Let $x_k$ be the solution of the recurrence equation $$x_{k+1} = \sum_{i=1}^m \theta_i x_{k+1-i} + r_{k+1}$$ where $(r_k)$ is a general sequence. I'm trying to find a explicit solution for $(x_k)$ of the form $$x_n = \sum_{k=1}^n b_{k,n} \cdot r_k$$
Let ${\bf y}_k= (x_k,x_{k-1},\dots,x_{k-m+1})^T$ where $k \geq m-1$.
Then $${\bf y}_{k+1} = A {\bf y}_k + {\bf z}_{k+1}$$ Where $A$ is a matrix consisting on $A_{1,{\bf :}} = (\theta_1, \dots, \theta_m) $ and $A_{i,{\bf :}} = Id_{(i-1),{\bf :}}$ for $i \geq 2$ ($Id$ is de identity matrix of size $m$). And $z_k = (r_k,0,\dots,0)^T$. Then is easy to show that $${\bf y}_n = \sum_{k=m}^n A^{n-k} {\bf z_k} + A^{n-(m-1)}{\bf y}_{m-1}$$ for $n \geq m$.
then by the definition of ${\bf y}$ and ${\bf z}$ we get that $$x_n = \sum_{k=m}^n a^{(n-k)}_{1,1} r_k + (A^{n-(m-1)}{\bf y_{m-1}})_1$$ for $n \geq m$.
So I almost got the representation I wanted ($b_{k,n} = a_{1,1}^{(n-k)}$ for $k \geq m$) but I can't get $b_{k,n}$ for $k \leq m-1$.
Any help will be appreciated
Thanks