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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?

Deep
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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 $\lceil x\rceil$.

pre-kidney
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  • +1 $f(x), g(x)$ are both defined for $x>0$ and are indeed continuous. Could you please a elaborate a little more on how the equation you have given follows by induction on Ceiling(x)? – Deep Mar 20 '17 at 09:30
  • Okay. The base case is when $\lceil x\rceil = 1$, in which case $0<x\leq 1$, and the formula is simply $g(x)=h(x)$. Now suppose the formula holds for $\lceil x\rceil = k$, that is, for $k-1<x\leq k$. This means that for all $x$ in this interval, $g(x)=h(x-k+1)\cdot f(x-1)^{-1} f(x-2)^{-1}\cdots f(x-k+1)^{-1}$. Substituting this into the formula $g(x+1)=g(x)/f(x)$ then yields $g(x+1)=h(x-k+1)\cdot f(x)^{-1}f(x-1)^{-1} \cdots f(x-k+1)^{-1}$, which shows that the formula holds for $x+1$, and thus for the entire interval $(k,k+1]$. This completes the induction.

    Any other assumptions on $f,g$?

     – pre-kidney Mar 21 '17 at 03:26
  • Awesome answer. Thanks a lot. I suppose the answer will not change if $f(x)\in [0,\infty)$. – Deep Mar 21 '17 at 06:31