Stability of a recurrence relation What I can say about the stability of the recursive formula
$$S_{r+1}(t)=at^{r}-bS_{r}(t), \ \ \ \ r\geq0$$
 for the determination of $S_{r}(t)$?
Assume: $S_{0}(t)=0$ and a,b>0 are constant.
Thanks
 A: If you consider $t$ as a fixed parameter, this is just a linear recurrence of first order, with constant coefficients. You can solve it explicitly. Define the generating function $g(z) = \sum_{n \ge 0} S_r(t) z^r$, multiply the recurrence by $z^r$, add up over $r \ge 0$ and recognize some sums:
$\begin{align*}
S_{r + 1}(t)
  &= a t^r - b S_r(t) \\
\sum_{r \ge 0} S_{r + 1}(t) z^r
  &= a \sum_{r \ge 0} (t z)^r - b \sum_{r \ge 0} S_r(t) z^r \\
\frac{g(z) - S_0(t)}{z}
  &= \frac{a}{1 - t z} - b g(z) \\
\end{align*}$
Solve for $g(z)$:
$\begin{align*}
g(z)
  &= \frac{S_0(t) - (t S_0(t) - a) z}{1 - (t - b) z + b t z^2} \\
  &= \frac{S_0(t) t + S_0(t) b - a}{t + b} \cdot \frac{1}{1 + b z}
       + \frac{a}{t + b} \cdot \frac{1}{1 - t z}
\end{align*}$
From here you get your solution as the coefficient of $z^r$, which is easy as it is just two geometric series:
$\begin{align*}
S_r(t)
  &= [z^r] g(z) \\
  &= \frac{S_0(t) t + S_0(t) b - a}{t + b} \cdot (-1)^r b^r
       + \frac{a}{t + b} \cdot t^r
\end{align*}$
