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SLU MATH135 recursion

SLU MATH135 recursion - Recurrence relations Discrete Math...

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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
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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 . . .
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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)
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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.
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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
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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
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Recurrence relations Discrete Math MATH 135 David Oury Homogeneous Nonhomogeneou Divide & conquer Linear homogeneous recurrence relations VI General solutions We know, if r is a root of the char. eqn. for a recurrence relation of degree d then r n is a solution to the recurrence relation. But there are more 1 If r is a root with multiplicity m
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