# Functional equation problem

I've been trying to find a function that satisfies this to solve a separate problem, but I'm finding it difficult and no polynomial seems to work.

$f(x) + \frac{1}{x+1} = f(x+1)$

• Looks like the harmonic series to me, but I am guessing you want to extend $x$ to the reals. – A.E Jul 11 '13 at 18:33
• You can almost take $\ln x$ because $\ln(x+1)-\ln x =\frac1{x+\theta}$ with $0<\theta<1$. – Hagen von Eitzen Jul 11 '13 at 18:38
• $f(x) = \psi(x+1)$ where $\psi(x)$ is the Digamma function will work. – achille hui Jul 16 '13 at 1:53

## 1 Answer

$f(x)+\dfrac{1}{x+1}=f(x+1)$

$f(x+1)-f(x)=\dfrac{1}{x+1}$

$f(x)=\sum\limits_x\dfrac{1}{x+1}+\Theta(x)$, where $\Theta(x)$ is an arbitrary periodic function with unit period

Since $\lim\limits_{x\to+\infty}\dfrac{1}{x+1}=0$, so according to http://en.wikipedia.org/wiki/Indefinite_sum#Mueller.27s_formula, the result can be further simplified to

$f(x)=\sum\limits_{n=0}^\infty\left(\dfrac{1}{n+1}-\dfrac{1}{x+n+1}\right)+\Theta(x)$, where $\Theta(x)$ is an arbitrary periodic function with unit period

• It might be better to rewrite the $\sum_x$ line with dummy indices and limits, that's a little odd notationally! – Sharkos Jul 16 '13 at 1:42