〈 Berger | Dillon 〉

But doesn’t that just say that f(f(x)) is is invertible? Pardon my analysis is a bit rusty

From the expression you have above in the box, if you assume f is smooth enough and defined for all R then f' can never vanish. And then it's inverse is found by just drawing the graph and reflecting on the x=y line (I mean there are theorems, but that's even easier :) )

Of course, the assumption that it is well-defined on all of R needs to be justified

Yeah, I think Picard-Lindelöf gives you that, no? I may be trying to extend it too far with the function composition but I think it’s still valid

I'm not actually sure. For example consider: f(f(x))=x f'(f(x))f'(x)=1 But f(x)=1/x solves this

I love this problem

So, it is perhaps that this problem got a family of solutions spanning on its "solution surface"?

Maybe! I guess I should look for a Lie group of symmetries, because that seems like a real possibility. I doubt it’s the case with f(f(x))=x, but it would make sense for exp(x) to be different in the right way for this to work