1 2 1 arc length is innite b the integral 1 2 d

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Unformatted text preview: +∞ 15. cos(1/k ) converges absolutely k2 ln k ln k √= √ because ln 1 = 0, k k k=2 k k k=1 14. (a) ∞ ∞ 1/k 2 ) 1 1 = 5k 4 · 599 1 = 0 (Divergence Test) 2 17. (a) p0 (x) = 1, p1 (x) = 1 − 7x, p2 (x) = 1 − 7x + 5x2 , p3 (x) = 1 − 7x + 5x2 − 4x3 , p4 (x) = 1 − 7x + 5x2 − 4x3 (b) If f (x) is a polynomial of degree n and k ≥ n then the Maclaurin polynomial of degree k is the polynomial itself; if k < n then it is the truncated polynomial. 18. ln(1 + x) = x − x2 /2 + · · · ; so |ln(1 + x) − x| ≤ x2 /2 by Theorem 11.7.2. 19. sin x = x − x3 /3! + x5 /5! − x7 /7! + · · · is an alternating series, so | sin x − x + x3 /3! − x5 /5!| ≤ x7 /7! ≤ π 7 /(47 7!) ≤ 0.00005 1 20. 0 x4 x6 1 − cos x x2 dx = − + − ··· x 2 · 2! 4 · 4! 6 · 6! 1 so 0 1 = 0 1 1 1 1 − + − · · ·, and < 0.0005, 2 · 2! 4 · 4! 6 · 6! 6 · 6! 1 1 1 − cos x dx = − = 0.2396 to three decimal-place accuracy. x 2 · 2! 4 · 4! 21. (a) ρ = lim k→+∞ 2k k! 1/k (b) ρ = lim uk k→+∞ 1/k = lim k→+∞ = lim k→+∞ 2 √ = 0, converges k k! k √ = e, diverges k k! 22. (a) 11 ≥ 1!; if k k ≥ k !, then (k +1)k+1 ≥ (k +1)k k ≥ (k +1)k ! = (k +1)!; mathematical induction (b) 1 ≤ kk 1 , converges k! Chapter 11 Supplementary Exercises (c) 1/k 1 kk lim k→+∞ = lim k→+∞ 100 k=1 2− uk = k=1 (b) u1 = 1; for k ≥ 2, uk = ∞ (c) 1 = 0, converges k 99 uk − 23. (a) u100 = 406 2− 1 100 1 k − 2− − 2− n uk = lim n→+∞ k=1 ∞ 2 3 −k 2k 3 24. (a) k=1 n 2− uk = lim n→+∞ k=1 ∞ = k=1 3 − 2k ∞ k=1 2 = 3k 1 , lim uk = 0 k (k − 1) k→+∞ 1 2 − 1 − (1/2) 3 3 2 1 = 2 (geometric series) 1 − (1/3) n n→+∞ k=1 n n→+∞ 1 2 1 1 − k k+2 = lim n→+∞ 1 2 1+ 1 1 1 − − 2 n+1 n+2 tan−1 (k + 1) − tan−1 k = lim lim n→+∞ [ln(k + 1) − ln k ] = lim ln(n + 1) = +∞, diverges k=1 lim k=1 (b) sin π = 0 π ππ −= 2 4 4 (d) e− ln 3 = 1/3 (c) cos e k+1 √ 1/2 1/4 1/2k ak = ak = ak−1 = · · · = a1 = c1/2 (a) If c = 1/2 then lim ak = 1. (b) if c = 3/2 then lim ak = 1. k→+∞ k→+∞ 27. Expand in a Maclaurin Series about x = 100 : e−[(x−100)/16] = 1 − 2 p≈ 3 4 = tan−1 (n + 1) − tan−1 (1) = n→+∞ 25. (a) e2 − 1 26. ak+1 = = 1 9900 ∞ k=1 (d) 1 k−1 = =2 [ln(k + 1) − ln k ] = ln(n + 1), so (b) (c) 1 n 1 99 1 √ 16 2π 110 1− 100 (x − 100)4 (x − 100)2 + 2 16 2 · 164 dx = 1 √ (x − 100)2 (x − 100)4 + , so 162 2 · 164 10 − 16 2π 105 103 + 2 3 · 16 2 · 5 · 164 ≈ 0.220678, or about 22.07%. 28. f (x) = xex = x + x2 + f (x) = (x + 1)ex = 1 + 2x + 29. Let A = 1 − ∞ x4 x3 + + ··· = 2! 3! 2 k=0 3 xk+1 , k! 4x 3x + + ··· = 2! 3! ∞ k=0 k+1 k x; k! ∞ k=0 k+1 = f (1) = 2e. k! 1 1 1 + 2 − 2 + · · · ; since the series all converge absolutely, 2 2 3 4 1 1 1 1 π2 − A = 2 2 + 2 2 + 2 2 + ··· = 6 2 4 6 2 1+ 1 1 + 2 + ··· 22 3 30. Compare with 1/k p : converges if p > 1, diverges otherwise. = 1 π2 π2 1 π2 , so A = = . 26 26 12 407 Chapter 11 1 3 3 k+1 |x| = 31. (a) x + x2 + x3 + x4 + · · ·; ρ = lim k→+∞ 3k + 1 2 14 35 1 converges if |x| < 1, |x| < 3 so R = 3. 3 25 27 8 k+1 2 3 |x| (b) −x + x − x + x9 − · · ·; ρ = lim k→+∞ 2k + 1 3 5 35 √ √ 1 converges if |x|2 < 1, |x|2 < 2, |x| < 2 so R = 2. 2 32. By the Ratio Test for absolute convergence, ρ = 1 |x|, 3 = 12 |x| , 2 |x − x0 | |x − x0 | = ; converges if b b lim k→+∞ ∞ |x − x0 | < b, diverges if |x − x0 | > b. If x = x0 − b, (−1)k diverges; if x = x0 + b, k=0 ∞ 1 diverges. The interval of convergence is (x0 − b, x0 + b). k=0 √ √ √ x x2 x3 ( x)2 ( x)4 ( x)6 + − +··· = 1− + − + · · ·; if x ≤ 0, then 2! 4! 6! 2! 4! 6! √ √ √ √ ( −x)4 ( −x)6 x x2 x3 ( −x)2 cosh( −x) = 1 + + + + ··· = 1 − + − + · · ·. 2! 4! 6! 2! 4! 6! 33. If x ≥ 0, then cos √ x = 1− 34. By Exercise 70 of Section 3.5, the derivative of an odd (even) function is even (odd); hence all odd-numbered derivatives of an odd function are even, all even-numbered derivatives of an odd function are odd; a similar statement holds for an even function. (a) If f (x) is an even function, then 2f (0) = lim+ h→0 f (0 + h) − f (0) f (0) − f (0 − h) f (0 + h) − f (0 − h) + lim = lim = 0, so + + h h h h→0 h→0 f (0) = 0. If f (x) is even then so is f (2k) , thus f (2k+1) (0) = 0, k = 0, 1, 2, . . . . Hence c2k+1 = f (2k+1) (0)/(2k + 1)! = 0. (b) If f (x) is an odd function, then f (2k−1) is even (k = 1, 2, . . .), and thus by Part (a), f (2k) (0) = 0, c2k = f (2k) (0)/(2k )! = 0. 35. 1− v2 c2 −1/2 +∞ 36. (a) n ≈1+ v2 , so K = m0 c2 2c2 1 1 − v 2 /c2 − 1 ≈ m0 c2 (v 2 /(2c2 ) = m0 v 2 /2 1 dx < (0.5)10−5 if n > 63.7; let n = 64. x3.7 (b) sn ≈ 1.10628342; CAS: 1.10628824 CHAPTER 12 Analytic Geometry in Calculus EXERCISE SET 12.1 1. (5, 8) (1, 6 ) (–3, i) (–5, @) (3, 3) (–1, r) (4, e) p/ 2 2. p/ 2 ( 3 , L) 2 (0, c) 0 (...
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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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