Download Exam II with Answers - Calculus and Analytic Geometry | MATH 222 and more Exams Analytical Geometry and Calculus in PDF only on Docsity! Math 222, Exam II, March 2, 2001 Answers I. (30 points.) Complete the following paragraph by writing in the correct formulas. The Taylor series of a function f(x) about the point a is the infinite series ∞∑ k=0 f (k)(a)(x− a)k k! It may be used to compute f(x) when x is sufficiently close to a. The nth degree Taylor polynomial of f(x) about the point a is the polynomial fn(x) of degree n which best approximates f(x) for x near a. It is fn(x) = n∑ k=0 f (k)(a)(x− a)k k! The nth remainder (error) is defined by f(x) = fn(x) +Rn(x, a). Three formulas for the remainder are Rn(x, a) = ∞∑ k=n+1 f (k)(a)(x− a)k k! Rn(x, a) = ∫ x a (x− t)n n! f (n+1)(t) dt and Rn(x, a) = f (n+1)(c)(x− a)n+1 (n+ 1)! for some number c with a < c < x or x < c < a. The last is called Lagrange’s formula for the remainder. II. (30 points.) Every power series ∑∞ k=0 akx k has a radius of convergence r. This number is charac- terized by the conditions that the series converges absolutely for |x| < r and diverges for |x| > r. Find the radius of convergence for each of the following series. Justify your answers. (i) ∞∑ n=1 nxn 2n Answer: The limit of the ratio of absolute values of successive terms is lim n→∞ |(n+ 1)xn+1/2n+1| |nxn/2n| = lim n→∞ |x|(n+ 1) 2n = |x| 2 . By the ratio test the series converges if this limit is less than one (i.e. if |x| < 2) and diverges if this limit is greater than one (i.e. if |x| > 2). Hence r = 2. (ii) ∞∑ n=0 (−1)nx2n+1 (2n+ 1)! Answer: The limit of the ratio of absolute values of successive terms is lim n→∞ |x2n+3/(2n+ 3)!| |x2n+1/(2n+ 1)!| = lim n→∞ x2 (2n+ 3)(2n+ 2) = 0. Hence by the ratio test the series converges for all x, i.e. r =∞. III. (30 points.) Find the 4th degree Taylor polynomial f4(x) about the point 0 for the function f(x) = cosx. Estimate the error | cosx− f4(x)| for |x| < 1. Answer: f(x) = f (4)(x) = cosx, f ′(x) = f (5)(x) = − sinx, f ′′(x) = − cosx, f (3)(x) = sinx, so f(0) = f (4)(0) = 1, f ′(0) = f (3)(0) = 0, f ′′(0) = −1. and hence the 4th degree Taylor polynomial is f4(x) = 4∑ n=0 f (n)(0)xn n! = 1− x 2 2! + x4 4! . The error is R4(x, 0) = f (5)(c)x5 5! = (cos c)x5 5! for some unknown c between 0 and x. As | cos c| ≤ 1 we have |R4(x, 0)| ≤ |x5| 5! < 1 5! for |x| < 1.
