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

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: 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...
View Full Document

Ask a homework question - tutors are online