tut06s - THE UNIVERSITY OF SYDNEY PURE MATHEMATICS Linear...

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Unformatted text preview: THE UNIVERSITY OF SYDNEY PURE MATHEMATICS Linear Mathematics 2012 Tutorial 6 — Solutions 1. Define a sequence of numbers { u n | n ≥ } by u = 0 , u 1 = 1 , u 2 = 2 and for all n ≥ , u n +3 = u n +2 + 1 4 u n +1- 1 4 u n . a ) Express the recurrence relation for u n in matrix form. b ) Find a formula for u n in terms of n . c ) Find lim n →∞ u n . Solution a ) Write u n = parenleftBig u n +2 u n +1 u n parenrightBig . Then u n +1 = parenleftBig u n +3 u n +2 u n +1 parenrightBig = parenleftbigg u n +2 + 1 4 u n +1 − 1 4 u n u n +2 u n +1 parenrightbigg = parenleftbigg 1 1 4 − 1 4 1 0 0 1 parenrightbigg parenleftBig u n +2 u n +1 u n parenrightBig . That is, u n +1 = A u n , where A = parenleftbigg 1 1 4 − 1 4 1 0 0 1 parenrightbigg . b ) If { v 1 , v 2 , v 3 } is a basis of R 3 consisting of eigenvectors for A , with A v i = λ i v i , then we know from lectures that the solution to the recurrence relation is of the form u n = r 1 λ n 1 v 1 + r 2 λ n 2 v 2 + r 3 λ n 3 v 3 , for some r 1 , r 2 , r 3 ∈ R . So we find the eigenvalues of A : If | A- λI | = λ 2 (1- λ )- 1 4 (1- λ ) = 0 , then λ = 1 , ± 1 2 , so the eigenvalues are 1 , 1 2 and- 1 2 . Since there are 3 distinct eigenvalues we know that there is a basis of R 3 consisting of eigenvectors of A . The eigenvectors corresponding to the eigenvalues 1 , 1 2 and- 1 2 are parenleftBig 1 1 1 parenrightBig , parenleftBigg 1 4 1 2 1 parenrightBigg and parenleftBigg 1 4 − 1 2 1 parenrightBigg respectively. So we have u n = r 1 (1) n parenleftBig 1 1 1 parenrightBig + r 2 ( 1 2 ) n parenleftBigg 1 4 1 2 1 parenrightBigg + r 3 (- 1 2 ) n parenleftBigg 1 4 − 1 2 1 parenrightBigg . Now, u = parenleftBig u 2 u 1 u parenrightBig = parenleftBig 2 1 parenrightBig , so putting n = 0 we have parenleftBig 2 1 parenrightBig = r 1 parenleftBig 1 1 1 parenrightBig + r 2 parenleftBigg 1 4 1 2 1 parenrightBigg + r 3 parenleftBigg 1 4 − 1 2 1 parenrightBigg = parenleftBigg 1 1 4 1 4 1 1 2 − 1 2 1 1 1 parenrightBigg parenleftBig r 1 r 2 r 3 parenrightBig . Math 2061: Tutorial 6 — Solutions A.M. 5/1/2012 Linear Mathematics Tutorial 6 — Solutions Page 2 Solving this system of equations gives r 1 = 8 3 , r 2 =- 3 and r 3 = 1 3 , and so u n = 8 3 parenleftBig 1 1 1 parenrightBig- 3 2 n parenleftBigg 1 4 1 2 1 parenrightBigg + ( − 1) n 3 · 2 n parenleftBigg 1 4 − 1 2 1 parenrightBigg . By definition, u n = parenleftBig u n +2 u n +1 u n parenrightBig , so this means that u n = 8 3- 3 2 n + ( − 1) n 3 · 2 n . c ) lim n →∞ u n = lim n →∞ parenleftBig 8 3- 3 2 n + ( − 1) n 3 · 2 n parenrightBig = 8 3 . 2. Let L = b 1 b 2 b 3 s 1 s 2 be the Leslie matrix for an animal population with 3 age groups....
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tut06s - THE UNIVERSITY OF SYDNEY PURE MATHEMATICS Linear...

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