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# sol6 - MATH239 Assignment 6 Solutions 1(4 points Solve the...

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MATH239 Assignment 6 Solutions 1. (4 points) Solve the recurrence a n = 3 a n - 1 - 4 a n - 3 , for n 3, subject to the initial conditions a 0 = 0, a 1 = 8, and a 2 = 14. Solution. The characteristic polynomial is: x 3 - 3 x 2 + 4 = ( x + 1)( x - 2) 2 . So there exist α , β and γ such that a n = α ( - 1) n + ( βn + γ )2 n . We require: a 0 = 0: so 0 = α + γ , a 1 = 8: so 8 = - α + 2 β + 2 γ , and a 2 = 14: so 14 = α + 8 β + 4 γ . Solving these equations gives: α = 2, β = - 2, and γ = 7. Hence a n = 2( - 1) n + (7 - 2 n )2 n , for all n 0 . 2. (4 points) Let f : N 0 N 0 be a function. Consider the following two recurrence relations: a n - q 1 a n - 1 - · · · - q k a n - k = f ( n ) (1) b n - q 1 b n - 1 - · · · - q k b n - k = 0 . (2) (a) Show that, if the sequence { a n } n 0 satisfies recurrence (1) and the sequence { b n } n 0 satisfies recurrence (2), then { a n + b n } n 0 satisfies recurrence (1). Solution. Adding (1) and (2) gives: ( a n + b n ) - q 1 ( a n - 1 + b n - 1 ) - · · · - q k ( a n - k + b n - k ) = f ( n ) . Therefore, if { a n } n 0 satisfies (1) and { b n } n 0 satisfies (2), then { a n + b n } n 0 satisfies (1). (b) Show that, if the sequences { a n } n 0 and { c n } n 0 both satisfy recurrence (1), then { c n - a n } n 0 satisfies recurrence (2). Solution. For each n k , we have a n - q 1 a n - 1 - · · · - q k a n - k = f ( n ) , and (3) c n - q 1 c n - 1 - · · · - q k c n - k = f ( n ) . (4) Subtracting (3) from (4) gives ( c n - a n ) - q 1 ( c n - 1 - a n - 1 ) - · · · - q k ( c n - k - a n - k ) = 0 .

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sol6 - MATH239 Assignment 6 Solutions 1(4 points Solve the...

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