# Functional equation, probably involving discrete differentiating?

I want to know how to solve this problem on functions.

Question: Find all functions $f: \mathbb{R}\rightarrow \mathbb{R}$ satisfying

$$f(x+1)-f(x)=nf(x)$$

where $\mathbb{R}$ is set of real number.

Thank you!

• Just bring the $f(x)$ over to the other side to get $(n+1)f(x)$...? Then there are uncountably many possible functions!? – user1537366 Jan 23 '15 at 4:14
• Can you please elaborate? I don't quite get it. Thank you! – Andy Lee Jan 23 '15 at 4:16
• $f(x+1)=(n+1)f(x)$. This extends arbitrary function on $[0,1)$ to all real numbers. – velut luna Jan 23 '15 at 4:20
• Then there are uncountably many functions not including same form? (Same form meaning change in C, or the constant) – Andy Lee Jan 23 '15 at 4:23
• They showed it holds for an arbitrary function. Note we don't need any special conditions on $f(x)$ – user76844 Jan 23 '15 at 4:30

By applying the recursion formula $k$ times we find that for any $k\in\mathbb{Z}$ we have

$$f(x+k)=(n+1)^k f(x)$$

Now define any function $f$ on $[0,1)$. For any other $x\in \mathbb{R}$ we can write

$$x = \lfloor x\rfloor + \left<x\right>$$

where $\left<x\right>\in [0,1)$ is the fractional part of $x$. From the formula above (with $k=\lfloor x \rfloor$ and taking $\left<x\right>$ for $x$) we have

$$f(x)=(n+1)^{\lfloor x\rfloor} f(\left<x\right>)$$

Thus by specifying $f(x)$ on $[0,1)$ the recursion formula extends $f$ uniquely to all $x$ and consequently there are infinitely many solutions.