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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?

Nosrati
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ftor
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  • 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. – callculus42 May 21 '19 at 15:21
  • Yes. That's a solution. I want to find other solutions. Or prove that is the only solution. – ftor May 21 '19 at 15:28
  • 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. – callculus42 May 21 '19 at 15:30
  • I checked again. That is a solution. – ftor May 21 '19 at 15:36
  • Yes. That is a critical equation in my research project. – ftor May 21 '19 at 15:38
  • One idea to start with is to try to simplify it by substituting $x \to f^{-1}(x)$ and using $f^{-1}(x)' = \frac{1}{f'(f^{-1}(x))}$ to write it on the form $\frac{g(x)-g\left(\frac{x}{g(x)}\right)}{x - \frac{x}{g(x)}} = g'(x)$ where $g(x) = f^{-1}(x)$. – Winther May 21 '19 at 15:39
  • Thanks. What should do the next step? – ftor May 21 '19 at 17:55
  • I don't know. My guess would be that linear functions are the only solutions, but this is just a guess. The only thing I can think off is to try to study the function along sequences on the form $x_{n+1} = x_n/g(x_n)$ and somehow try to use this local information to obtain some constraints on the global behavior of your function. It's a very vague idea and don't know if it would work. Hopefully someone will come up with a better idea. – Winther May 21 '19 at 18:22

1 Answers1

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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...

Cosmas Zachos
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