University:
Whitman College
Course:
MATH-126 | Calculus II
Academic year:

2024
Views:

404

Pages:

2
Author:

Kianna Hurley

Homework Hints, 11.9 (Series as Geo Series) 1, 3, 7, 13(a,b), 40(a,b) 1. The radius of convergence for the derivative of a power series is the same as the radius of convergence for the series itself (Theorem 2, p. 748). 3. Use the Geo series with r = −x: ∞ X 1 1 = = (−x)n 1+x 1 − (−x) n=0 and that converges if |x| < 1. 7. We want to manipulate the given expression until it resembles the sum of a geometric series. For example: x x 1 x 1 = · = · 2 2 9+x 9 1 + (x /9) 9 1 − (−x2 /9) We treat x/9 as a constant (with respect to n and the ratio r), and the series this is a sum for is given by: !n ∞ ∞ X X x −x2 (−1)n x2n+1 · = 9 9n+1 n=0 9 n=0 The series converges when x2 < 9, or −3 < x < 3. The series diverges for all other x. 13(a). The “trick” is to notice that −1 d 1 = dx 1 + x (1 + x)2 Replacing the left side by its series (from #3) we have: ∞ ∞ X d X −1 n n n(−1)n xn−1 = (−1) x = (1 + x)2 dx n=0 n=1 ! Multiply by (−1), and we get: ∞ X 1 = n(−1)n+1 xn−1 2 (1 + x) n=1 and the series converges absolutely when |x| < 1. It is fine to leave your answer this way, but if you want to see how to make this the same as the back of the text, let m = n − 1. Then if n = 1, m = 0, and substitute into the sum. Notice that the sum should start with +1, then alternate. ∞ X 1 = (m + 1)(−1)m xm (1 + x)2 m=0 1 13(b). Differentiate again: ∞ X −2 = m(m + 1)(−1)m xm−1 (1 + x)3 m=1 Multiply both sides by −1/2 (the resulting sum begins with a positive term) and re-write the sum on the right by substituting n = m − 1 to get what’s in the text: ∞ 1 1X (n + 1)(n + 2)(−1)n xn = (1 + x)3 2 n=0 40. These are good practice manipulating the geometric series. (a) Find the sum of the series ∞ X nxn1 n=1 SOLUTION: Start with the standard geometric series: ∞ X 1 xn = 1 − x n=0 and differentiate both sides with respect to x: ∞ X 1 nxn−1 , = (1 − x)2 n=1 |x| < 1 (b) Similarly, for the first of the two sums, note that ∞ X nxn = ∞ X x · nxn−1 n=1 n=1 so multiply our previous expression by x: ∞ X x = nxn , 2 (1 − x) n=1 |x| < 1 For the second sum (in (ii)), set x = 1/2 from this sum so that ∞ X 1/2 n = 2 n (1 − (1/2)) n=1 2 This expression simplifies to 2. 2

Calculus II Final Exam Review Solutions

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