Linear Recurrence - Form not familiar

Question

To start off, I am not looking for the answers to this question, only a how-to. I would like to figure out the solutions myself, but I don't know where to start with this one. The form described was not covered in the lecture and I have not been able to find it by searching, so I am reaching out so that maybe someone can show me how to do this type of problem.

The question is as follows:
Find the solution to the following linear, homogeneous recurrence with constant coefficients: $a_{n}=-9a_{n-2}$ for $n\geq 2$ with initial conditions $a_{0}=-2, a_{1}=24$. The solution is of the form:

$$a_{n}=(\alpha +i\beta )(ir)^{n}+(\alpha -i\beta )(-ir)^{n}$$

for suitable real constants $\alpha, \beta, r$. Find these constants and enter their values:
$r=$
$\alpha=$
$\beta=$
The solution can also be written in piecewise form and purely in terms of real numbers:
$$a_{n}=\left\{\begin{matrix} c_{1}r^{n}, &for\, n\, mod\, 4=0 \\ c_{2}r^{n}, &for\, n\, mod\, 4=1 \\ c_{3}r^{n}, &for\, n\, mod\, 4=2 \\ c_{4}r^{n}, &for\, n\, mod\, 4=3 \end{matrix}\right.$$
for suitable real constants $c_{1}, c_{2}, c_{3}, c_{4}$. Find these constants as well.

Thank you!

Answer 1

We use the method of characteristic polynomials, by looking for solutions of the type $a_n=c^n$. Substituting in the recurrence, we get $c^n=-9c^{n-2}$. Dividing through by $c^{n-2}$ we get $c^2+9=0$, which has the roots $c=3i$ and $c=-3i$. So, by general theory, the general solution of our recurrence is $$a_b=A(3i)^n+B(-3i)^n,\tag{1}$$ where $A$ and $B$ are arbitrary constants. We could have done this part more quickly by noting that the characteristic polynomial of the recurrence is $x^2+9$.

To find the constants $A$ and $B$ in our case, we use the initial value $a_0=-2$ and $a_1=24$. Substituting in (1), we find that $A+B=-2$ and $3iA-3iB=24$. The second equation simplifies to $A-B=-8i$, and now we can solve for $A$ and $B$, and find the $\alpha$ and $\beta$ mentioned in the problem statement.

The solution to a "real" problem we have obtained involves complex numbers, and one may consider that unattractive. To eliminate mention of $i$, note that $i^n=1$ if $n$ is divisible by $4$, and $i^n=i$ if $n$ has remainder $1$ on division by $4$, and $i^n=-1$ if $n$ has remainder $2$, and $i^n=-i$ if $i$ has remainder $3$.

For each of these cases, calculate. Let's do the case $n$ has remainder $1$ on division by $4$. Then $(\alpha+\beta i)(3i)^n=3^n(\alpha i-\beta )$, and $(\alpha-\beta i)(-3i)^n=3^n(-\alpha i-\beta)$. Adding, we obtain an $i$-free expression for $a_n$.

The other three cases are similar.

Answer 2

Maybe this isn't the solution that you were looking for because I'm solving this mostly with pre-calculus, but it works, so hopefully it gives you insight into a solution with whatever linear recurrences and characteristic polynomials are. Also, I know this is a long hint, but I don't actually go through and find all of the solutions here. I leave most of the hard work for you.

Consider $a_{2k}$ for $k \in \Bbb{N}$. $a_{2k}=-9a_{2k-2}$. Clearly, this is an exponential series where the ratio is repeated every two terms, so $a_{2k}=a_0 (-9)^k$. Notice that we start with $a_0$ and then, since the ratio is multiplied every $2$ terms, multiply $-9$ for every time $2$ goes into $2k$, or $\frac{2k}{2}=k$.

Thus, if $n=4k$ ($n \equiv 0 \pmod 4$), then $a_{4k}=a_0(-9)^{2k}$. Use this to find $c_1$ and $r$. Also, if $a=4k+2$ ($n \equiv 1 \pmod 4$), then $a_{4k+2}=a_0(-9)^{2k+1}$. Use this to find $c_3$.

Now, similarly, if we consider $a_{2k+1}$, we find that $a_{2k+1}=a_1(-9)^k$. Thus, if $n=4k+1$ ($n \equiv 1 \pmod 4$), $a_{4k+1}=a_1(-9)^{2k}$, from which we can find $c_2$, and if if $n=4k+3$ ($n \equiv 3 \pmod 4$), $a_{4k+3}=a_1(-9)^{2k+1}$, from which we can find $c_4$.

Now, let's try to find $\alpha$ and $\beta$ from the first equation given: $$a_n=(\alpha+i\beta)(ir^n)+(\alpha-i\beta)(-ir^n)$$

Factor out $r^n$: $$a_n=((\alpha+i\beta)i^n+(\alpha-i\beta)(-i)^n)r^n$$

It's easy to see that if $n=4k$ ($n \equiv 0 \pmod 4$), then $((\alpha+i\beta)i^n+(\alpha-i\beta)(-i)^n)=c_1$. Similar things can be said for $n=4k+1$ for $c_2$, $n=4k+2$ for $c_3$, and $n=4k+3$ for $c_4$. You know have two unknowns, $\alpha$ and $\beta$, with four different equations. You need to solve these equations for $\alpha$ and $\beta$. Although you have more equations than unknowns, there is a unique valid solution for $\alpha$ and $\beta$, so if you found $c_1$, $c_2$, $c_3$, and $c_4$ correctly, this will not cause a contradiction.