Solving recurrence relation with boundary condition at infinity

Question

I want to solve the following recurrence relation: \begin{align} g(x+1)=\frac{g(x)}{f(x)} \end{align} for a known function $f(x)$, defined for $x>0$ and $f(x)\in [0,1]$. I am aware that an initial condition on $g$ is required, but all I know is $g(x)\to 0$ as $x\to\infty$. Is it possible to solve this in general?

Answer 1

Unless some more properties are imposed on $g(x)$ and $f(x)$ (e.g., continuity) then the restriction of $g(x)$ to the interval $(0,1]$ may be chosen arbitrarily.

Thus we may describe the general solution as follows. Given any function $h(x)$ on $(0,1]$ and any function $f(x)>0$ on $(0,\infty)$, the unique function $g(x)$ on $(0,\infty)$ solving your equation is $$ g(x)=\frac{h(x-\lceil x\rceil+1)}{\prod_{i=1}^{\lceil x\rceil-1} f(x-i)}. $$ Indeed, this equation follows by induction on $\lceil x\rceil$.