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Unformatted text preview: 1 Question 1 (10 points) (a) (2 points) Give an example of a linear secondorder 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 nonhomogeneous 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.
 Spring '09
 MarniMishna

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