I'm trying to solve the recurrence relation
$$a_n = (\lambda +\mu)a_{n+2}+\mu a_{n+3}.$$
with initial relationships of:
$\lambda a_1 = \mu a_2$
$(\lambda +\mu)a_2 = \mu a_3.$
I found a site online which suggests using the roots of a, in this case, cubic polynomial to solve it, but the roots are of course really complex and since I don't know what $\lambda$ and $\mu$ are they don't simplify very much. Is there another way to go about this? Thanks.
On
The site online probably wanted you to write $$ \mu \left[ \begin{array}{c} a_{n+3} \\ a_{n+2} \\ a_{n+1} \end{array} \right] = \left[ \begin{array}{ccc} -\lambda - \mu & 0 & 1 \\ \mu & 0 & 0 \\ 0 & \mu & 0 \end{array} \right] \left[ \begin{array}{c} a_{n+2} \\ a_{n+1} \\ a_{n} \end{array} \right] $$ and then diagonalize the matrix using eigenvalues to compute a general formula. This is extremely gross, but I don't see any better way.
For completeness, add $a_0$. Setting up the recurrence for $n = 0$ gives: $$ \begin{align*} \mu a_3 + (\lambda + \mu) a_2 &= a_0 \\ 2 (\lambda + \mu) a_2 &= a_0 \\ \end{align*} $$ This gives enough to get the recurrence going: $$ \begin{align*} a_1 &= \frac{a_2}{\lambda} = \frac{a_0}{2 \lambda (\lambda + \mu)} \\ a_2 &= \frac{a_0}{2 (\lambda + \mu)} \end{align*} $$ Whatever you do, the result will be the same, unless you change the constants to simplify the recurrence. Use Wilf's "generatingfunctionology" techniques. Define $A(z) = \sum_{n \ge 0} a_n z^n$. From the recurrence: $$ \mu \frac{A(z) - a_0 - a_1 z - a_2 z^2}{z^3} + (\lambda + \mu) \frac{A(z) - a_0 - a_1 z}{z^2} = A(z) $$ Solving for $A(z)$ gives a rather horrible expression: $$ A(z) = \frac{a_0}{2 \lambda (\lambda + \mu)} \frac{2 \lambda^2 \mu + 2 \lambda \mu^2 + (2 \lambda^3 + 4 \lambda^2 \mu + 2 \lambda \mu^2 + \lambda \mu + \mu) z + (\lambda + \mu) z^2} {\mu + (\lambda + \mu) z - z^3} $$