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mt2Soln

# mt2Soln - 1 Question 1(10 points(a(2 points Give an example...

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1 Question 1 (10 points) (a) (2 points) Give an example of a linear second-order homogeneous re- currence with constant coefficients, different from any other such re- currences appearing in this exam. Answer : Many possible answers, one that keeps things simple is a n +2 = 4 a n +1 - 3 a n . (b) (2 points) If a 0 , a 1 , a 2 , . . . is a sequence satisfying this recurrence and a 0 = a 1 = 1, what is the value of a 5 ? Answer : a 2 = 4 · 1 - 3 · 1 = 1, and similarly a 3 = a 4 = a 5 = 1. (c) (3 points) What is the general solution to the recurrence you provided? Answer : The characteristic polynomial is λ 2 - 4 λ +3 = ( λ - 1)( λ - 3), thus a n = c 1 + c 2 3 n . (d) (3 points) Find a non-homogeneous linear recurrence whose general solution is a n = c 1 2 n + c 2 ( - 1) n - 5. Answer : The characteristic polynomial should have roots 2 and - 1, hence it should be ( λ - 2)( λ + 1) = λ 2 - λ - 2 and the homogeneous part of the recurrence should be a n +2 - a n +1 - 2 a n = 0. Therefore, we look for a recurrence of form a n +2 - a n +1 - 2 a n = f ( n ). Furthermore, a n = - 5 should be a solution, thus f ( n ) = ( - 5) - ( - 5) - 2( - 5) = 10. The reccurence is a n +2 - a n +1 - 2 a n = 10.

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2 Question 2 (11 points) Solve the following recurrences: (a) (3 points) a n = 5 a n 1 , a 0 = 1. Answer : a n = 5 n . (b) (4 points) a n +1 - 4 - 5 a n = 0 (find the general solution). Answer : We guess a solution for the non-homogeneous recurrence as a constant, a n = k . We get k - 4 - 5 k = 0, thus k = - 1. The general solution is a n = c 5 n - 1.
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