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$$\frac{1-f^{-1}\left(\frac{f(x)}{x}\right)}{1-x} = 1- \frac{f(x)}{xf'(x)}$$

I know $f(x) = ax+b$ is a solution. How can I find other solutions?

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  • $\begingroup$ Have you checked if $f(x)=ax+b \ \ \forall \ \ a,b\neq 0$ is a solution? This is the first what I would do if I´m facing this problem. $\endgroup$ – callculus May 21 at 15:21
  • $\begingroup$ Yes. That's a solution. I want to find other solutions. Or prove that is the only solution. $\endgroup$ – ftor May 21 at 15:28
  • $\begingroup$ I have to admit that the solution doesn´t work for me. Maybe I´m wrong, but maybe I´m right. A calculation that shows that $f(x)=ax+b$ is a solution would indicate that you are really interested in that exercise. $\endgroup$ – callculus May 21 at 15:30
  • $\begingroup$ I checked again. That is a solution. $\endgroup$ – ftor May 21 at 15:36
  • $\begingroup$ Yes. That is a critical equation in my research project. $\endgroup$ – ftor May 21 at 15:38
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This is an extended comment, not an answer, but I'd need the space... Inverse functions are stressful, so you might eliminate yours by $$ f^{-1}\left(\frac{f(x)}{x}\right) =1 -\left (1- \frac{f(x)}{xf'(x)}\right ) (1-x),$$ so that $$ \frac{f(x)}{x} =f\left (1 -\left (1- \frac{f(x)}{xf'(x)}\right ) (1-x)\right ),$$ so that $$ f(x)=xf\left( \frac{f(x)/f’(x)}{ x}+\frac{xf'(x) -f(x)}{f'(x)} \right),$$ manifestly possessing your leading binomial solution. Near the origin, only the leading term in the r.h.side argument is singular.

After doing this, you might well start plugging in series solutions and devising creative coefficient recursions...

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