f(x) can be found in terms of the Lambert W function. I haven't been able to find this anywhere else so I'm not sure if I discovered it (highly unlikely, I know) or re-discovered it.
Anyways...
Let's say f is of the following form,
\( f(x)=e^{bx} \)
Then the original definition becomes
\( f(f(x))=e^{be^{bx}}=e^x \)
Taking natural log of both sides,
\( be^{bx}=x\\\Rightarrow bxe^{bx}=x^2\\\Rightarrow W(x^2)=bx\\\Rightarrow f(x)=e^{W(x^2)} \)
which, by definition, can also be written as
\( f(x)=x^2/W(x^2) \)
Perhaps someone can verify that I'm using the W function properly, as I'm rather new to applying it. There isn't a closed-form expression for W as far as I know, so perhaps writing f(x) in this way doesn't actually answer any of your questions.
EDIT: So I just noticed that I posted this somewhere else on the forum several months back. Sorry about the double-post! I completely for got about it, but I guess I never did find out why it doesn't work (if that is indeed the case).
Anyways...
Let's say f is of the following form,
\( f(x)=e^{bx} \)
Then the original definition becomes
\( f(f(x))=e^{be^{bx}}=e^x \)
Taking natural log of both sides,
\( be^{bx}=x\\\Rightarrow bxe^{bx}=x^2\\\Rightarrow W(x^2)=bx\\\Rightarrow f(x)=e^{W(x^2)} \)
which, by definition, can also be written as
\( f(x)=x^2/W(x^2) \)
Perhaps someone can verify that I'm using the W function properly, as I'm rather new to applying it. There isn't a closed-form expression for W as far as I know, so perhaps writing f(x) in this way doesn't actually answer any of your questions.
EDIT: So I just noticed that I posted this somewhere else on the forum several months back. Sorry about the double-post! I completely for got about it, but I guess I never did find out why it doesn't work (if that is indeed the case).