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Unformatted text preview: MATH 239 Assignment 4 This assignment is due at noon on Friday, June 26, 2009, in the drop boxes opposite the Tutorial Centre, MC 4067. 1. You are given that x 3 3 x + 2 = ( x 1) 2 ( x + 2) . (a) Find c n explicitly where c n satisfies the nonhomogeneous recurrence c n 3 c n 2 + 2 c n 3 = 2 n for n ≥ 3 , with initial conditions c = 1 ,c 1 = 7 ,c 2 = 5. (b) Find c n explicitly where c n satisfies the nonhomogeneous recurrence c n 3 c n 2 + 2 c n 3 = 6 for n ≥ 3 , with initial conditions c = 2 ,c 1 = 5 ,c 2 = 1. Solution. (a) Suppose that for n ≥ 3, a n = α · 2 n is a solution to the recurrence for some constant α . Then the recurrence gives α · 2 n 3 α · 2 n 2 + 2 α · 2 n 3 = ( α 3 4 α + 1 4 α ) · 2 n = 1 2 α · 2 n . Therefore, 1 2 α = 1, and α = 2. So a n = 2 · 2 n satisfies the recurrence, and b n = c n a n satisfies the homogeneous recurrence b n 3 b n 2 + 2 b n 3 = 0 . Since the roots of the characteristic polynomial are 1, 1, 2, we know that c n = 2 · 2 n + ( A + Bn )(1) n + C ( 2) n for some constants A,B,C . We can solve for A,B,C using the initial conditions. 2 + A + C = 1 4 + A + B 2 C = 7 8 + A + 2 B + 4 C = 5 . This gives A = 3 ,B = 4 ,C = 2. Therefore, an explicit formula for c n is c n = 2 · 2 n + (3 + 4 n ) 2( 2) n , n ≥ . (b) When one tries to find a solution to the recurrence by guessing a n = α or a n = αn for some constant α , the LHS of the recurrence becomes 0 and cannot equal the RHS. So suppose that a n = αn 2 for some constant α . Then the recurrence gives αn 2 3( α ( n 2) 2 ) + 2( α ( n 3) 2 ) = 6 α. 1 Therefore, α = 1, and a n = n 2 is a solution to the recurrence. Then (as in part (a)), c n has the form c n = n 2 + ( A + Bn )(1) n + C ( 2) n for some constants A,B,C . We can solve for....
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This note was uploaded on 08/09/2009 for the course MATH 239 taught by Professor M.pei during the Spring '09 term at Waterloo.
 Spring '09
 M.PEI
 Math, Combinatorics

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