All


I have been looking around for a general way to solve the problem of $f(x+1) - f(x) = g(x)$, where $g(x)$ is given. Has this problem been studied before?


If there does not exist such a general way, could you please solve the following problem for me? I urgently want to know what $f(x)$ is.


$$ f(x+1) - f(x) = \frac{(1+\mu)^x}{1+(1+\nu)^{x-p}(1+\mu)^x} $$


where $\mu$, $\nu$ and $p$ are constants.


Your help is very much appreciated, thank you!



Generic way to solve f(x+1) - f(x) = g(x) when g(x) is given
Depending on the growth of $g$ at $\infty$ (if defined till there) you can try telescoping sums to obtain


$f(x+n)-f(x) = \sum\_{k=0}^{n-1} g(x+k)$


If the series in the right hand side converges as $k\to \infty$ then you obtain a solution $f\_0$ vanishing at $\infty$. This is so in your example (assuming your parameters are positive). You're not going to get explicit formulas though, but you can deduce asymptotics:


$f\_0(x)\sim\_{x\to+\infty} \frac{(1+\nu)^{(p+1-x)}}{\nu}$


This solution is unique up to the addition of a $1$-periodic function, so if you have conditions at $\infty$ (say a given limit $\ell$) the unique solution will be 
$f\_0+\ell$.


