I was wondering if I could ask for your help with the following question. I get down to a certain point and get so lost I just can't progress and was wondering if I had made an obvious and horrendous mistake or if you had any pointers on where to go next.
The Question:
We have the recurrence relation
\begin{align*} a_{n} & =2a_{n-1}-5a_{n-2}\\ \end{align*}
with the initial conditions
\begin{align*} a_{0} & =0,a_{1}=1\\ \end{align*}
Find the closed form
My (maybe poor) Attempt:
Using $\underset{k\geq0}{f(x)=\sum}a_{k}x^{k}$ we get the following:
\begin{align*} \underset{k\geq2}{\sum}a_{k} & =\underset{k\geq2}{\sum}2a_{k-1}x^{k}+\underset{k\geq2}{\sum}5a_{k-2}x^{k}\\ \end{align*}
\begin{align*} & =\underset{1}{\underbrace{\underset{k\geq2}{\sum}a_{k}}}=2\underset{2}{\underbrace{\underset{k\geq2}{\sum}a_{k-1}x^{k}}}+5\underset{3}{\underbrace{\underset{k\geq2}{\sum}a_{k-2}x^{k}}}\\ \end{align*}
Working on 1
\begin{align*} \underset{k\geq2}{\sum}a_{k} & =\underset{k\geq0}{\sum}a_{k}-a_{1}x-a_{0}=f(x)-x\\ \end{align*}
Working on 2 \begin{align*} \underset{k\geq2}{\sum}a_{k-1}x^{k} & =x\underset{k\geq1}{\sum}a_{k}x^{k}=x\left(\underset{k\geq1}{\sum}a_{k}x^{k}-a_{0}\right)=xf\left(x\right)\\ \end{align*} Working on 3
\begin{align*} \underset{k\geq2}{\sum}a_{k-2}x^{k} & =x^{2}\left(\underset{k\geq0}{\sum}a_{k}x^{k}\right)=x^{2}\left(f\left(x\right)\right)\\ \end{align*} Putting it back together
\begin{align*} f\left(x\right)-x & =2\left[x\left(f\left(x\right)\right)\right]-5x^{2}f\left(x\right)\\ \end{align*}
\begin{align*} & \therefore f\left(x\right)=\frac{x}{1-2x+5x^{2}}\\ \end{align*}
From here, I am not too sure of where to go as I can see that the denominator is irriducible over the real numbers and I have the feeling that partial fractions with complex values is going to get a bit nasty rather quickly. Hence my believe that I have messed something up along the line and turning to you guys for a hand.
Thank you and kindest regards!
$$ \dfrac{x}{1 - 2x + 5x^2} = \left(\dfrac{1}{2} - \dfrac{1}{4}i\right)\dfrac{1}{5x - 1 - 2i} + \left(\dfrac{1}{2} + \dfrac{1}{4}i\right)\dfrac{1}{5x - 1 + 2i} $$
The first fraction can be expressed as: $$ \dfrac{1}{5x - 1 - 2i} = -\dfrac{1 - 2i}{5}\sum_{n = 0}^\infty (1 - 2i)^n x^n $$ Similarly for the second fraction: $$ \dfrac{1}{5x - 1 + 2i} = -\dfrac{1 + 2i}{5}\sum_{n = 0}^\infty (1 + 2i)^n x^n $$
Thus, $$ \dfrac{x}{1 - 2x + 5x^2} = \dfrac{1}{4}\sum_{n = 0}^\infty x^n i[(1-2i)^n - (1 + 2i)^n] $$
which means $$ a_n = \dfrac{1}{4}i[(1 - 2i)^n - (1 + 2i)^n] $$
By the De Moivre's formula, $$ a_n = \dfrac{5^{n/2}}{2}\sin(n \alpha) \ \forall n \ge 0 $$
where $\alpha$ is the angle such that $$ \cos\alpha = \dfrac{1}{\sqrt{5}}, \sin \alpha = \dfrac{2}{\sqrt{5}} $$