I am trying to understand the general term of following recurrence relation
I rewrote it as:
$$g(j) = e^{-h}[p g(j+1) +q g(j-1)] $$
which yields:
$$ g(j+1) = \frac{e^h}{p} g(j) - \frac{q}{p}g(j-1) $$
Then we can write the following relation:
$$ \begin{bmatrix}0 & 1 \\ \frac{e^h}{p} & \frac{q}{p} \end{bmatrix}\begin{bmatrix} g(j-1) \\ g(j) \end{bmatrix} = \begin{bmatrix} g(j) \\ g(j+1) \end{bmatrix}$$
Which can be rewritten as
$$A v_j = v_{j+1} $$
where
$$ A = \begin{bmatrix}0 & 1 \\ \frac{e^h}{p} & \frac{q}{p} \end{bmatrix} \\ v_j=\begin{bmatrix} g(j-1) \\ g(j) \end{bmatrix} $$
this gives us
$$A^j v(1) = v(j+1) $$
Now, we want to find $D$ diagonal such that $B A B^{-1} = D$. That is, we want to diagonalize the matrix.
First we go for the eigen values $\lambda_+$ and $\lambda_-$
Which are the roots of $\det (A - \lambda I ) = 0$
Once we have the roots, we find the eigenvectors:
$$A w_- = \lambda_- w_-\\ A w_+ = \lambda_- w_+$$
Choose $w_- = (1, y_-)$ and $ w_+ = (1, y_+)$ this gives
$$y_- = \lambda_-\\ y_+ = \lambda_+ $$
Now the basis change matrix are
$$ B = \begin{bmatrix}1 & 1 \\ \lambda_- & \lambda_+ \end{bmatrix} \\ B^{-1} = \begin{bmatrix}1 & 1 \\ \lambda_- & \lambda_+ \end{bmatrix}^{-1}$$
and
$$ B^-1 A B = D $$
or equivalently
$$ A = B D B^{-1}$$
Now we come back to the relation
\begin{align}A^j v(1) &= v(j+1) \\ B D^j B^{-1} v(1) &= v(j+1) \end{align}
To make the computation one needs an explicit expression of $B^{-1}$ :
$$B^{-1} = \begin{bmatrix}\lambda_+ & -1 \\ -\lambda_- & 1 \end{bmatrix} \det(B)^{-1} $$
Finally:
$$ \begin{bmatrix}0 & 1 \\ \frac{e^h}{p} & \frac{q}{p} \end{bmatrix}^j\begin{bmatrix} g(0) \\ g(1) \end{bmatrix} = \begin{bmatrix} g(j) \\ g(j+1) \end{bmatrix}$$ $$ \begin{bmatrix}1 & 1 \\ \lambda_- & \lambda_+ \end{bmatrix}\begin{bmatrix}\lambda_-^j & 0 \\ 0 & \lambda_+^j \end{bmatrix}\begin{bmatrix}\lambda_+ & -1 \\ -\lambda_- & 1 \end{bmatrix} \det(B)^{-1}\begin{bmatrix} g(0) \\ g(1) \end{bmatrix} = \begin{bmatrix} g(j) \\ g(j+1) \end{bmatrix}$$
the computations now follow:
\begin{align}g(j)& = \frac{1}{\det(B)} \begin{bmatrix}1 & 1 \end{bmatrix}\begin{bmatrix}\lambda_-^j & 0 \\ 0 & \lambda_+^j \end{bmatrix}\begin{bmatrix} \lambda_+g(0) - g(1) \\ g(1)-\lambda_-g(0) \end{bmatrix} \\ & = \frac{1}{det(B)}\lambda_-^j(\lambda_+g(0) - g(1)) + \lambda_+^j(g(1)-\lambda_-g(0)) \end{align}
Now we remember that $g(0) = e^{-h}g(1)$ substituting we obtain:
\begin{align}g(j)& = \frac{1}{det(B)} \big(\lambda_+^j(1-\lambda_-e^{-h}) - \lambda_-^j(1-\lambda_+ e^{-h} ) \big)g(1) \end{align}
This is almost
But how do we show that $\frac{g(1)}{det(B)} = \big(\lambda_+^z(1-\lambda_-e^{-h}) - \lambda_-^z(1-\lambda_+ e^{-h} )\big)^{-1}$ ?
It suffices to write
$$1 = g(z) = \frac{g(1)}{det(B)} \big(\lambda_+^z(1-\lambda_-e^{-h}) - \lambda_-^z(1-\lambda_+ e^{-h} ) \big) $$
and conclude that
$$\frac{g(1)}{det(B)} = \big(\lambda_+^z(1-\lambda_-e^{-h}) - \lambda_-^z(1-\lambda_+ e^{-h} ) \big)^{-1} $$