# 1 3 5 2k 1 2k1 x 2k k 2k 1 c r 1 36 a

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Unformatted text preview: ges. k=1 k=1 ∞ ∞ (b) diverges because 1/k 3/2 converges. 1/(3k + 2) diverges and k=1 k=1 ∞ (c) converges because both k=2 ∞ 1 (Exercise 25) and k (ln k )2 uk , then S − sn = uk and sn = k=1 1/k 2 converge. k=2 ∞ n 30. (a) If S = ∞ uk . Interpret uk , k = n + 1, n + 2, . . ., as k=1 k=n+1 the areas of inscribed or circumscribed rectangles with height uk and base of length one for the curve y = f (x) to obtain the result. n (b) Add sn = uk to each term in the conclusion of part (a) to get the desired result: k=1 +∞ +∞ sn + f (x) dx < n+1 +∞ uk < sn + f (x) dx n k=1 31. (a) In Exercise 30 above let f (x) = 1 . Then x2 +∞ f (x) dx = − n 1 x +∞ = n 1 ; n use this result and the same result with n + 1 replacing n to obtain the desired result. (b) s3 = 1 + 1/4 + 1/9 = 49/36; 58/36 = s3 + (d) 1/11 < 12 π − s10 < 1/10 6 32. Apply Exercise 30 with f (x) = (a) 1 < 46 ∞ k=1 ∞ k=1 (c) 12e−11 < ∞ k=1 +∞ n (b) 1 , (2x + 1)2 +∞ f (x) dx = n 1 : 2(2n + 1) 1 1 − s10 < (2k + 1)2 42 (b) π/2 − tan−1 (11) < 33. (a) 1 1 1 < π 2 < s3 + = 61/36 4 6 3 1 − s10 < π/2 − tan−1 (10) k2 + 1 k < 11e−10 ek 1 1 dx = 2 ; use Exercise 30(b) 3 x 2n 1 1 − < 0.01 for n = 5. 2 2n 2(n + 1)2 (c) From Part (a) with n = 5 obtain 1.1995 < S < 1.2057, so S ≈ 1.203. 375 Chapter 11 +∞ 34. (a) n 1 1 1 1 dx = 3 ; choose n so that − < 0.005, n = 4; S ≈ 1.084 x4 3n 3n3 3(n + 1)3 n 1 1 , then dx = ln n and x x 1 ln(n + 1) < sn < 1 + ln n. n+1 35. (a) Let F (x) = 1 1 dx = ln(n + 1), u1 = 1 so x (b) ln(1, 000, 001) < s1,000,000 < 1 + ln(1, 000, 000), 13 < s1,000,000 < 15 (c) s109 < 1 + ln 109 = 1 + 9 ln 10 < 22 (d) sn > ln(n + 1) ≥ 100, n ≥ e100 − 1 ≈ 2.688 × 1043 ; n = 2.69 × 1043 36. p-series with p = ln a; convergence for p > 1, a > e 37. x2 e−x is decreasing and positive for x > 2 so the Integral Test applies: ∞ x2 e−x dx = (x2 + 2x + 2)e−x 1 ∞ = 5e−1 so the series converges. 1 38. (a) f (x) = 1/(x3 + 1) is decreasing and continuous on the interval [1, +∞], so the Integral Test applies. (c) n 10 20 30 40 50 sn 0.681980 0.685314 0.685966 0.686199 0.686307 n 60 70 80 90 100 sn 0.686367 0.686403 0.686426 0.686442 0.686454 +∞ (e) Set g (n) = n √ √ 1 n3 + 1 1 3 3 dx = π+ tan−1 − x3 + 1 6 6 (n + 1)3 3 g (n) − g (n + 1) ≤ 0.001; s10 decimal places. 2n − 1 √ ; for n ≥ 10, 3 + (g (10) + g (11))/2 ≈ 0.6865, so the sum ≈ 0.6865 to three EXERCISE SET 11.5 1. (a) f (k) (x) = (−1)k e−x , f (k) (0) = (−1)k ; e−x ≈ 1 − x + x2 /2 (quadratic), e−x ≈ 1 − x (linear) (b) f (x) = − sin x, f (x) = − cos x, f (0) = 1, f (0) = 0, f (0) = −1, cos x ≈ 1 − x2 /2 (quadratic), cos x ≈ 1 (linear) (c) f (x) = cos x, f (x) = − sin x, f (π/2) = 1, f (π/2) = 0, f (π/2) = −1, sin x ≈ 1 − (x − π/2)2 /2 (quadratic), sin x ≈ 1 (linear) (d) f (1) = 1, f (1) = 1/2, f (1) = −1/4; √ √ 1 1 1 x = 1 + (x − 1) − (x − 1)2 (quadratic), x ≈ 1 + (x − 1) (linear) 2 8 2 2. (a) p2 (x) = 1 + x + x2 /2, p1 (x) = 1 + x 1 1 1 (x − 9)2 , p1 (x) = 3 + (x − 9) (b) p2 (x) = 3 + (x − 9) − 6 216 6 √ √ √ 7 π π 3 3 (x − 2) − (x − 2) 3(x − 2)2 , p1 (x) = + (c) p2 (x) = + 3 6 72 3 6 (d) p2 (x) = x, p1 (x) = x Exercise Set 11.5 376 1 −1/2 1 1 1 x , f (x) = − x−3/2 ; f (1) = 1, f (1) = , f (1) = − ; 2 4 2 4 √ 1 1 x ≈ 1 + (x − 1) − (x − 1)2 2 8 √ 1 1 (b) x = 1.1, x0 = 1, 1.1 ≈ 1 + (0.1) − (0.1)2 = 1.04875, calculator value ≈ 1.0488088 2 8 3. (a) f (x) = 4. (a) cos x ≈ 1 − x2 /2 (b) 2◦ = π/90 rad, cos 2◦ = cos(π/90) ≈ 1 − π2 ≈ 0.99939077, calculator value ≈ 0.99939083 2 · 902 5. f (x) = tan x, 61◦ = π/3 + π/180 rad; x0 = π/3, f (x) = sec2 x, f (x) = 2 sec2 x tan x; √ √ √ √ f (π/3) = 3, f (π/3) = 4, f (x) = 8 3; tan x ≈ 3 + 4(x − π/3) + 4 3(x − π/3)2 , √ √ tan 61◦ = tan(π/3 + π/180) ≈ 3 + 4π/180 + 4 3(π/180)2 ≈ 1.80397443, calculator value ≈ 1.80404776 √ 1 −1/2 1 x , f (x) = − x−3/2 ; 2 4 1 1 1√ 1 , f (36) = − ; x ≈ 6 + (x − 36) − (x − 36)2 ; f (36) = 6, f (36) = 12 864 12 1728 √ 0.03 (0.03)2 − ≈ 6.00249947917, calculator value ≈ 6.00249947938 36.03 ≈ 6 + 12 1728 6. f (x) = x, x0 = 36, f (x) = 1 7. f (k) (x) = (−1)k e−x , f (k) (0) = (−1)k ; p0 (x) = 1, p1 (x) = 1 − x, p2 (x) = 1 − x + x2 , 2 ∞ 1 1 1 1 1 (−1)k k p3 (x) = 1 − x + x2 − x3 , p4 (x) = 1 − x + x2 − x3 + x4 ; x 2 3! 2 3! 4! k! k=0 8. f (k) (x) = ak eax , f (k) (0) = ak ; p0 (x) = 1, p1 (x) = 1 + ax, p2 (x) = 1 + ax + p3 (x) = 1 + ax + a2 2 a3 3 a2 a3 a4 x + x , p4 (x) = 1 + ax + x2 + x3 + x4 ; 2 3! 2 3! 4! ∞ k=0 a2 2 x, 2 ak k x k! 9. f (k) (0) = 0 if k is odd, f (k) (0) is alternately π k and −π k if k is even; p0 (x) = 1, p1 (x) = 1, p2 (x) = 1 − π2 2 π2 2 π2 2 π4 4 x ; p3 (x) = 1 − x , p4 (x) = 1 − x+ x; 2! 2! 2! 4! ∞ k=0 (−1)k π 2k 2k x (2k )! 10. f...
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