Recurrence Relations

Transcription

1 Recurrence Relations Simplest Case of General Solutions Ioan Despi University of New England September 13, 2013

2 Outline Ioan Despi AMTH140 2 of 1

3 Simplest Case of General Solutions Ioan Despi AMTH140 3 of 1 c m a n+m + c m 1 a n+m c 1 a n+1 + c 0 a n = g(n), n 0 (* * *) m c k a n+k = g(n) Definition k=0 A solution a n of a recurrence relation (* * *) is said to be a general solution, typically containing some arbitrary constants in the solution expression for a n, if any particular solution of the recurrence relation (* * *) can be obtained as a special case of the general solution.

4 Example Ioan Despi AMTH140 4 of 1 Example Verify that a n = A2 n + B3 n for arbitrary constants A and B solves the recurrence relation a n+2 5a n+1 + 6a n = 0. Likewise we can show that a n = 5 2 n is also a (particular) solution. Obviously the particular solution a n = 5 2 n is included in the more general solution expression a n = A2 n + B3 n if we choose A = 5 and B = 0. In fact one can show that all the solutions of a n+2 5a n+1 + 6a n = 0 are embraced by the solution expression a n = A2 n + B3 n, which is hence the general solution.

5 Simplest Case of General Solutions Ioan Despi AMTH140 5 of 1 An alternative way to determine if a solution is the general solution of an m-th order linear, constant coefficient recurrence relation is to see if the solution expression contains exactly m independent arbitrary constants. The word independent here roughly means that none of the arbitrary constants can be made redundant. From this perspective, we can also conclude that a n = A2 n + B3 n is the general solution of the second order recurrence relation a n+2 5a n+1 + 6a n = 0 because the solution contains A and B as the two independent arbitrary constants.

6 Recall Ioan Despi AMTH140 6 of 1 A polynomial f(λ) has λ 0 as one of its roots precisely when f(λ 0 ) = 0. For example, if f(λ) = λ 2 5λ + 6, then λ 0 = 3 is one of its roots because f(λ 0) = f(3) = = 0. In other words, 3 is a root of the equation λ 2 5λ + 6 = 0. In general, a polynomial equation of order n will have exactly n roots, some of which may be distinct while others may be repeated, some of which may be real while others may be complex numbers. A second order polynomial equation has two roots, λ 1 and λ 2, given by aλ 2 + bλ + c = 0, a 0 λ 1 = b + b 2 4ac 2a, λ 2 = b b 2 4ac 2a If λ 1 = λ 2, then the two roots are the repeated roots and λ 1, which is the same as λ 2, is a root of multiplicity 2. If b 2 4ac 0 the two roots are real numbers, else if b 2 4ac < 0 the two roots are complex numbers.,

7 Recall Ioan Despi AMTH140 7 of 1 For a polynomial equation f(λ) = 0, a value λ 0 is said to be a root of multiplicity m if f(λ) can be written as f(λ) = (λ λ 0 ) m g(λ) such that g(λ) is again a polynomial with g(λ 0 ) 0. For example, the polynomial equation λ 2 6λ + 9 = 0 has a root 3 of multiplicity 2. This is because λ 2 6λ + 9 = (λ 3) 2 1, in which f(λ) = λ 2 6λ + 9, g(λ) = 1, λ 0 = 3 and m = 2. If we solve the equation λ 2 6λ + 9 = 0 through the use of the above root formula for λ 1 and λ 2, we see that both λ 1 and λ 2 are equal to the same value 3. This also explains why 3 is a root of multiplicity 2 for the equation λ 2 6λ + 9 = 0

8 Recall Ioan Despi AMTH140 8 of 1 A polynomial equation f(λ) = 0 has a root λ 0, i.e., f(λ 0 ) = 0, if and only if f(λ) = (λ λ 0 ) g(λ) for another nonzero polynomial g(λ). If λ 0 is furthermore a root of multiplicity m > 1 of f(λ) = 0, then λ 0 must be a root of multiplicity m 1 of g(λ) = 0. A root of multiplicity 1 is called a simple root. A simple root is thus not a repeated root.

9 Simplest Case of General Solutions Ioan Despi AMTH140 9 of 1 Theorem If the characteristic equation of an m th order homogeneous, linear, constant coefficient recurrence relation c m a n+m + c m 1 a n+m c 1 a n+1 + c 0 a n = 0, c m c 0 0, n 0 has m distinct roots λ 1, λ 2,, λ m, then a n = A 1 λ n 1 + A 2 λ n A m λ n m with arbitrary constants A 1,, A m is the general solution of the recurrence relation.

10 Ioan Despi AMTH of 1 Simplest Case of General Solutions Proof. m First we show a n = A i λ n i is a solution. For this purpose we substitute the i=1 expression for a n into the recurrence relation and obtain c m a n+m + + c 1 a n+1 + c 0 a 0 = = c m (A 1 λ n+m A m λ n+m m ) + + c 1 (A 1 λ n A m λ n+1 m ) + c 0 (A 1 λ n A m λ n m) = A 1 (c m λ n+m c 1 λ n c 0 λ n 1 ) + A 2 (c m λ n+m c 1 λ n c 0 λ n 2 ) + + A m (c m λ n+m m + + c 1 λ n+1 m + c 0 λ n m) m = A i λ n i (c m λ m i + + c 1 λ i + c 0 ) = 0. i=1 m Hence a n = A i λ n i is a solution. Since the solution involves m arbitrary i=1 constants A 1,, A m, it is in fact a general solution.

11 Simplest Case of General Solutions Ioan Despi AMTH of 1 Proof. Alternatively, we can also argue that for any initial values of a 0,, a m 1, since the linear system of m equations m A i λ k i = a k, k = 0, 1,, m 1 i=1 m has a (unique) solution for A 1,, A m, the expression a n = A i λ n i is indeed a general solution. If the roots λ 1,, λ m are not distinct, i.e., there are repeated roots, then m a n = A i λ n i is still a solution but is not a general solution. The general i=1 solution for such cases will be dealt with in the next section. i=1

12 Corollary (1) Simplest Case of General Solutions Given two solutions {x n } and {y n } of an m th order homogeneous, linear, constant coefficient recurrence relation, any linear combination of them z n = Ax n + By n where A, B are constants, is also a solution of the same recurrence relation. Proof. Given c m a n+m + c m 1 a n+m c 1 a n+1 + c 0 a n = 0, c m c 0 0, n 0, if {x n } and {y n } are solutions, then c m x n+m + c m 1 x n+m c 1 x n+1 + c 0 x n = 0 A c m y n+m + c m 1 y n+m c 1 y n+1 + c 0 y n = 0 B If we multiply the first relation by A, the second by B and then sum them up, we get c m [Ax n+m + By n+m ] + c m 1 [Ax n+m 1 + By n+m 1 ] + + c 1 [Ax n+1 + By n+1 ] + c 0 [Ax n + By n ] = 0 that is c m z n+m + c m 1 z n+m c 1 z n+1 + c 0 z n = 0 which shows that z n is a solution for the same recurrence relation. Ioan Despi AMTH of 1

13 Simplest Case of General Solutions Ioan Despi AMTH of 1 Corollary (2) If the recurrence relation is a non-homogeneous one, then the difference of any two solutions is a solution of the homogeneous version of the recurrence relation. Proof. Given c m a n+m + c m 1 a n+m c 1 a n+1 + c 0 a n = g(n), c m c 0 0, n 0, if {x n } and {y n } are solutions, then they satisfy c m x n+m + c m 1 x n+m c 1 x n+1 + c 0 x n = g(n) and their difference gives c m y n+m + c m 1 y n+m c 1 y n+1 + c 0 y n = g(n) c m [x n+m y n+m ] + c m 1 [x n+m 1 y n+m 1 ] + + +c 1 [x n+1 y n+1 ] + c 0 [x n y n ] = g(n) g(n) = 0 c m z n+m + c m 1 z n+m c 1 z n+1 + c 0 z n = 0

14 Example Ioan Despi AMTH of 1 Example Find the general solution of a n+2 5a n+1 + 6a n = 0, n 0. Give also the particular solution satisfying a 0 = 0 and a 1 = 1. Solution. Since the associated characteristic equation is λ 2 5λ + 6 = 0 and has 2 distinct roots λ 1 = 2 and λ 2 = 3, the general solution for the recurrence relation, according to the theorem earlier on, is a n = A 1 2 n + A 2 3 n, n 0, where A 1 and A 2 are 2 arbitrary constants. To find the particular solution, we need to determine A 1 and A 2 explicitly using the initial conditions a 0 = 0 and a 1 = 1. Hence we require a 0 = A A = 0 a 1 = A A = 1 i.e., A 1 + A 2 = 0 2A 1 + 3A 2 = 1 which has the solution A 1 = 1, A 2 = 1 and, thus, the particular solution is a n = 2 n + 3 n for n 0.

15 Note Ioan Despi AMTH of 1 It is often a good practice to check your answers, even if just partially. To check the above particular solution, we see a 0 = = 0 a 1 = = 1 a n+2 5a n+1 + 6a n = ( 2 n n+2 ) 5( 2 n n+1 ) + 6( 2 n + 3 n ) = 2 n ( ) +3 n ( ) = i.e., all conditions are satisfied.