Assignment 4 Soln

# Assignment 4 Soln - MATH 239 Assignment 4 This assignment...

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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 0 = - 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 0 = 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 0 . (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

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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 A, B, C using the initial conditions.
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