2nd order homogenous linear recurrence

Question

What will be the 2nd-order homogeneous linear recurrence of:

$$x_{n+1}+ x_n+ 6x_{n-1}=0 \text{?}$$

Answer 1

As alreay mentioned by Lemoine, start by solving the characteristic equation $x^2+x+6=0$.

This has the complex solutions $z_{1/2}=-\frac12\pm\frac{\sqrt{23}}2i$. So the general complex solution is $az_1^n+bz_2^n$ with $a,b\in\mathbb C$.

To transform it in real form note that $z_{1/2}=re^{\mp\varphi i}$ where $r=\sqrt6$ and $\varphi=\arctan\sqrt{23}$. Hence, the general real solution is $cr^n\cos(n\varphi)+dr^n\sin(n\varphi)$ with $c,d\in\mathbb R$.

(Of course, $cr^n\cos(n\varphi)+dr^n\sin(n\varphi)$ with $c,d\in\mathbb C$ is another representation of the general complex solution.)

Answer 2

Let $ {\left( x_i\right)}_{i=0}^{\infty} $ be your sequence and consider the power series $$ h(z)=\sum_{j=1}^{\infty}z^jx_j$$ So your recurrence formulates to $$ \dfrac{h(z)-a_1z -a_0}{z^2} + \dfrac{h(z)-a_0}{z} + 6h(z) = 0$$ So $$ h(z)=\dfrac{x_1z+x_0z+x_0}{1+z+6z^2} $$ I hope you can extract the coefficient now or tell me I would add it.