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$$\begin{equation} S(r,k) = f(r)S(0,k-1) + g(r)S(r+1,k-1) \end{equation}\ ,$$ where $r=0,1,2,\dots$ and $k=1,2,3,\dots$ . Also, $0<f(r)<1$ is an increasing function and $0<g(r)<1$ is a decreasing function, while $f(r)+g(r)<1$.

The base case is $S(r,1) = f(r) + g(r)\ \forall\ r$.

Do you think a closed-form expression of $S(r,k)$ is achievable?

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  • $\begingroup$ The conditions on $f$ and $g$ seem irrelevant. $\endgroup$
    – Martin Rubey
    Commented Mar 11, 2023 at 20:46
  • $\begingroup$ Possibly. But I think they imply that S(r,k) is bounded. $\endgroup$
    – K. Bountrogiannis
    Commented Mar 11, 2023 at 20:49
  • $\begingroup$ You could try the approach with power series from here math.stackexchange.com/questions/2065067/… $\endgroup$
    – Thomas Kojar
    Commented Mar 11, 2023 at 22:43
  • $\begingroup$ Just trying it a bit, in order to solve it that way, I needed extra information such as the boundary data $S_{0,k}$ and some relation between $g(r)$ and $g(r+1)$. The boundary data can in principle be determined from the recursion but is tricky. Unless we at least have a known compatible boundary data $S_{0,k}$ , this problem is likely a regular numerical discrete-pde problem with no closed forms. But perhaps there is some other approach or pattern I missed. $\endgroup$
    – Thomas Kojar
    Commented Mar 11, 2023 at 23:17
  • $\begingroup$ @ThomasKojar If we assume that $S_{0,k}$ is just a variable, could you at least get an analytic expression involving $S_{0,k}$, $f(r)$ and $g(r)$? $\endgroup$
    – K. Bountrogiannis
    Commented Mar 12, 2023 at 11:17

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