Business Calc Homework w answers_Part_75

Business Calc Homework w answers_Part_75 - Section 9.2 55....

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Unformatted text preview: Section 9.2 55. Given an 0, by definition of convergence there and corresponds an N such that for all n N, L1 an L 2 an . (There is one such number for each series, and we may let N be the larger of the two numbers.) Now L2 L1 L2 an an L1 an L1 L 2 an 1 L1, lim ai(n) L2, and L1 n→ ak(n) L1 , and an N2 such that for i(n) ai(n) L2 0 N1, max {N1, N2}. Then for n an L1 and an implies that lim an L2 N2, N, we have that for infinitely many n. This L1 and lim an n→ n→ L2 where L1 (b) The line y L2. 1. f (x) f (x) f (x) f (x) f (n)(x) 1 (x f (x) 3x x 1 , which means lim f (x) 1 x→ 3. an for all positive integers n, n→ s Section 9.2 Taylor Series (pp. 469–479) Exploration 1 Designing a Polynomial to Specifications 1, we know that the constant coefficient is 1. Since P (0) 1) 4 ( 1)nn!(x 1) 1)! x f (x) nx n f (x) n(n 1)x n (The 2 in the denominator is needed to cancel the factor of f (x) n(n 1)(n 2 that results from differentiating x 2.) Similarly, we find the f (k)(x) 4 5 coefficients of x 3 and x 4 to be and . 6 24 32 43 54 Thus, P(x) 1 2x x x x. 2 6 24 n for n 1 xn 5. f (x) 3 3, we know that the coefficient of x is . 2 (n 1) 3x 3x ln 3 3x(ln 3)2 3x(ln 3)3 3x(ln 3)n 4. f (x) ln x f (x) x 1 f (x) x2 f (x) 2x 3 f (4)(x) 6x 4 f (n)(x) ( 1)n 1(n 2, we know that the coefficient of x is 2. Since P (0) 3. f (x) f (x) f (x) f (x) f (n)(x) 3 1) 6(x f (n)(x) 2 1) 2(x f (x) it follows that lim an must also be 3. 1. Since P(0) x f (x) 3 is a horizontal asymptote of the graph of Because f (n) e2x 2e2x 4e2x 8e2x 2ne2x 1 2. f (x) 3 the function f (x) Approximating sin 13 4. 20 terms. an does not converge and hence diverges. 1 1 x6 6! x6 , and the Taylor series is 6! 2n … ( 1)n x …. (2n)! 1. 0.4201670368… Since the limit of a sequence is unique (by Exercise 55), 3n 57. (a) lim n→ n x4 4! x4 4! Quick Review 9.2 . Assume an converges. Let N x2 2 x2 2 Exploration 3 L2. Given an there corresponds an N1 such that for k(n) 1 2. A clever shortcut is simply to differentiate the previouslydiscovered series for sin x term-by-term! 56. Consider the two subsequences ak(n) and ai(n), where lim ak(n) A Power Series for the Cosine 1. cos (0) 1 cos (0) sin (0) 0 cos (0) cos (0) 1 cos(3) (0) sin (0) 0 etc. The pattern 1, 0, 1, 0 will repeat forever. Therefore, P6(x) 2. L2 L1 2 says that the difference between two fixed values is smaller than any positive number 2 . The only nonnegative number smaller than every positive number is 0, so L2 L1 0 or L1 L2. n→ Exploration 2 371 2 f (n)(x) 1 2 2)x n 3 n! xn k (n k)! n! 0 x n! 0! 6. dy dx d xn dx n! nx n n! 7. dy dx d 2n(x a)n n! dx 1 xn 1 (n 1)! 2nn(x a)n n! 1 2n(x (n a)n 1) ! 1 372 8. 9. 10. Section 9.2 d x 2n 1 ( 1)n dx (2n 1)! dy dx 2n dy dx d (x a) dx (2n)! dy dx 2n(x d (1 x)n n! dx n(1 ( 1)n (2n 1)x 2n (2n 1)! 2n 1 ( 1)nx 2n (2n)! 5. cos (x (x a) (2n 1)! x)n 1( 1) n! (1 x)n 1 ( n 1) ! (cos 2) at the end of Section 9.2. (2x) 3! 4x 3 3 2x 2x 5 2n 1 (2x) … ( 1)n (2x) 5! (2n 1)! 4x 5 ( 1)n(2x)2n 1 … … 15 (2n 1)! … x for x in the Maclaurin series for ln (1 x) x) x5 … 5! (cos 2)x 2 2! (sin 2)x ( x)2 ( x)3 … 2 3 n ( x) … ( 1)n 1 n 2 3 n x x …x … x 2 3 n 1 x 1, so the interval … ( 1)n(cos 2) where 2n (2n)! 2n 1 k. Thus the coefficient is B sin 2 if n is even and cos 2 if n is odd. 1 cos 2 sin 2 x , 2 cos 2 cos (2 and so on, so the general term is 2 2n 1 (x 2)3 (x 2)5 … ( 1)n (x ) 3 5 2n 1 n 4n 2 x6 x 10 … ( 1) x … 3 5 2n 1 x2 x2 This series converges when x 2 7x(1 7x x 7x 2 … n x2 …x …) 2! n! 3 n1 7x … 7x 2! n! This series converges for all real x. ), n 2 1 cos 2 n! x n. The series converges for all real x. 6. x 2 cos x x2 2! x2 1 x4 2 x2 1, so the interval of convergence is [ 1, 1]. 4. 7xe x k. Thus the coefficient is shown at the end of Section 9.2. x2 x 2n 1 … (2n 1)! (sin 2)x 3 (cos 2)x 4 3! 4! ( 1)n Another way to handle the general term is to observe that 3. Substitute x 2 for x in the Maclaurin series for tan 1 … We need to write an expression for the coefficient of x k. and B of convergence is [ 1, 1). tan x 2n (2n)! ( 1)int[(k 1)/2](cos 2) ( 1)(k 1)/2(sin 2) , which is the same as . k! (2n 1)! An ( 1) Bx n1 , where A int , Hence the general term is n! 2 ( x) This series converges when ( 1)n ( 1)int[(k 1)/2](cos 2) ( 1)k/2(cos 2) , which is the same as . k! k! n1 ( 1) (sin 2) where If k is odd, the coefficient is (2n 1)! shown at the end of Section 9.2. ln (1 (sin 2)(sin x) … If k is even, the coefficient is This series converges for all real x. 2. Substitute x4 4! x3 3! (sin 2)x 5 5! 1. Substitute 2x for x in the Maclaurin series for sin x shown sin 2x x2 2! (sin 2) x Section 9.2 Exercises 3 (cos 2)(cos x) (cos 2) 1 2n a) (2n)! 2) x6 24 … x4 … 4! ( 1)nx 2n (2n)! ( 1)n 1 x 2n (2n)! … … The series converges for all real x. 7. Factor out x and substitute x 3 for x in the Maclaurin series for 1 1 x shown at the end of Section 9.2. x 1 1 x3 x[1 1 x3 x x3 x x4 (x 3)2 x7 … The series converges for x 3 convergence is ( 1, 1). … x 3n (x 3)n 1 …] … 1, so the interval of Section 9.2 2x for x in the Maclaurin series for e x shown at 8. Substitute x1/2 12. f(4) the end of Section 9.2. 2x 1 2x ( 2x)2 … 2! nnn … ( 1) 2 x n! ( 2x) 1 e 2x 2 ( 2x)n n! … f (4) 9. f (2) 1 xx 2 f (2) f (2) 2x f (2) P0(x) P1(x) P2(x) P3(x) 10. f f f 4 1 3 x2 4 4 6x x2 1 2 1 2 1 2 1 2 x f (2) 1 2! 8 3 f (2) , so 8 3! 4 2 8 x 2 (x 4 2)3 16 (x (b) f (1) 2 /4 2 x cos x 2 /4 f 2 x 2 /4 4 , so 2 2! f 4 , so 2 12 3! 2 P1(x) 2 P2(x) P3(x) 2 2 x 2 2 2 2 2 2 4 f f f cos x 4 4 4 P0(x) 4 x3 2x 6x f (1) 6 P3(x) 3 4 4 2 1 x1 f (1) 2! 3 6, so f (1) 3! 1 1) 3(x 1)2 6, so x1 x1 4 (x 4 2x 3 x2 3x 6x 2 2x 12x 2 f (1) 12 P3(x) 3 2 8 14, so f (1) 2! f (1) 12, so 3! 2 x1 x1 11(x 1) 2 /4 2 x 2 /4 2 /4 2 P2(x) 2 P3(x) 2 7 1)2 7(x 2(x f , so 4 2 2! 4 (b) f (1) x4 1 x1 f , so f (1) 3! 12x 2 24x 1 4(x 1) 16. (a) P3(x) 2 12 4x 3 P3(x) 4 f (1) 4 5x 82 x 2! 4 5x x1 x1 4 12, so f (1) 2! 2 2 2 2 2 2 2 2 2 x 12 x 4 2 x 4 4 4 4 2 x 3 4 x1 6 24, so f (1) 3! 4 2 x 4 2 x 6(x 4x 2 4 f (0.2) P3(0.2) 4.848 1)2 63 x 3! x3 1)3 0, the Taylor 2 x 2 2 11 x1 2 P1(x) 2 x1 15. (a) Since f (0) f (0) f (0) f (0) polynomial of order 3 is P3(0) 0. 2 1)3 (x 14. (a) Since f is a cubic polynomial, it is its own Taylor polynomial of order 3. P3(x) 2x 3 x 2 3x 8 or 8 3x x 2 2x 3 f (1) 2 x 4 2 x 4)3 512 3 f (1) sin x (x x1 4 3 /4 cos x 4 4 f (1) 2 x sin x 4 x 2 x 2 2 2 2 x 12 11. f 4 x 3x 2 (b) f (1) 4 x 2 2 (x 4 2 P0(x) 4)2 64 (x 4)2 64 4 f (1) 2 sin x x 2 /4 x 4 4 f (1) 2 cos x 4 2)2 8 x 2 x 1 512 13. (a) Since f is a cubic polynomial, it is its own Taylor polynomial of order 3. P3(x) x 3 2x 4 or 4 2x x 3 2)2 (x 4 4 1 16 2 P3(x) , so 2 x sin x 4 f x2 2 P2(x) 1 4 2 x f (4) 3 , so 3! 256 1 64 2 P1(x) 1 2 f (4) 1 , so 2! 32 3 5/2 x 8 x4 P0(x) The series converges for all real x. 1 4 1 3/2 x 4 x4 f (4) … 2 x4 1 1/2 x 2 x4 f (4) 373 4(x 1)3 374 Section 9.2 16. continued (x)(e x) d ex 1 x dx xe x e x 1 x2 ee1 1 1 (c) g (x) (b) Since the Taylor series of f (x) can be obtained by differentiating the terms of the Taylor series of f (x), the g (1) second order Taylor polynomial of f (x) is given by 5 3x 2. Evaluating at x 8x f (0.2) g (x) 3 (x 2! 4 ( 1)(x 1) 4 (x 3 (x 2 f (1.2) 1) P3(1.2) 2 (x 3! 1)2 1 (x 3 1)2 1)3 3.863 ∑ second order Taylor polynomial of f (x) is given by 1) f (1.2) 1)2. Evaluating at x (x 1.2, n1 n1 n (n x ex g(x) (c) g(x) x2 2! x3 3! x4 4! f (t) t2 … … x for x in the Maclaurin series for e x shown 2 x 2 x 2 1 ex (b) g(x) x2 8 3 n! … 2 1 xn 2 … 2 G(x) … x 2! … (t 2)3 2t 4 2x 3 3 x)1/2 (1 f (0) f (0) xn 2n n! 1 (1 2 … 2t 6 P4(x) 2x 5 5 (t 2)n 2t 2n …] … x) 5/2 x0 x2 8 x 2 2x 2n 1 2n 1 1 f (0) , so 4 2! 3/2 x0 x) … … 1 2 1/2 x0 3 (1 8 1 2x 7 7 1 x0 x) 1 (1 4 3 f (0) , so 8 3! 1 8 1 16 x3 16 x x2 2! x 2! x3 3! x2 3! x2 3! f (x 2), the first four terms are (b) Since g(x) n x 3! ... x n! … xn n! … xn 1 n! … … … xn (n 1)! … 1 1 x2 2 x4 8 (c) Since h (0) x6 . 16 5, the constant term is 5. The next three terms are obtained by integrating the first three terms of This can also be written as 1 , which means 1 x 2! 1 1)! 0, the constant term is zero and we may 2x f (0) x 1 1 x 1 x x … find G(x) by integrating the terms of the series for f (x). xn 1 (n 1)! at the end of Section 9.2 e x/2 n (n (t 2)2 2t 2 (b) Since G(0) 21. (a) f (0) x2 2 nx n 1 (n 1)! … … 1)! 1 at the end of Section 9.2. 1x 2 1 t2 1 2 1 t2 1 19. (a) Substitute (n series for 2 (b) Multiply each term of f (x) by x. xn … 1. 1)! 2[1 1 . 11 x3 4! 20. (a) Factor out 2 and substitute t 2 for x in the Maclaurin 0.36 1 x 1 18. (a) Since f (0)x , f (0) . 2 2! 2! f (10)(0) 10 10! x 10 (10) Since x , f (0) 11! 10 ! 11! ∑ Therefore, g (1) differentiating the terms of the Taylor series of f (x), the 3(x x2 d x 1 3! dx 2! 2x 3x 2 1 3! 4! 2! nxn 1 1)! n 1 (n ∑ 1)3 (b) Since the Taylor series of f (x) can be obtained by 1 1)(1) From the series, 0.2, 3.52 17. (a) P3(x) (e x x2 the answer to part (b). The first four terms of the series …. for h(x) are 5 x x3 6 x5 . 40 375 Section 9.2 22. (a) a0 1 3 a 10 3 a 21 3 a 32 a1 a2 a3 31 3 3 2 9 2 3 (x 2 a2 we have f (0) 3 n 3 . n! 1 92 x 2 3x … 93 x 2 (b) Since the series can be written as ∑ n0 3 and 3e 3x 3n x n! … L2(x) 1 (x 2 3 n (3x) , it represents n! 1 x 2 3) 3 2 4x and 3, respectively. e. 3e 3 x1 23. First, note that cos 18 ∑( n 3x the function f (x) Using cos x 3) 4, f ( 1 , so the linearizations are L1(x) 2 f ( 3) n0 (c) f (1) 0, f (0) n term by , an n 1)(4) (4x)(2x) (x 2 1)2 4 4x 2 , (x 2 1)2 9 2 Since each term is obtained by multiplying the previous ∑ anx 4x d dx x 2 1 29. (a) Since f (x) 0.6603. 1)n n0 x 2n , enter the following two-step (2n)! commands on your home screen and continue to hit [ 2, 4] by [ 3, 3] (b) f (a) must be 0 because of the inflection point, so the second degree term in the Taylor series of f at x a is zero. ENTER. 30. The series represents tan tan 1 1 4 . When x tan 1( 1) The sum corresponding to N 25 is about 0.6582 (not withing 0.001 of exact value), and the sum corresponding to N 26 is about 0.6606, which is within 0.001 of the exact value. Since we began with N 0, it takes a total of 27 terms (or, up to and including the 52nd degree term). 24. One possible answer: Because the end behavior of a polynomial must be unbounded and sin x is not unbounded. Another: Because sin x has an infinite number of local extrema, but a polynomial can only have a finite number. 25. (1) sin x is odd and cos x is even (2) sin 0 0 and cos 0 1 (3x)5 35 81 so = . 5! 5! 40 x. When x 1, it converges to 1, it converges to . 1 (sin x) x 2n 1 1 x3 x5 … ( 1)n x …) (x x 3! 5! (2n 1)! 2 4 2n x x … ( 1)n x … 1 3! 5! (2n 1)! 31. (a) f (x) (b) Because f is undefined at x (c) k 0. 1 32. Note that the Maclaurin series for 1 26. Replace x by 3x in series for sin x. Therefore, we have 4 1 … x2 x 1 1 x is …. If we differentiate this series xn and multiply by x, we obtain the desired Maclaurin series x 2x 2 … 3x 3 …. Therefore, the desired nx n function is 3 27. Since is 1 4 3! d ln x dx 3 2x 3 , which is 1 at x 4 2, the coefficient 1 . 24 28. The linearization of f at a is the first order Taylor polynomial generated by f at x a. f (x) x d 1 dx 1 x 33. (a) f (x) (1 x 1 (x 1)2 . x)m f (x) m(1 x)m f (x) m( m 1)(1 f (x) x x)2 (1 m(m 1 1)(m x)m 2 2)(1 x)m (b) Differentiating f (x) k times gives f (k)(x) m(m 1)(m 2) … (m Substituting 0 for x, we have f (k)(0) m(m 1)(m 2) … (m 3 k 1)(1 k 1). x)m k. ...
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