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Unformatted text preview: Recurrence relations Discrete Math MATH 135 David Oury Homogeneous Nonhomogeneou Divide & conquer Recurrence relations Discrete Math MATH 135 David Oury December 8, 2011 Chapter 8, Sections 1, 2, 3 of Discrete Mathematics and its Applications, by Rosen Recurrence relations Discrete Math MATH 135 David Oury Homogeneous Nonhomogeneou Divide & conquer Linear homogeneous recurrence relations I Definition A linear homogeneous recurrence relation of order d is an equation of the form a n = c 1 a n 1 + c 2 a n 2 + ... + c d a n d where c 1 ,...,c d are constants and c d 6 = 0 . A solution to the recurrence relation is an infinite sequence { a n } n ∈ N which we also write as { a n } or as a n . . . Recurrence relations Discrete Math MATH 135 David Oury Homogeneous Nonhomogeneou Divide & conquer Linear homogeneous recurrence relations II Examples A ( n ) = 1 . 06 A ( n 1) has order 1 B ( n ) = 3 B ( n 4) has order 4 F ( n ) = F ( n 1) + F ( n 2) has order 2 Nonexamples Q ( n ) = Q ( n 1) + 1 R ( n ) = R ( n 1) 2 S ( n ) = n S ( n 1) Recurrence relations Discrete Math MATH 135 David Oury Homogeneous Nonhomogeneou Divide & conquer Linear homogeneous recurrence relations III Solving linear homogeneous recurrence relations Suppose that { a n = r n } n ∈ N is a solution for a n = c 1 a n 1 + c 2 a n 2 + ... + c d a n d Then r n = c 1 r n 1 + c 2 r n 2 + ... + c d r n d and equivalently r n c 1 r n 1 c 2 r n 2 ... c d r n d = 0 divide by r n d to get r d c 1 r d 1 c 2 r d 2 ... c d = 0 this last equation is called the characteristic equation it has a solution (for r ) if and only if { a n = r n } is a solution of the recurrence relation solutions to the characteristic equation are called characteristic roots and are often denoted r 1 ,r 2 , etc. Recurrence relations Discrete Math MATH 135 David Oury Homogeneous Nonhomogeneou Divide & conquer Linear homogeneous recurrence relations IV Example: P ( n ) = 1 . 06 P ( n 1) The relation has degree 1 The characteristic equation is r 1 . 06 = 0 The only characteristic root is r = 1 . 06 So { a n = 1 . 06 n } is a solution to the recurrence relation Example: F ( n ) = F ( n 1) + F ( n 2) The relation has degree 2 The char. eqn. is r 2 r 1 = 0 The char. roots are 1 ± √ 5 2 So { a n = 1+ √ 5 2 n } and { a n = 1+ √ 5 2 n } are solutions to the recurrence relation Recurrence relations Discrete Math MATH 135 David Oury Homogeneous Nonhomogeneou Divide & conquer Linear homogeneous recurrence relations V Example: B ( n ) = 6 B ( n 1) 9 B ( n 2) The relation has degree 2 The char. eqn. is r 2 6 r + 9 = 0 The char. root 3 has multiplicity 2 So { a n = 3 n } is a solution to the recurrence relation Recurrence relations Discrete Math MATH 135 David Oury Homogeneous Nonhomogeneou Divide & conquer Linear homogeneous recurrence relations VI...
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This note was uploaded on 02/11/2012 for the course MATH 135 taught by Professor Dr.oury during the Spring '12 term at MO St. Louis.
 Spring '12
 Dr.Oury
 Math

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