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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!

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closed as off-topic by Andrés E. Caicedo, Will Jagy, Pietro Majer, Ramiro de la Vega, Michael Renardy Oct 12 '13 at 11:24

This question appears to be off-topic. The users who voted to close gave this specific reason:

  • "This question does not appear to be about research level mathematics within the scope defined in the help center." – Andrés E. Caicedo, Will Jagy, Pietro Majer, Ramiro de la Vega, Michael Renardy
If this question can be reworded to fit the rules in the help center, please edit the question.

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) – 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
up vote 2 down vote accepted

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 $f_0+\ell$.

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thanks for your answer. Why does u eleminated from the asymptotic formula? And what do you think of this – 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
@Bloodmoon: Why did you unaccept my answer? – Loïc Teyssier Oct 16 '13 at 6:31
Oh, I am so sorry! It was a misoperation. – Bloodmoon Oct 16 '13 at 7:19

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