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Unformatted text preview: MATH 681 Notes Combinatorics and Graph Theory I 1 Recurrence relations, continued continued Linear homogeneous recurrences are only one of several possible ways to describe a sequence as a recurrence. Here are several other situations which may arise. 1.1 Linear nonhomogeneous recurrence relations A linear nonhomogeneous recurrence relation is one which has a linear homogeneous form as well as a nonhomogeneous term. here’s an example: Question 1: How many strings of n numbers from the set { , 1 , 2 , 3 , 4 } are there so that there is at least one “1” and the first “1” occurs before any “0”? Answer 1: Let us call the above number a n , and a string satisfying the above description “valid”. To produce a valid string of length n , we can consider every possible first number in the string as an individual case. A string beginning with “0” is necessarily invalid. If a string begins with “1”, everything that could follow that is valid. Thus, of the 5 n 1 strings beginning with “1”, all are valid. If a string begins with a “2”, “3”, or “4”, then it is valid if and only if the remaining n 1 terms form a valid string, which could happen in a n 1 ways. Thus, a n = 3 a n 1 + 5 n 1 . This is quite similar to a firstorder linear homogeneous differential equation, except it has that inconvenient 5 n 1 term. Unfortunately, we don’t have the tools to solve that, yet! In addition, note the initial condition a = 0 , since the string of length zero contains no “1”s and is thus invalid. We could, if we choose, work out a few small values from this: a 1 = 1 , a 2 = 8 , a 3 = 49 , and so forth. We’ll define the class of problems to which this belongs, and discuss solution techniques: Definition 1. A sequence { a n } is given by a linear nonhomogeneous recurrence relation of order k if a n = c 1 a n 1 + c 2 a n 2 + c 3 a n 3 + ··· + c k a n k + p ( n ) for all n ≥ k . The recurrence relation b n = c 1 b n 1 + c 2 b n 2 + c 3 b n 3 + ··· + c k b n k is referred to as the associated linear homogeneous recurrence relation One result is as easy to show for LNRRs as for LHRRs; the following can be proven as a very slight variation of the similar proof for LHRRs. Proposition 1. A sequence is uniquely determined by an LNRR of order k and the initial values a ,a 1 ,a 2 ,...,a k 1 . However, there is one very important difference between LNRRs and LHRRs: linear combinations of LNRRsatisfying sequences do no, in general, satisfy the LNRR. Howver, we do have the result: Proposition 2. If { a n } satisfies an LNRR, and { b n } satisfies the associated LHRR, then { a n + b n } satisfies the LNRR. Proof. We know that a n = c 1 a n 1 + c 2 a n 2 + c 3 a n 3 + ··· + c k a n k + p ( n ) Page 1 of 5 October 1, 2009 MATH 681 Notes Combinatorics and Graph Theory I and b n = c 1 b n 1 + c 2 b n 2 + c 3 b n 3 + ··· + c k b n k Adding these two equations will give ( a n + b n ) = c 1 ( a n 1 + b n 1 ) +...
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This note was uploaded on 01/12/2012 for the course MATH 681 taught by Professor Wildstrom during the Fall '09 term at University of Louisville.
 Fall '09
 WILDSTROM
 Combinatorics, Graph Theory

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