# P d 40 60 40 cos 60 20 p p directrix 43 use an xy

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Unformatted text preview: π/36)2 ≈ 0.99619, calculator value 0.99619 . . . 2 8. −175◦ = −π + π/36 rad; x0 = −π, x = −π + π/36, cos x = −1 + |Rn | ≤ (x + π )4 (x + π )2 − − ···; 2 4! (π/36)2 (π/36)n+1 ≤ 0.00005 for n = 3; cos(−π + π/36) = −1 + ≈ −0.99619, (n + 1)! 2 calculator value −0.99619 . . . Exercise Set 11.9 394 9. f (k) (x) = cosh x or sinh x, |f (k) (x)| ≤ cosh x ≤ cosh 0.5 = so |Rn (x)| &lt; 1 0.5 e + e−0.5 2 &lt; 1 (2 + 1) = 1.5 2 1.5(0.5)n+1 (0.5)3 ≤ 0.5 × 10−3 if n = 4, sinh 0.5 ≈ 0.5 + ≈ 0.5208, calculator (n + 1)! 3! value 0.52109 . . . 10. f (k) (x) = cosh x or sinh x, |f (k) (x)| ≤ cosh x ≤ cosh 0.1 = 1 0.1 e + e−0.1 &lt; 1.06 so |Rn (x)| &lt; 2 (0.1)2 1.06(0.1)n+1 ≤ 0.5 × 10−3 for n = 2, cosh 0.1 ≈ 1 + = 1.005, calculator value 1.0050 . . . (n + 1)! 2! 11. f (x) = sin x, f (n+1) (x) = ± sin x or ± cos x, |f (n+1) (x)| ≤ 1, |Rn (x)| ≤ |x − π/4|n+1 , (n + 1)! |x − π/4|n+1 = 0; by the Squeezing Theorem, lim |Rn (x)| = 0 n→+∞ n→+∞ (n + 1)! lim so lim Rn (x) = 0 for all x. n→+∞ 12. f (x) = ex , f (n+1) (x) = ex ; if x &gt; 1 then |Rn (x)| ≤ |Rn (x)| ≤ ex |x − 1|n+1 ; if x &lt; 1 then (n + 1)! e |x − 1|n+1 |x − 1|n+1 . But lim = 0 so lim Rn (x) = 0. n→+∞ (n + 1)! n→+∞ (n + 1)! 13. (a) Let x = 1/9 in series (17). (b) ln 1.25 ≈ 2 1/9 + (1/9)3 3 = 2(1/9 + 1/37 ) ≈ 0.223, which agrees with the calculator value 0.22314 . . . to three decimal places. 14. (a) Let x = 1/2 in series (17). (b) ln 3 ≈ 2 1/2 + (1/2)3 3 = 2(1/2 + 1/24) = 13/12 ≈ 1.083; the calculator value is 1.099 to three decimal places. 15. (a) (1/2)9 /9 &lt; 0.5 × 10−3 and (1/3)7 /7 &lt; 0.5 × 10−3 so tan−1 1/2 ≈ 1/2 − (1/2)5 (1/2)7 (1/2)3 + − ≈ 0.4635 3 5 7 tan−1 1/3 ≈ 1/3 − (1/3)5 (1/3)3 + ≈ 0.3218 3 5 (b) From Formula (21), π ≈ 4(0.4635 + 0.3218) = 3.1412 1 1 (c) Let a = tan−1 , b = tan−1 ; then |a − 0.4635| &lt; 0.0005 and |b − 0.3218| &lt; 0.0005, so 2 3 |4(a + b) − 3.140| ≤ 4|a − 0.4635| + 4|b − 0.3218| &lt; 0.004, so two decimal-place accuracy is guaranteed, but not three. 16. (27+ x)1/3 = 3(1+ x/33 )1/3 = 3 1 + √ 1 3·2 &lt; 0.0005, 28 ≈ 3 1 + 4 82 3 3 1 1·2 1·2·5 x − 8 x2 + 12 x3 + . . . , alternates after ﬁrst term, 4 3 32 3 3! ≈ 3.0370 395 Chapter 11 17. (a) sin x = x − |R4 (x)| ≤ x3 + 0 · x4 + R4 (x), 3! 0.0001 (b) |x|5 &lt; 0.5 × 10−3 if |x|5 &lt; 0.06, 5! |x| &lt; (0.06)1/5 ≈ 0.569, (−0.569, 0.569) -0.7 0.7 0 18. (a) f (k) (x) = ex ≤ eb , |R2 (x)| ≤ 0.0004 (b) eb b3 &lt; 0.0005, 3! eb b3 &lt; 0.003 if b &lt; 0.137 (by trial and error with a hand calculator), so [0, 0.136]. -0.2 0.2 0 19. (a) cos x = 1 − |R5 (x)| ≤ x4 x2 + + (0)x5 + R5 (x), 2! 4! 0.000000005 (b) (0.2)6 |x|6 ≤ &lt; 9 × 10−8 6! 6! -0.2 0.2 0 20. (a) f (x) = −1/(1 + x)2 , 0.00005 (b) f (x) &lt; 1/(0.99)2 ≤ 1.03, |R1 (x)| ≤ 1.03(0.01)2 1.03|x|2 ≤ 2 2 ≤ 5.11 × 10−5 for − 0.01 ≤ x ≤ 0.01 -0.01 0.01 0 21. (a) (1 + x)−1 = 1 − x + −1(−2) 2 −1(−2)(−3) 3 −1(−2)(−3) · · · (−k ) k x+ x + ··· + x + ··· 2! 3! k! ∞ (−1)k xk = k=0 (b) (1 + x)1/3 = 1 + (1/3)x + + (1/3)(−2/3) 2 (1/3)(−2/3)(−5/3) 3 x+ x + ··· 2! 3! (1/3)(−2/3) · · · (4 − 3k )/3 k x + · · · = 1+ x/3+ k! ∞ (−1)k−1 k=2 (c) (1 + x)−3 = 1 − 3x + ∞ = (−1)k k=0 2 · 5 · · · (3k − 4) k x 3k k ! (−3)(−4) 2 (−3)(−4)(−5) 3 (−3)(−4) · · · (−2 − k ) k x+ x + ··· + x + ··· 2! 3! k! (k + 2)! k x= 2 · k! ∞ (−1)k k=0 (k + 2)(k + 1) k x 2 Exercise Set 11.9 m 22. (1 + x) 23. (a) 396 ∞ m 0 = + k=1 ∞ k x= k=0 m k xk (k − 1)! 1 dk (k − 1)! d d ln(1 + x) = , k ln(1 + x) = (−1)k−1 ln(1 − x) = − ; similarly , k dx 1 + x dx (1 + x) dx (1 − x)k 1 (−1)n + . (1 + x)n+1 (1 − x)n+1 so f (n+1) (x) = n! (b) m k f (n+1) (x) ≤ n! (−1)n 1 1 1 + + n! = n! n+1 n+1 n+1 (1 + x) (1 − x) (1 + x) (1 − x)n+1 (c) If f (n+1) (x) ≤ M on the interval [0, 1/3] then |Rn (1/3)| ≤ M (n + 1)! (d) If 0 ≤ x ≤ 1/3 then 1 + x ≥ 1, 1 − x ≥ 2/3, f (n+1) (x) ≤ M = n! 1 + M (n + 1)! (e) 0.000005 ≥ 1 3 n+1 = 1 n+1 1 3 n+1 + (1/3)n+1 (2/3)n+1 = 1 n+1 n+1 1 3 . 1 . (2/3)n+1 n+1 1 3 + 1 2 n+1 24. Set x = 1/4 in Formula (17). Follow the argument of Exercise 23: parts (a) and (b) remain n+1 M 1 ≤ 0.000005 for x in the interval unchanged; in part (c) replace (1/3) with (1/4): (n + 1)! 4 [0, 1/4]. From part (b), together with 0 ≤ x ≤ 1/4, 1 + x ≥ 1, 1 − x ≥ 3/4, follows part (d): = M = n! 1 + 1 n+1 1 4 n+1 + 1 3 1 . (3/4)n+1 Part (e) now becomes 0.000005 ≥ M (n + 1)! 1 4 n+1 n+1 , which is true for n = 9. 25. f (x) = cos x, f (n+1) (x) = ± sin x or ± cos x, |f (n+1) (x)| ≤ 1, set M = 1, 1 |x − a|n+1 |x − a|n+1 , lim = 0 so lim Rn (x) = 0 for all x. n→+∞ (n + 1)! n→+∞ (n + 1)! |Rn (x)| ≤ 26. f (x) = sin x, f (n+1) (x) = ± sin x or ± cos x, |f (n+1) (x)| ≤ 1, follow Exercise 25. π 27. (a) From Machin’s formula and a CAS, ≈ 0.7853981633974483096156609, accurate to the 25th 4 decimal place. (b) n 0 1 2 3 1/p sn 0.3183098 78 . . . 0.3183098 861837906 067 . . . 0.3183098 861837906 7...
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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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