Solving advancement operator equations

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 \(p(A) = A^2+A-6 = (A+3)(A-2)\). With \(p(A)\) factored like this, we realize that we've already solved part of this problem in ! In that example, the polynomial of \(A\) we encountered was (while not explicitly stated as such there) \(A-2\). The solutions to \((A-2)f_1=0\) are of the form \(f_1(n) = c_12^n\). What happens if we try such a function here? We have \[\begin{aligned}\end{aligned}\] 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 . Since \((A+3)(A-2) = (A-2)(A+3)\), we see right away that \(f_2\) is also a solution of .

Now we've got two infinite families of solutions to . 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 \[\begin{aligned}(A+3)(A-2)f(n) \amp = (A+3)(c_1 2^{n+1} + c_2 (-3)^{n+1} - 2(c_12^n + c_2(-3)^n)) \\ \amp = (A+3)(-5c_2(-3)^{n}) \\ \amp = -5c_2(-3)^{n+1}-15c_2(-3)^n \\ \amp = 15c_2(-3)^n - 15c_2(-3)^n \\ \amp =0.\end{aligned}\]

It's not all that hard to see that since \(f\) gives a two-parameter family of solutions to , it gives us all the solutions, as we will show in detail in .

The corresponding homogeneous equation here is the same as in the previous example, so its general solution is again \(f_1(n) = c_1(-2)^n + c_2 6^n\). Thus, the real work here is finding a particular solution \(f_0\) to the given advancement operator equation. Let's just try what our work on the previous example would suggest here, namely \(f_0(n) = d6^n\). Applying the advancement operator polynomial \((A+2)(A-6)\) to \(f_0\) then gives, uh, well, zero, since \((A-6)(d6^n) = d6^{n+1}-6d6^n =0\). Huh, that didn't work out so well. However, we can take a cue from how we tackled homogeneous advancement operator equations with repeated roots and introduce a factor of \(n\). Let's try \(f_0(n) = dn6^n\). Now we have \[\begin{aligned}(A+2)(A-6)(dn6^n) \amp = (A+2)(d(n+1)6^{n+1}-6dn6^n) \\ \amp = (A+2)d6^{n+1} \\ \amp = d6^{n+2} + 2d 6^{n+1} \\ \amp = 6^n(36d+12d) = 48d6^n.\end{aligned}\] We want this to be equal to \(6^n\), so we have \(d = 1/48\). Therefore, the general solution is \[\begin{aligned}\end{aligned}\]

All that remains is to use our initial conditions to find the constants \(c_1\) and \(c_2\). We have that they satisfy the following pair of equations: \[\begin{aligned}1 \amp = c_1 + c_2 \\ 5 \amp = \frac{1}{8} -2c_1+6c_2\end{aligned}\] Solving these, we arrive at the desired solution, which is \[\begin{aligned}\end{aligned}\]