Functional Square Root

[tommy1729]

Idk Tommy. In my attempt I tried to set it up as \(Dg=C_fg\) i.e. \[g'(x)=g(f(x))\] where we solve in the \(g\)

I did that only because pre-composition by \(f\) is linear. Only for that reason. In fact \(g'(x)-g(f(x))=0\) can be expressed in linear operators as \((D-C_f)[g(x)]=0\)...

Anyways I remember I read something about \(g^{-1}(g'(x))\)... somewhere... maybe something about the legendre transform? Too bad I'm ignorant about that.

Just a note: define \(\mathcal F[g]=g^{-1}\circ Dg\) then what we are loking for are fixed points \[\mathcal F[f]=f\] So maybe we could unlock some iterative method: consider
\[f_{n+1}(x)=f_n^{-1}(f_n'(x))\]

and take the limit \(\lim f_n\)

I think it all solves to or all converges to my solution.

Consider that taking many iterations of inverses of analytic functions usually gives singularities that keep making the radius smaller.
So if the radius goes to 0 or the derivative is no longer defined the concept f ' (x) is getting dubious.
But not in my example.


regards

tommy1729