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

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Unformatted text preview: 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 , a 1 , a 2 , . . . is a sequence satisfying this recurrence and a = 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 2- 4 +3 = ( - 1)( - 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 ( - 2)( + 1) = 2- - 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. 2 Question 2 (11 points) Solve the following recurrences: (a) (3 points) a n = 5 a n 1 , a = 1. Answer : a n = 5 n . (b) (4 points) a n +1- 4- 5 a n = 0 (find the general solution)....
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This note was uploaded on 07/15/2010 for the course MACM 201 taught by Professor Marnimishna during the Spring '09 term at Simon Fraser.

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

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