Linear Algebra Solutions 13

Linear Algebra Solutions 13 - Now by Question 7, with A =...

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Now by Question 7, with A = 1 r 1 1 , A n = k n A - λ 1 λ 2 k n - 1 I 2 = k n A - (1 - r ) k n - 1 I 2 , where λ 1 = 1 + r and λ 2 = 1 - r are the roots of the polynomial x 2 - 2 x + (1 - r ) and k n = λ n 1 - λ n 2 2 r . Hence x n y n = ( k n A - (1 - r ) k n - 1 I 2 ) a b = µ• k n k n r k n k n - (1 - r ) k n - 1 0 0 (1 - r ) k n - 1 ‚¶• a b = k n - (1 - r ) k n - 1 k n r k n k n - (1 - r ) k n - 1 ‚• a b = a ( k n - (1 - r ) k n - 1 ) + bk n r ak n + b ( k n - (1 - r ) k n - 1 ) . Hence, in view of the fact that k n k n - 1 = λ n 1 - λ n 2 λ n - 1 1 - λ n - 1 2 = λ n 1 (1 -{ λ 2 λ 1 } n ) λ n - 1 1 (1 -{ λ 2 λ 1 } n - 1 ) λ 1 , as n → ∞ , we have x n y n = a ( k n - (1 - r ) k n - 1 ) + bk n r ak n + b ( k n - (1 - r ) k n - 1 ) = a ( k n k n - 1 - (1 - r )) + b k n k n - 1 r a k n
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This note was uploaded on 12/19/2011 for the course MAS 3105 taught by Professor Dreibelbis during the Fall '10 term at UNF.

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