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Unformatted text preview: Timothy Eng tje32@cornell.edu Dis 202,210,212 Week 10: Chapters 3.8,8.1,8.2 Math 2930: Discussion Worksheet 8 On Radius of Convergence: When facing c n +1 = ...c n , you can make use the equation by simply dividing by c n and to evaluating the limit using the following equation. = lim n  c n c n +1  but when facing recurrence relations with higher than first order, for example c n +2 = ...c n +1 + ...c n , you will be required to use Theorem 1 on p.519. 1) Solve y 00 + y 2 y = 0; y (0) = 1 , y (0) = 2 X n =2 n ( n 1) c n x n 2 + X n =1 nc n x n 1 2 X n =0 c n x n = 0 Shifting summations to start at n = 0 X n =0 ( n + 2)( n + 1) c n +2 x n + X n =0 ( n + 1) c n +1 x n 2 X n =0 c n x n = 0 Because 0 = n =0 0: ( n + 2)( n + 1) c n +2 + ( n + 1) c n +1 2 c n = 0 Solving for recurrence relation, the relation between new and old coefficients: c n +2 = 2 c n ( n + 1) c n +1 ( n + 2)( n + 1) To solve the equation, we know the solution is in the form of an infinite series of polynomials with unknown...
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 Fall '07
 TERRELL,R
 Math

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