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

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


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

Thank you!

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


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.