Review Power Series Example 3 Example Since X n 0 n 2 n 1 a n 2 a n t n 0 it

# Review power series example 3 example since x n 0 n 2

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Review Power Series Example 3 Example: Since X n =0 [( n + 2)( n + 1) a n +2 - a n ] t n = 0 , it follows that ( n + 2)( n + 1) a n +2 - a n = 0 The first two coefficients, a 0 and a 1 are arbitrary , then all other coefficients are specified by the recursive relation : a n +2 = a n ( n + 2)( n + 1) Thus, with a 0 arbitrary a 2 = a 0 2! , a 4 = a 2 4 · 3 = a 0 4! , ..., a 2 n = a 0 (2 n )! Joseph M. Mahaffy, h [email protected] i Lecture Notes – Power Series Ordinary Point — (7/24) Introduction Series Solutions of Differential Equations Example Review Power Series Example 4 Example: Similarly, with a 1 arbitrary a 3 = a 1 3 · 2 , a 5 = a 3 5 · 4 = a 1 5! , ..., a 2 n +1 = a 1 (2 n + 1)! It follows that we have two linearly independent solutions y 1 ( t ) = X n =0 t 2 n (2 n )! and y 2 ( t ) = X n =0 t 2 n +1 (2 n + 1)! , with the general solution y ( t ) = a 0 y 1 ( t ) + a 1 y 2 ( t ) Note: y 1 ( t ) = cosh( t ) and y 2 ( t ) = sinh( t ) Joseph M. Mahaffy, h [email protected] i Lecture Notes – Power Series Ordinary Point — (8/24) Subscribe to view the full document.

Introduction Series Solutions of Differential Equations Example Review Power Series Review Power Series Review Power Series: Consider the power series: X n =0 a n ( x - x 0 ) n The series converges at x if lim k →∞ k X n =0 a n ( x - x 0 ) n exists for x . It clearly converges at x 0 , but may or may not for other values of x The series converges absolutely if the following converges: X n =0 | a n ( x - x 0 ) n | Joseph M. Mahaffy, h [email protected] i Lecture Notes – Power Series Ordinary Point — (9/24) Introduction Series Solutions of Differential Equations Example Review Power Series Ratio Test Ratio Test: For the power series: X n =0 a n ( x - x 0 ) n The ratio test provides a means of showing absolute convergence . If a n 6 = 0, x fixed, and lim n →∞ a n +1 ( x - x 0 ) n +1 a n ( x - x 0 ) n = | x - x 0 | lim n →∞ a n +1 a n = | x - x 0 | L, then the power series converges absolutely at x , if | x - x 0 | L < 1. If | x - x 0 | L > 1, then the series diverges . The test is inconclusive with | x - x 0 | L = 1. Joseph M. Mahaffy, h [email protected] i Lecture Notes – Power Series Ordinary Point — (10/24) Introduction Series Solutions of Differential Equations Example Review Power Series Example Example: For the power series: X n =1 ( - 1) n +1 n ( x - 2) n The ratio test gives: lim n →∞ ( - 1) n +2 ( n + 1)( x - 2) n +1 ( - 1) n +1 n ( x - 2) n = | x - 2 | lim n →∞ n + 1 n = | x - 2 | . This converges absolutely for | x - 2 | < 1. It diverges for | x - 2 | L 1. Joseph M. Mahaffy, h [email protected] i Lecture Notes – Power Series Ordinary Point — (11/24) Introduction Series Solutions of Differential Equations Example Review Power Series Radius of Convergence Radius of Convergence: For the power series: X n =0 a n ( x - x 0 ) n , typically, there is a positive number ρ , called the radius of convergence , such that the series converges absolutely for | x - x 0 | < ρ and diverges for | x - x 0 | > ρ Generally, we are not concerned about convergence at the endpoints Joseph M. Mahaffy, h [email protected] i Lecture Notes – Power Series Ordinary Point — (12/24) Introduction Series Solutions of Differential Equations Example Review Power Series Properties of Series 1 Consider the series X n =0 a n ( x - x 0 ) n = f ( x ) and X n =0 b n ( x - x 0 ) n = g ( x ) converging for | x - x 0 | < ρ with Subscribe to view the full document. • Fall '08
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