# A r sin 3 b r c r2 4r cos 0 r 4 cos d

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Unformatted text preview: ternately −1 1 and 1 if k is even; p0 (x) = −1, p2 (x) = −1 + (x − π )2 , 2 1 1 p4 (x) = −1 + (x − π )2 − (x − π )4 , 2 24 1 1 1 (x − π )6 p6 (x) = −1 + (x − π )2 − (x − π )4 + 2 24 720 32. f (0) = 0; for k ≥ 1, f (k) (x) = i 1.25 o 0 -1.25 (−1)k−1 (k − 1)! , (x + 1)k 1.5 f (k) (0) = (−1)k−1 (k − 1)!; p0 (x) = 0, p1 (x) = x, 1 1 1 p2 (x) = x − x2 , p3 (x) = x − x2 + x3 2 2 3 -1 1 -1.5 33. p(0) = 1, p(x) has slope −1 at x = 0, and p(x) is concave up at x = 0, eliminating I, II and III respectively and leaving IV. 34. Let p0 (x) = 2, p1 (x) = p2 (x) = 1 − 3x, p3 (x) = 1 − 3x + x3 , and, for any arbitrary integer n ≥ 4 n and constants c4 , c5 , . . . , cn , let pn (x) = 2 − 3(x − 1) + (x − 1) + ck (x − 1)k ; then any of the 3 k=4 polynomials p0 , p1 , . . . , pn is a possible Taylor polynomial for f about x = 1. 35. f (k) (ln 4) = 15/8 for k even, f (k) (ln 4) = 17/8 for k odd, which can be written as f (k) (ln 4) = 16 − (−1)k ; 8 ∞ k=0 16 − (−1)k (x − ln 4)k 8k ! Exercise Set 11.6 380 36. (a) cos α ≈ 1 − α2 /2; x = r − r cos α = r(1 − cos α) ≈ rα2 /2 (b) In Figure Ex-36 let r = 4000 mi and α = 1/80 so that the arc has length 2rα = 100 mi. 4000 = 5/16 mi. Then x ≈ rα2 /2 = 2 · 802 37. From Exercise 2(a), p1 (x) = 1 + x, p2 (x) = 1 + x + x2 /2 3 (a) -1 1 -1 (b) x –1.000 –0.750 –0.500 –0.250 0.000 0.250 0.500 f (x) 0.431 0.506 0.619 0.781 1.000 1.281 1.615 p1(x) 0.000 0.250 0.500 0.750 1.000 1.250 1.500 p2(x) 0.500 0.531 0.625 0.781 1.000 1.281 1.625 (c) |esin x − (1 + x)| < 0.01 for − 0.14 < x < 0.14 0.750 1.977 1.750 2.031 (d) 1.000 2.320 2.000 2.500 |esin x − (1 + x + x2 /2)| < 0.01 for − 0.50 < x < 0.50 0.01 0.015 -0.15 0.15 -0.6 0 0.6 0 EXERCISE SET 11.6 1. (a) (b) 2. (a) 3. (a) 4. (a) 1 1 1 ≤2 = 2, 5k 2 − k 5k − k 2 4k 3 3 >, k − 1/4 k ∞ k=1 1 converges 4k 2 3/k diverges k=1 k k+1 1 > 2= , 2−k k k k 1 1 < k, k +5 3 3 ∞ ∞ k=1 ∞ 1/k diverges (b) k=2 1 converges 3k ln k 1 > for k ≥ 3, k k ∞ k=1 1 diverges k k k 1 (b) > 3/2 = √ ; 3/2 − 1/2 k k k ∞ 1 √ diverges k k=1 (b) 2 2 < 4, 4+k k k 5 5 sin2 k <, k! k! ∞ k=1 ∞ k=1 2 converges k4 5 converges k! 381 Chapter 11 ∞ 4k 7 − 2k 6 + 6k 5 = 1/2, converges k→+∞ 8k 7 + k − 8 1/k 5 , ρ = lim 5. compare with the convergent series k=1 ∞ 6. compare with the divergent series k = 1/9, diverges k→+∞ 9k + 6 1/k , ρ = lim k=1 ∞ 3k = 1, converges k→+∞ 3k + 1 5/3k , ρ = lim 7. compare with the convergent series k=1 ∞ 8. compare with the divergent series k 2 (k + 3) = 1, diverges k→+∞ (k + 1)(k + 2)(k + 5) 1/k , ρ = lim k=1 ∞ 9. compare with the divergent series k=1 ρ = lim k→+∞ (8k 2 1 k 2/3 , k 2/3 1 = lim = 1/2, diverges k→+∞ (8 − 3/k )1/3 − 3k )1/3 ∞ 1/k 17 , 10. compare with the convergent series k=1 17 ρ = lim k→+∞ k 1 = lim = 1/217 , converges k→+∞ (2 + 3/k )17 (2k + 3)17 3k+1 /(k + 1)! 3 = lim = 0, the series converges k /k ! k→+∞ k→+∞ k + 1 3 11. ρ = lim 4k+1 /(k + 1)2 4k 2 = lim = 4, the series diverges k→+∞ k→+∞ (k + 1)2 4k /k 2 12. ρ = lim 13. ρ = lim k→+∞ k = 1, the result is inconclusive k+1 (k + 1)(1/2)k+1 k+1 = 1/2, the series converges = lim k k→+∞ k→+∞ 2k k (1/2) 14. ρ = lim (k + 1)!/(k + 1)3 k3 = lim = +∞, the series diverges k→+∞ k→+∞ (k + 1)2 k !/k 3 15. ρ = lim (k + 1)/[(k + 1)2 + 1] (k + 1)(k 2 + 1) = lim = 1, the result is inconclusive. 2 + 1) k→+∞ k→+∞ k (k 2 + 2k + 2) k/(k 16. ρ = lim 17. ρ = lim k→+∞ 3k + 2 = 3/2, the series diverges 2k − 1 18. ρ = lim k/100 = +∞, the series diverges k→+∞ k 1/k = 1/5, the series converges k→+∞ 5 19. ρ = lim 20. ρ = lim (1 − e−k ) = 1, the result is inconclusive k→+∞ Exercise Set 11.6 382 21. Ratio Test, ρ = lim 7/(k + 1) = 0, converges k→+∞ ∞ 22. Limit Comparison Test, compare with the divergent series 1/k k=1 (k + 1)2 = 1/5, converges k→+∞ 5k 2 23. Ratio Test, ρ = lim 24. Ratio Test, ρ = lim (10/3)(k + 1) = +∞, diverges k→+∞ 25. Ratio Test, ρ = lim e−1 (k + 1)50 /k 50 = e−1 < 1, converges k→+∞ ∞ 1/k 26. Limit Comparison Test, compare with the divergent series k=1 ∞ 27. Limit Comparison Test, compare with the convergent series converges 28. 4 4 <k, 2 + 3k k 3k ∞ k=1 4 converges (Ratio Test) so 3k k ∞ k=1 1/k 5/2 , ρ = k=1 lim 4 converges by the Comparison Test 2 + k 3k ∞ 29. Limit Comparison Test, compare with the divergent series diverges 30. 3 2 + (−1)k ≤ k, 5k 5 ∞ ∞ k 3/5 converges so k=1 k=1 1/k , ρ = k=1 2 + (−1)k converges 5k ∞ 1/k 5/2 , 31. Limit Comparison Test, compare with the convergent series k=1 k 3 + 2k 5/2 ρ = lim 3 = 1, converges k→+∞ k + 3k 2 + 3k 32. 5 4 + | cos k | < 3, 3 k k ∞ ∞ 5/k 3 converges so k=1 k=1 ∞ 33. Limit Comparison Test, compare with 4 + | cos k | converges k3 √ 1/ k k=1 34. Ratio Test, ρ = lim (1 + 1/k )−k = 1/e < 1, converges k→+∞ ln(k + 1) k = li...
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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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