advertisement

Wednesday, March 23 Announcements If you submitted your Quiz #4 for a regrade, come pick it up from me (before or after class) WeBWorK #10 is due at 9pm Covers material from “Week 10” No class on Friday March 25 or Monday March 28 WeBWorK #11 is due next Wednesday (March 30) at 9pm Covers material from “Week 11”—see syllabus on course web page Solutions to (all versions of) Quiz #5 are online Quiz #5 papers will be available in the MLC starting tomorrow Grades posted in Connect by Friday Wednesday, March 23 Clicker Questions Clicker Question 1 What about here? . . . what about here? Suppose the power series ∞ X cn (x − 3)n converges when n=0 x = −2 and diverges when x = −5. Of the values x = −6, x = −4, x = 1, x = 4, x = 7, x = 9, x = 12, where can we be sure that the series converges? A. at x = 1, x = 4, and x = 7 Close enough to 3 B. at x = −4, x = 1, and x = 4 The series converges at x = −2, so the radius of convergence is at least |(−2) − 3| = 5. The series diverges at x = −5, so the radius of convergence is at most |(−5) − 3| = 8. . . . C. at x = −4, x = 1, x = 4, x = 7, and x = 9 (maybe) D. only at x = 1 E. none of the above Clicker Question 2 Power series representation x5 Find a power series that represents 2 on the interval x +3 √ √ (− 3, 3). A. B. C. D. ∞ X (−1)2n+5 n=0 ∞ X n=0 ∞ X n=0 ∞ X n=0 32n+6 xn (−1)n 5n √ x ( 3)n+1 (−1)n n √ x ( 3)n+1 (−1)n 3n+1 x2n+5 E. none of the above Using a previous example ∞ X (−1)n 1 = xn x+3 3n+1 (|x| < 3) n=0 1 = x2 + 3 x5 x2 + 3 = ∞ X (−1)n n=0 3n+1 ∞ X (−1)n n=0 3n+1 (x2 )n (x2 )n x5 (|x2 | < 3) (|x| < √ 3)