I have the following recursive relations:
$$ \left \{ \begin{array}{ccc} x_{t+1} &=& x_t \cdot (1-2c) + 0.5 v_t\\ v_{t+1}&=& 0.5v_t - 2c\cdot x_t \end{array} \right. $$
$c$ is just a constant.
Initial conditions are:
$x_0 = x_0$, $x_1 = (1-2c)x_0$, $v_0 = 0$.
I solved the following succeeding orders:
$v_1 = -2c\cdot x_0$
$x_2 = x_0 \cdot (4c^2 - 5c+1)$
$v_2 = -cx_0 ( 3-4c)$
I want to generalize the higher orders, but of course we don't like to solve all of them. How do we make something like a generating function for these?
Notice that $x_1 = x_0 (1-2c)$, $x_2 = x_0 ( 4c^2 -5c+1)$, $x_3 = x_0 (-8c^3 +12c^2 - \frac{17}{2}c + 1)$ .
So I want to say that $x_n = x_0 {P^n(c)}$ where $n$ is the degree of the polynomial.
Thanks for your insights
If $z_n=\begin{pmatrix}x_n\\v_n\end{pmatrix}$ then $z_{n+1}=Az_n$ where $$A=\begin{pmatrix}1-2c&0.5\\-2c&0.5\end{pmatrix}$$
Thus, we have: $$z_n = A^nz_0$$
This shows why both are multiples of $x_0$. Since $z_0=x_0\begin{pmatrix}1\\0\end{pmatrix}$, so $z_n = x_0A^n\begin{pmatrix}1\\0\end{pmatrix}$. So we can first solve for $x_0=1$.
The trick is to diagonal $A$, so we can compute $A^n$ more quickly.
The eigenvalues of the matrix are the solutions to $\lambda^2 -(1.5-2c)\lambda + 0.5=0$, so $$\lambda_1,\lambda_2 = \frac{1.5-2c\pm \sqrt{0.25-6c+4c^2}}{2}$$. Let's just write $D:=0.25-6c+4c^2$.
You'll end up with (assuming $D\neq 0$):
$$x_n=\alpha_1\lambda_1^n + \alpha_2\lambda_2^n\\v_n=\beta_1\lambda_1^n+\beta_2\lambda_2^n$$
For some values $\alpha_1,\alpha_2,\beta_1,\beta_2$.
We know $v_0=0$ so $\beta_1+\beta_2=0$. We know $v_1=2cx_0=2c=\beta_1(\lambda_1-\lambda_2)$. So $\lambda_1-\lambda_2=\sqrt{D}$ and $$\beta_1=\frac{2c}{\sqrt{D}}$$
So $$v_n=\frac{2c}{\sqrt{D}}\left(\lambda_1^n-\lambda_2^n\right)$$.
This might not look like it, but it will be a polynomial in $c$ of degree $n$ in general.
The solution for $x_n$ will be similar, or you can realize the recursion means that $x_t=x_{t-1}+v_t$, so $x_t=x_0+\sum_{s=1}^t v_t$, and then use the closed form for $v_t$ to get a closed form for $x_t$.