Math 116 — Practice for Exam 3.pdf

Math 116 / Final (April 19, 2012) page 14 9. [10 points] A second way to approximate the function Z x√ 1 + 9t4 dt. F (x) = 0 is by using its Taylor polynomials. a. [2 √ points] Find the first three nonzero terms in the Taylor series for the function 1 + u about u = 0. Solution: √ 1 1 1 + u ≈ 1 + u − u2 2 8 b. [2 points] Find the first three nonzero terms in the Taylor series for t = 0. √ 1 + 9t4 about Solution: Let u = 9t4 then √ 9 81 1 + 9t4 ≈ 1 + t4 − t8 2 8 c. [2 points] Find the first three nonzero terms in the Taylor series for F (x) about x = 0. Solution: Z x 81 9 9 9 F (x) ≈ 1 + t4 − t8 dt = x + x5 − x9 2 8 10 8 0 d. [2 points] For which values of x do you expect the Taylor series for F (x) about x = 0 to converge? Justify your answer. Solution: We substituted u = 9t4 into the Binomial series. The interval of convergence for the Binomial series is −1 < u < 1. Then we expect the series to converge for 0 ≤ 9x4 < 1. Hence the Taylor series for F (x) about x = 0 converges if − √13 < x < √13 . e. [2 points] Use the fifth degree Taylor polynomial for F (x) about x = 0 to approximate the value of F ( 21 ).


9 5 9 1 5 Solution: P5 (x) = x + 10 x , then F 12 ≈ P5 21 = 12 + 10 = 0.528 2 Math 116 / Final (April 28, 2014) page 4 3. [10 points] For each of the following questions circle the correct answer. a. [2 points] What is the value of the series ∞ X (−1)n 22n n! n=0 cos(2) cos(4) e−2 b. [2 points] What is the value of the series ∞ X 22n (−1)n n=1 1 2 sin(2) ? sin(2) − 2 (2n + 1)! e−4 ? 1 2 (sin(2) sin(2) − 2) 1 3 x is the 3rd degree Taylor polynomial for a c. [2 points] Suppose that 1 + x − 14 x2 + 10 function f (x). Which of the following pictures could be a graph of f (x)? f (x) 3 f (x) 3 2 2 1 1 I II x 1 −3 −2 −1 −1 2 3 x 1 −3 −2 −1 −1 −2 −2 −3 −3 f (x) 3 f (x) 3 2 2 1 1 III x −3 −2 −1 −1 1 2 3 IV 2 3 x 1 −3 −2 −1 −1 −2 −2 −3 −3 2 3 2 d. [2 points] What is the Taylor series of 2xex centered at x = 0? ∞ X 2x2n+1 ∞ X 2x2n−1 n=0 n=1 n! ∞ X 2x2n+1 (n − 1)! n! ∞ X 2x2n−1 n=1 e. [2 points] The radius of convergence of the Taylor series n=0 ∞ X (x + 5)n 5−n n=1 n+5 n! is R = 5. What is the interval of convergence of the series? [−10, 0) (−10, 0) (0, 10] [−10, 0] [0, 10) Math 116 / Final (December 12, 2014) page 4 2 2. [6 points] Let f (x) = xe−x . a. [4 points] Find the Taylor series of f (x) centered at x = 0. Be sure to include the first 3 nonzero terms and the general term. 2 Solution: We can use the Taylor series of ey to find the Taylor series for e−x by substituting y = −x2 . e −x2 = ∞ X (−x2 )n n=0 n! = 1 + (−x2 ) + (−x2 )2 (−x2 )n + ··· + + ··· 2! n! 2 Therefore the Taylor series of xe−x is 2 xe−x = ∞ X (−1)n x2n+1 n=0 n! = x − x3 + (−1)n x2n+1 x5 + ··· + + ··· 2! n! b. [2 points] Find f (15) (0). (15) Solution: We know that f 15!(0) will appear as the coefficient of the degree 15 term of the Taylor series. Using part (a), we see that the degree 15 term has coefficient −1 7! . Therefore −15! f (15) (0) = = −259, 459, 200 7! 3. [3 points] Determine the exact value of the infinite series 1− 2 4 8 (−1)n 2n + − + ··· + + ··· 1! 2! 3! n! Solution: Notice that this is the Taylor series for ey applied to y = −2. Therefore, the series has exact value e−2 . Math 116 / Final (December 12, 2014) page 13 12. [8 points] Franklin, your friendly new neighbor, is building a large chicken sanctuary. You decide to help Franklin build a special chicken coop with volume (in cubic km) given by the integral Z 1 p x 1 − cos(x2 ) dx. 0 This integral is difficult to evaluate precisely, so you decide to use the methods you’ve learned this semester to help out Franklin. Your friend and president-elect, Kazilla, stops by to give you a hand. She suggests finding the 4th degree Taylor polynomial, P4 (x), for the function 1 − cos(x2 ) near x = 0. a. [4 points] Find P4 (x). Solution: We can use the Taylor series expansion for cos(x2 ) cos(y) = ∞ X (−1)n y 2n (2n)! n=0 so 2 cos(x ) = ∞ X (−1)n (x2 )2n n=0 (2n)! y2 (−1)n y 2n + ··· + + ··· 2! (2n)! =1− =1− (x2 )2 (−1)n (x2 )2n + ··· + + ··· 2! (2n)! Therefore P4 = 1 − (1 − x4 x4 )= 2! 2! b. [4 points] Substitute P4 (x) for 1−cos(x2 ) in the integral and compute the resulting integral by hand, showing all of your work. Solution: Z 1 0 Z 1 r 4 p x x P4 (x)dx = x 2 0 Z 1 3 x √ = 2 0 x4 = √ 4 2 1 = √ 4 2 1 x=0 2. [8 points] Let f (x) = x2x . The first two derivatives of f are given below. f ′ (x) = 2(1 + ln x)x2x f ′′ (x) = 2x2x−1 + 4(1 + ln x)2 x2x a. [4 points] Find the 2nd degree Taylor polynomial P2 (x) of f centered at x = 1. Solution: Using the formula for Taylor polynomials, f ′′ (1) f ′ (1) (x − 1) + (x − 1)2 1! 2! = 1 + 2(x − 1) + 3(x − 1)2 P2 (x) = f (1) + 1 + 2(x − 1) + 3(x − 1)2 P2 (x) = b. [4 points] Find x2x − 1 . x→1 3x − 3 lim Clearly show your reasoning. Your answer from part (a) may be helpful. Solution: 1 + 2(x − 1) + 3(x − 1)2 − 1 x2x − 1 = lim x→1 x→1 3x − 3 3(x − 1) 2 + 3(x − 1) = lim x→1 3 2 = 3 lim 10. [8 points] The Taylor series centered at x = 0 for a function F (x) converges to F (x) for all x and is given below. F (x) = ∞ X (−1)n n=0 x4n+1 (2n)!(4n + 1) a. [3 points] What is the value of F (101) (0)? Make sure your answer is exact. You do not need to simplify. Answer: F (101) (0) = b. [3 points] Find P9 (x), the 9th degree Taylor polynomial that approximates F (x) near x = 0. c. [2 points] Use your Taylor polynomial from part b. to compute F (x) − x x→0 2x5 lim 6. [8 points] Values of a function f and some of its derivatives are given in the table on the right. Use this information to answer the questions that follow. x 0 π f (x) −6 2π f ′ (x) 6 2 f ′′ (x) 1 −3 f ′′′ (x) −1 0 f ′′′′ (x) 5 −9/2 a. [4 points] Find a formula for the Taylor polynomial of degree 4 for f about x = π. b. [4 points] Find the first three nonzero terms of the Taylor series for x = 0. Z x f (t2 ) dt about 0