# When x 11 and 1 the series diverges by divergence

• 129

This preview shows page 95 - 99 out of 129 pages.

When x = - 11 and 1, the series diverges by divergence test, so the interval of convergence ( - 11 , 1). Example 8.5.8. Find the interval and radius of convergence for the following series: X n =1 ( x - 2) n ln( n + 4) Using the ratio test, we have lim n →∞ ( x - 2) n +1 ln( n + 5) · ln( n + 4) ( x - 2) n = | x - 2 | converges when this limit is less than 1, and so | x - 2 | < 1 tells us that the radius of convergence is 1. The open interval of convergence is thus (2 - 1 , 2 + 1) = (1 , 3). Checking the endpoints, when x = 1 the series converges by alternating series test, and when x = 3 the series diverges by comparing to the series 1 n +4 . Thus, the interval of convergence is [1 , 3). 95
Section 8.5 Exercises Find the radius and interval of convergence for the given power series. 1. X n =0 x n n ! 2. X n =0 ( x - 2) n n 2 + 1 3. X n =1 x n n 3 n 4. X n =1 n !(2 x - 1) n 5. X n =1 x n 1 · 3 · 5 · · · (2 n - 1) 96
8.6 Representing Functions as Power Series We know that, when | x | < 1, we have 1 1 - x = X n =0 x n What this tells us is that the function f ( x ) = 1 1 - x can be approximated to any degree of accuracy (for values of x with - 1 < x < 1 anyway) by just looking at polynomials 1 + x + x 2 + · · · + x n for as large n as we require. This is fantastic as polynomials are well-studied and extremely easy to evaluate. Example 8.6.1. Express f ( x ) = 1 1+8 x 3 as a power series and find its interval of convergence. Notice that we have f ( x ) = 1 1 + 8 x 3 = 1 1 - ( - 8 x 3 ) = X n =0 ( - 8 x 3 ) n = X n =0 ( - 1) n (8 x 3 ) n . This converges when | 8 x 3 | < 1, i.e., when | x | < 1 2 . The interval of convergence is thus ( - 1 2 , 1 2 ) . Theorem 8.6.2. If the power series X n =0 c n ( x - a ) n has radius of convergence R > 0 , then the function f ( x ) = X n =0 c n ( x - a ) n is differentiable on the interval ( a - R, a + R ) and 1. d dx f ( x ) = X n =0 d dx [ c n ( x - a ) n ] = X n =1 nc n ( x - a ) n - 1 2. Z f ( x ) dx = X n =0 Z c n ( x - a ) n dx = C + X n =0 c n n + 1 ( x - a ) n +1 Example 8.6.3. Given the power series for f ( x ) = 1 1 - x , use differentiation to express g ( x ) = 1 (1 - x ) 2 as a power series. Find the radius of convergence of this new power series. We notice that f 0 ( x ) = 1 (1 - x ) 2 = g ( x ) , and so g ( x ) = d dx f ( x ) = X n =0 d dx [ x n ] = X n =1 nx n - 1 . Using the Ratio test, we have lim n →∞ a n +1 a n = lim n →∞ ( n + 1) x n nx n - 1 = | x | , which converges when | x | < 1, so the radius of convergence is 1, which is exactly the same as the radius of convergence for the series representation of f ( x ). 97
Proposition 8.6.4. Given the series with radius of convergence R > 0 and the function f ( x ) defined in the premise for Theorem 8.6.2, the new series obtained from d dx [ f ( x )] and R f ( x ) dx both have radius of convergence R . Example 8.6.5. Given the power series for f ( x ) = 1 1 + x , find a power series representation for g ( x ) = ln(1 + x ). Notice that g 0 ( x ) = 1 1 + x = f ( x ) , so we have g ( x ) = Z f ( x ) dx = Z 1 1 - ( - x ) dx = Z " X n =0 ( - 1) n x n # dx = X n =0 Z ( - 1) n x n dx = C + X n =0 ( - 1) n n + 1 x n +1 . To determine C , we set x = 0 and get that g (0) = 0 = C , so our power series representation is just g ( x ) = X n =0 ( - 1) n n + 1 x n +1 = X n =1 ( - 1) n - 1 n x n .