x2k+1 , it follows that 22k−1 (k − 1)!k ! x2 y + xy + (x2 − 1)y ∞ = k=1 (−1)k (2k + 1)x2k+1 + 22k (k − 1)!(k + 1)! ∞ k=0 ∞ − k=0 = xx −+ 2 2 ∞ k=1 (−1)k x2k+1 − 1)!k ! 22k−1 (k (−1)k (2k + 1)x2k+1 + 22k+1 (k !)(k + 1)! ∞ k=1 (−1)k−1 x2k+1 22k−1 (k − 1)!k ! (−1)k x2k+1 22k+1 k !(k + 1)! 2k + 1 1 2k + 1 + −1− 2(k + 1) 4k (k + 1) 4k (k + 1) = 0. Chapter 11 Supplementary Exercises 404 ∞ (c) From part (a), J0 (x) = k=0 ∞ ∞ k=0 ∞ bk xk for x in (−r, r) then ak xk = 43. If (−1)k+1 x2k+1 = −J1 (x). 22k+1 (k + 1)!k ! k=0 ∞ (ak −bk )xk = 0 so by Theorem 11.10.6 k=0 (ak −bk )xk k=0 is the Taylor series for f (x) = 0 about 0 and hence ak − bk = 0, ak = bk for all k . CHAPTER 11 SUPPLEMENTARY EXERCISES ∞ 4. (a) k=0 9. (a) (b) (c) (d) (e) (f ) (g) (h) (i) (j) (k) (l) ∞ f (k) (0) k x k! (b) k=0 f (k) (x0 ) (x − x0 )k k! always true by Theorem 11.4.2 sometimes false, for example the harmonic series diverges but sometimes false, for example f (x) = sin πx, ak = 0, L = 0 always true by Example 3(d) of Section 11.1 1 1 sometimes false, for example an = + (−1)n 2 4 sometimes false, for example uk = 1/2 always false by Theorem 11.4.3 sometimes false, for example uk = vk = (2/k ) always true by the Comparison Test always true by the Comparison Test sometimes false, for example (−1)k /k sometimes false, for example (−1)k /k (1/k2 ) converges 10. (a) false, f (x) is not diﬀerentiable at x = 0, Deﬁnition 11.5.4 (b) true: sn = 1 if n is odd and s2n = 1 + 1/(n + 1); lim sn = 1 n→+∞ (c) false, lim ak = 0 (b) 1/(5k + 1) < 1/5k , converges 11. (a) geometric, r = 1/5, converges (c) 9 9 9 √ √ = √, ≥√ k+1 k+ k 2k ∞ k=1 9 √ 2k diverges 12. (a) converges by Alternating Series Test ∞ (b) absolutely convergent, k=1 (c) 13. (a) k −1 1 k −1/2 > = , 2 2+1 3k 2 + sin k 1 1 < 3, k 3 + 2k + 1 k k+2 3k − 1 ∞ k=1 k converges by the Root Test. 1 diverges 3k ∞ ∞ 1/k 3 converges, so k=1 k=1 1 converges by the Comparison Test k 3 + 2k + 1 ∞ (b) Limit Comparison Test, compare with the divergent series k=1 1 k 2/5 , diverges 405 Chapter 11 cos(1/k ) 1 < 2, k2 k (c) ∞ ∞ k=1 1 converges, so k2 2 4 2 ln x ln x dx = lim − 1/2 − 1/2 3/2 →+∞ x x x = ∞ (c) absolutely convergent, k=1 1 − 5k 99 k=0 1 = 5k 16. no, lim ak = k→+∞ ∞ k=100 k2 1 1 = 100 5k 5 √ 2(ln 2 + 2) so 2 ln k converges k 3/2 k=2 1 diverges 14k 2/3 k=1 1 converges (compare with +1 ∞ k=0 ∞ ∞ k 4/3 k 4/3 1 ≥2 = , 2 + 5k + 1 8k 8k + 5k 2 + k 2 14k 2/3 (b) k=0 k=1 ∞...
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This document was uploaded on 02/23/2014 for the course MANAGMENT 2201 at University of Michigan.

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