Functional Square Root

[tommy1729]

This thread seems a wastebin, full of not related ides.
But I'll take it as a place where to discuss interesting functional eqns.

I'd like to say something on the equation

\[Df=f\circ f\]

It seem interesting. Idk if this is fruitfull. Sure there is Tommy's derivation from the hypothesis \(f(x)=ax^b\)... but I'd like to be see if we can be more general: define the function \(C_fg:=g\circ f\)
Now I'd like to look at the equation
\[Dg=C_fg\]

So assume that \(g\in A\) for \(A\) an algebra over the ring \(R\). We can look for abstract derivations \(d:A\to A\), i.e. a \(R\)-linear algebra morphism satisfying Leibniz rule.
Now, assume that \(A\) is an \(R\)-algebra of functions and \(f\in A\) can be precomposed with functions in \(A\).

fact. Precomposition is always linear: let \(f,g,h\in A\), \(k\in R\) then \(C_f(g+h)=(g+h)\circ f=gf+hf=C_f(g)+C_f(h)\) and \(C_f(\lambda g)=(\lambda g)\circ f=\lambda (gh)=\lambda C_f(g)\).

So both precompositions \(C:A\to A\) and derivation \(d:A\to A\) are \(R\)-linear operators. They form a vector space, we can add, subtract and scale them.

 Given a derivation \(d\) and a precomposition \(C\) over \(A\). We now look for \(g\in A\) that satisfies \(dg=Cg\), i.e. solutions to the linear equation \(dg-Cg=0\)... in other words we study the kernel of the operator
\[(d-C)g=0\]

The question can be reformulated as follows: find all the \(f\in A\) s.t. \[f\in {\rm ker}(d-C_f)\]

Question for experts in linear algebra... what can we deduce by studying the kernels of operators that are subtraction of a derivation minus a precomposition?

Hmm 

Basically we could look at it as a difffential equation

F ‘ (x) = G(F(x))

This has solution

F(x) =inverse ( integral dx/ G(x) + C ) + C2

For Some appropriate constants C and C2.

Now G = F itself.

I think this ultimately forces F to be slower than exp and 
Reduces finally to my solution.



Using f(x) = a x^b + h(x) seems to imply h(x) = 0.

Maybe I made a mistake.

But I even think I made this exercise before…

regards

tommy1729