Page 540 - Discrete Mathematics and Its Applications
P. 540

8.2 Solving Linear Recurrence Relations  519


                                                     Solution: The characteristic polynomial of this recurrence relation is

                                                         3
                                                               2
                                                        r − 6r + 11r − 6.
                                                                                                                  3    2
                                                     The characteristic roots are r = 1, r = 2, and r = 3, because r − 6r + 11r − 6 =
                                                     (r − 1)(r − 2)(r − 3). Hence, the solutions to this recurrence relation are of the form

                                                                  n
                                                                                  n
                                                                          n
                                                        a n = α 1 · 1 + α 2 · 2 + α 3 · 3 .
                                                     To find the constants α 1 , α 2 , and α 3 , use the initial conditions. This gives
                                                        a 0 = 2 = α 1 + α 2 + α 3 ,
                                                        a 1 = 5 = α 1 + α 2 · 2 + α 3 · 3,
                                                        a 2 = 15 = α 1 + α 2 · 4 + α 3 · 9.

                                                     When these three simultaneous equations are solved for α 1 , α 2 , and α 3 , we find that α 1 = 1,
                                                     α 2 =−1, and α 3 = 2. Hence, the unique solution to this recurrence relation and the given initial
                                                     conditions is the sequence {a n } with

                                                                         n
                                                                  n
                                                        a n = 1 − 2 + 2 · 3 .                                                       ▲
                                                        We now state the most general result about linear homogeneous recurrence relations with
                                                     constant coefficients, allowing the characteristic equation to have multiple roots. The key point
                                                     is that for each root r of the characteristic equation, the general solution has a summand of the
                                                              n
                                                     form P (n)r , where P (n) is a polynomial of degree m − 1, with m the multiplicity of this root.
                                                     We leave the proof of this result as Exercise 51.


                                     THEOREM 4        Let c 1 ,c 2 ,...,c k be real numbers. Suppose that the characteristic equation


                                                          k
                                                         r − c 1 r k−1  − ··· − c k = 0
                                                      has t distinct roots r 1 ,r 2 ,...,r t with multiplicities m 1 ,m 2 ,...,m t , respectively, so
                                                      that m i ≥ 1 for i = 1, 2,...,t and m 1 + m 2 + ··· + m t = k. Then a sequence {a n } is a
                                                      solution of the recurrence relation


                                                         a n = c 1 a n−1 + c 2 a n−2 + ··· + c k a n−k

                                                      if and only if

                                                         a n = (α 1,0 + α 1,1 n + ··· + α 1,m 1 −1 n m 1 −1 )r n
                                                                                               1
                                                               + (α 2,0 + α 2,1 n + ··· + α 2,m 2 −1 n m 2 −1 )r n
                                                                                                  2
                                                               + ··· + (α t,0 + α t,1 n + ··· + α t,m t −1 n m t −1 )r n
                                                                                                      t
                                                      for n = 0, 1, 2,..., where α i,j are constants for 1 ≤ i ≤ t and 0 ≤ j ≤ m i − 1.


                                                        Example 7 illustrates how Theorem 4 is used to find the general form of a solution of a
                                                     linear homogeneous recurrence relation when the characteristic equation has several repeated
                                                     roots.
   535   536   537   538   539   540   541   542   543   544   545