# Generic way to solve f(x+1) - f(x) = g(x) when g(x) is given [closed]

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!

• Where did this question arise? Why is it urgent? – David Roberts Oct 12 '13 at 6:53
• @Bloodmoon. A problem for you: prove e.g.: If $g$ is decreasing and non-negative on $[0,+\infty)$ then $f(x)=\sum_{k=0}^\infty\big(g(k)-g(k+x)\big)$ is the only increasing solution on $[0,+\infty)$ s.t. $f(0)=0$. Any other solution differ by a $1$-periodic function. – Pietro Majer Oct 12 '13 at 9:03
• See another question (with no answer) mathoverflow.net/questions/42484 – Gerald Edgar Oct 12 '13 at 11:44
• Then you shouldn't have asked it here, as homework is off-topic for MO. – David Roberts Oct 16 '13 at 0:31

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 $f_0+\ell$.

• thanks for your answer. Why does u eleminated from the asymptotic formula? And what do you think of this answer:upload.wikimedia.org/wikipedia/commons/6/62/Sum_of_i.pdf – Bloodmoon Oct 15 '13 at 13:28
• $\mu$ disappears because the formula is only an asymptotic one, and your $g$ is asymptotic to a funtcion not depending on $\mu$. As for your link, this matter is all pretty standard anyway, which is why your question has been put on hold in the first place. I shouldn't even have answered you since this site is not for homework... – Loïc Teyssier Oct 15 '13 at 15:51
• thanks anyway. I would also want to apply this solution to my research. And the assignment is not actually the same to everyone, but specific to my research topic. I should have clarified this first. – Bloodmoon Oct 16 '13 at 3:33
• Oh, I am so sorry! It was a misoperation. – Bloodmoon Oct 16 '13 at 7:19
• And I want to vote up for you, but I don't even have enough credit. – Bloodmoon Oct 16 '13 at 7:25