Example 9.9.
Find all solutions to the advancement operator equation
\begin{equation}
(A^2+A-6)f = 0.\tag{9.4.1}
\end{equation}
Solution.
Before focusing on finding all solutions as we’ve been asked to do, let’s just try to find some solution. We start by noticing that here \(p(A) = A^2+A-6 =
(A+3)(A-2)\text{.}\) With \(p(A)\) factored like this, we realize that we’ve already solved part of this problem in Example 9.8! In that example, the polynomial of \(A\) we encountered was (while not explicitly stated as such there) \(A-2\text{.}\) The solutions to \((A-2)f_1=0\) are of the form \(f_1(n) = c_12^n\text{.}\) What happens if we try such a function here? We have
\begin{equation*}
(A+3)(A-2)f_1(n) = (A+3)0 = 0,
\end{equation*}
so that \(f_1\) is a solution to our given advancement operator equation. Of course, it can’t be all of them. However, it’s not hard to see now that \((A+3)f_2 = 0\) has as a solution \(f_2(n)
= c_2(-3)^n\) by the same reasoning that we used in Example 9.8. Since \((A+3)(A-2) = (A-2)(A+3)\text{,}\) we see right away that \(f_2\) is also a solution of (9.4.1).
Now we’ve got two infinite families of solutions to (9.4.1). Do they give us all the solutions? It turns out that by combining them, they do in fact give all of the solutions. Consider what happens if we take \(f(n) = c_1 2^n +
c_2 (-3)^n\) and apply \(p(A)\) to it. We have
\begin{align*}
(A+3)(A-2)f(n) \amp = (A+3)(c_1 2^{n+1} + c_2 (-3)^{n+1} - 2(c_12^n +
c_2(-3)^n))\\
\amp = (A+3)(-5c_2(-3)^{n})\\
\amp = -5c_2(-3)^{n+1}-15c_2(-3)^n\\
\amp = 15c_2(-3)^n - 15c_2(-3)^n\\
\amp =0.
\end{align*}
It’s not all that hard to see that since \(f\) gives a two-parameter family of solutions to (9.4.1), it gives us all the solutions, as we will show in detail in Section 9.5.