I've already derived it, replaced it with other variables, applied the function in the form of a power series, but we always end up with an even more complex problem. With a simple look we can see that $f(x) = \pm x$ is a trivial solution, but how do we arrive at the $f(x) = \ln(e^x - 1)$ result, which is also a solution of the equation? Are there still other solutions?
Asked
Active
Viewed 114 times
3
-
I believe you can show $f$ is monotonic, which could be helpful. – eyeballfrog Jul 21 '23 at 13:43
1 Answers
1
There are in fact many solutions. We can construct different solutions by combining $f(x) = x$ for some $x$ and $f(x) = -x$ for others. For example $$ f(x) = \begin{cases} x; &x \in \mathbb Q \\ -x; &x \in \mathbb R\setminus \mathbb Q \end{cases}. $$ You can also construct very complex such functions with the axiom of choice.
Jan
- 66