Advanced High-School Mathematics

98 CHAPTER 2 Discrete Mathematics

0 = u 0 =A 120 +A 2 (−1)^0 =A 1 +A 2

1 = u 1 =A 121 +A 2 (−1)^1 = 2A 1 −A 2

all of which implies thatA 1 =^13 , A 2 =−^13 .The particular solution

of the above linear difference equation is therefore

un =

2 n

3

−

(−1)n

3

, n= 0, 1 , 2 ,...

Quadratic—repeated factor over the reals.

Here we assume that our polynomialC(x) is quadratic with a mul-

tiple factor: C(x) =x^2 − 2 kx−k^2 = (x−k)^2 , wherek∈R. As in

the above case, one solution has the formun=Aknn= 0, 1 , 2 ,....

However, a second solution has the formun=Bnkn, n= 0, 1 , 2 ,....

We check this by direct substitution:

un+2− 2 kun+1+k^2 un = B(n+ 2)kn+2− 2 kB(n+ 1)kn+1+nBk^2 kn

= Bkn+2((n+ 2)−2(n+ 1) +n) = 0.

Likewise, one than then show that the sum of these solutions is a

solution of the second-order homogeneous difference equation:

un = Akn+Bnkn, n= 0, 1 , 2 ,....

Quadratic—irreducible. In this case we consider the second-order lin-

ear homogeneous difference equation whose characteristic equation

is irreducible (over the reals). Thus the discriminant of the charac-

teristic polynomial is negative (and has complex conjugate zeros).

A simple example of such would be the difference equation