Solution of Linear Nonhomogeneous Recurrence Relations Ioan Despi despi@turing.une.edu.au University of New England September 25, 2013
Outline 1 A Case for Thought 2 Method of Undetermined Coefficients Notes Examples Ioan Despi – AMTH140 2 of 16
A Case for Thought We already mentioned that finding a particular solution for a nonhomogeneous problem can be more involved than those exemplified in the previous lecture. Let us first highlight our point with the following example. Example Solve a n +2 + a n +1 − 6 a n = 2 n for n ≥ 0. Ioan Despi – AMTH140 3 of 16
A Case for Thought We already mentioned that finding a particular solution for a nonhomogeneous problem can be more involved than those exemplified in the previous lecture. Let us first highlight our point with the following example. Example Solve a n +2 + a n +1 − 6 a n = 2 n for n ≥ 0. Solution. First we observe that the homogeneous problem u n +2 + u n +1 − 6 u n = 0 has the general solution u n = A 2 n + B ( − 3) n for n ≥ 0 because the associated characteristic equation λ 2 + λ − 6 = 0 has 2 distinct roots λ 1 = 2 and λ 2 = − 3. Ioan Despi – AMTH140 3 of 16
A Case for Thought We already mentioned that finding a particular solution for a nonhomogeneous problem can be more involved than those exemplified in the previous lecture. Let us first highlight our point with the following example. Example Solve a n +2 + a n +1 − 6 a n = 2 n for n ≥ 0. Solution. First we observe that the homogeneous problem u n +2 + u n +1 − 6 u n = 0 has the general solution u n = A 2 n + B ( − 3) n for n ≥ 0 because the associated characteristic equation λ 2 + λ − 6 = 0 has 2 distinct roots λ 1 = 2 and λ 2 = − 3. Since the r.h.s. of the nonhomogeneous recurrence relation is 2 n , if we formally follow the strategy in the previous lecture, we would try v n = C 2 n for a particular solution. Ioan Despi – AMTH140 3 of 16
A Case for Thought We already mentioned that finding a particular solution for a nonhomogeneous problem can be more involved than those exemplified in the previous lecture. Let us first highlight our point with the following example. Example Solve a n +2 + a n +1 − 6 a n = 2 n for n ≥ 0. Solution. First we observe that the homogeneous problem u n +2 + u n +1 − 6 u n = 0 has the general solution u n = A 2 n + B ( − 3) n for n ≥ 0 because the associated characteristic equation λ 2 + λ − 6 = 0 has 2 distinct roots λ 1 = 2 and λ 2 = − 3. Since the r.h.s. of the nonhomogeneous recurrence relation is 2 n , if we formally follow the strategy in the previous lecture, we would try v n = C 2 n for a particular solution. But there is a difficulty: C 2 n fits into the format of u n which is a solution of the homogeneous problem. Ioan Despi – AMTH140 3 of 16
A Case for Thought We already mentioned that finding a particular solution for a nonhomogeneous problem can be more involved than those exemplified in the previous lecture. Let us first highlight our point with the following example. Example Solve a n +2 + a n +1 − 6 a n = 2 n for n ≥ 0. Solution. First we observe that the homogeneous problem u n +2 + u n +1 − 6 u n = 0 has the general solution u n = A 2 n + B ( − 3) n for n ≥ 0 because the associated characteristic equation λ 2 + λ − 6 = 0 has 2 distinct roots λ 1 = 2 and λ 2 = − 3. Since the r.h.s. of the nonhomogeneous recurrence relation is 2 n , if we formally follow the strategy in the previous lecture, we would try v n = C 2 n for a particular solution. But there is a difficulty: C 2 n fits into the format of u n which is a solution of the homogeneous problem. ◮ In other words, it can’t be a particular solution of the nonhomogeneous problem. Ioan Despi – AMTH140 3 of 16
A Case for Thought This is really because 2 happens to be one of the two roots λ 1 and λ 2 . However, we suspect that a particular solution would still have to have 2 n as a factor, so we try v n = Cn 2 n . Substituting it to v n +2 + v n +1 − 6 v n = 2 n , we obtain C ( n + 2)2 n +2 + C ( n + 1)2 n +1 − 6 Cn 2 n = 2 n , i.e., 10 C 2 n = 2 n or C = 1 10 . Hence a particular solution is v n = n 102 n and the general solution of our nonhomogeneous recurrence relation is a n = A 2 n + B ( − 3) n + n 102 n , n ≥ 0 . In general, it is important that a correct form, often termed ansatz in physics, for a particular solution is used before we fix up the unknown constants in the solution ansatz. The following theorem can help. Ioan Despi – AMTH140 4 of 16
A Case for Thought Theorem (The Rational Roots Test.) Let P ( x ) = a n x n + a n − 1 x n − 1 + · · · + a 1 x + a 0 be a polynomial with real coefficients and a n ̸ = 0 . If P ( x ) has rational roos, they are of the form ± p q where p | a 0 and q | a n . The choice of the form of a particular solution, covering the cases in this current lecture as well as the previous one, can be summarized below. Ioan Despi – AMTH140 5 of 16
Method of Undetermined Coefficients Consider a linear, constant coefficient recurrence relation of the form c m a n + m + · · · + c 1 a n +1 + c 0 a n = g ( n ) , c 0 c m ̸ = 0 , n ≥ 0 . ( * ) Ioan Despi – AMTH140 6 of 16
Method of Undetermined Coefficients Consider a linear, constant coefficient recurrence relation of the form c m a n + m + · · · + c 1 a n +1 + c 0 a n = g ( n ) , c 0 c m ̸ = 0 , n ≥ 0 . ( * ) Suppose function g ( n ), the nonhomogeneous part of the recurrence relation, is of the following form g ( n ) = µ n ( b 0 + b 1 n + · · · + b k n k ) , ( ** ) where k ∈ N , µ, b 0 , · · · , b k are constants, and µ is a root of multiplicity M of the associated characteristic equation c m λ m + · · · + c 1 λ + c 0 = 0 . Then a particular solution v n of ( * ) should be sought in the form k [︃ ]︃ µ n n M = µ n (︁ B 0 + B 1 n + · · · + B k − 1 n k − 1 + B k n k )︁ ∑︂ B i n i n M v n = i =0 ( * * * ) where constants B 0 , · · · , B k are to be determined from the requirement that a n = v n should satisfy the recurrence relation ( * ). Ioan Despi – AMTH140 6 of 16
Method of Undetermined Coefficients B 0 + B 1 n + · · · + B k − 1 n k − 1 + B k n k )︁ v n = µ n (︁ n M ( * * * ) Obviously, the v n in ( * * * ) is composed of two parts: Ioan Despi – AMTH140 7 of 16
Method of Undetermined Coefficients B 0 + B 1 n + · · · + B k − 1 n k − 1 + B k n k )︁ v n = µ n (︁ n M ( * * * ) Obviously, the v n in ( * * * ) is composed of two parts: µ n ( B 0 + B 1 n + · · · + B k n k ) (which is of the same form of g ( n ) in ( ∗∗ )) ◮ Ioan Despi – AMTH140 7 of 16
Method of Undetermined Coefficients B 0 + B 1 n + · · · + B k − 1 n k − 1 + B k n k )︁ v n = µ n (︁ n M ( * * * ) Obviously, the v n in ( * * * ) is composed of two parts: µ n ( B 0 + B 1 n + · · · + B k n k ) (which is of the same form of g ( n ) in ( ∗∗ )) ◮ ◮ n M , which is a necessary adjustment for the case when µ , appearing in g ( n ) in ( ∗∗ ), is also a root (of multiplicity M ) of the characteristic equation of the associated homogeneous recurrence relation. Ioan Despi – AMTH140 7 of 16
Method of Undetermined Coefficients B 0 + B 1 n + · · · + B k − 1 n k − 1 + B k n k )︁ v n = µ n (︁ n M ( * * * ) Obviously, the v n in ( * * * ) is composed of two parts: µ n ( B 0 + B 1 n + · · · + B k n k ) (which is of the same form of g ( n ) in ( ∗∗ )) ◮ ◮ n M , which is a necessary adjustment for the case when µ , appearing in g ( n ) in ( ∗∗ ), is also a root (of multiplicity M ) of the characteristic equation of the associated homogeneous recurrence relation. If µ is not a root of the characteristic equation, then just choose M = 0, implying alternatively that µ is a ”root” of 0 multiplicity. Ioan Despi – AMTH140 7 of 16
Notes � v n = µ n (︂∑︁ k + M j =0 A j n j )︂ 1 We can also try ˜ . Ioan Despi – AMTH140 8 of 16
Notes � v n = µ n (︂∑︁ k + M j =0 A j n j )︂ 1 We can also try ˜ . j = M A j n j + µ n ∑︁ M − 1 v n as µ n ∑︁ k + M ◮ If we rewrite ˜ j =0 A j n j , then Ioan Despi – AMTH140 8 of 16
Notes � v n = µ n (︂∑︁ k + M j =0 A j n j )︂ 1 We can also try ˜ . j = M A j n j + µ n ∑︁ M − 1 v n as µ n ∑︁ k + M ◮ If we rewrite ˜ j =0 A j n j , then ⋆ the first part is essentially ( * * * ), while Ioan Despi – AMTH140 8 of 16
Notes � v n = µ n (︂∑︁ k + M j =0 A j n j )︂ 1 We can also try ˜ . j = M A j n j + µ n ∑︁ M − 1 v n as µ n ∑︁ k + M ◮ If we rewrite ˜ j =0 A j n j , then ⋆ the first part is essentially ( * * * ), while ⋆ the second part is just a solution of the homogeneous problem. Ioan Despi – AMTH140 8 of 16
Notes � v n = µ n (︂∑︁ k + M j =0 A j n j )︂ 1 We can also try ˜ . j = M A j n j + µ n ∑︁ M − 1 v n as µ n ∑︁ k + M ◮ If we rewrite ˜ j =0 A j n j , then ⋆ the first part is essentially ( * * * ), while ⋆ the second part is just a solution of the homogeneous problem. ◮ It is however obvious that v n in ( ∗ ∗ ∗ ) is simpler than ˜ v n . Ioan Despi – AMTH140 8 of 16
Notes 3 We briefly hint why v n is chosen in the form of ( * * * ). � Ioan Despi – AMTH140 9 of 16
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