Solutions to f ' (x) = f(f(x)) ?

[tommy1729]

We all know f ' (x) = f (x) for f(x) := C exp(x).

But what is the general solution to f ' (x) = f(f(x)) ?
Of course f(x) = id(0) will do.

I was thinking about its fixpoints : y = f ' (y) = f(f(y))
If y is also a fixpoint of f then
f '' (y) = f ' ( f(y) ) * f ' (y) = [f ' (y)] ^2 = y^2
If y is also a fixpoint of f '' (y) then y^2 = y !
Now since f ' = f(f) we get f^[3](y) * f^[2](y) = y^2

Let f ' (x) = f(f(x)) then f '' (x) = f ' (f(x)) * f ' (x).
Since f ' (x) = f(f(x)) we get f '' (x) = f^[3](x) * f ' (x).

Rearranging gives f '' (x)/f ' (x) = f^[3](x).
Integrating gives ln( f ' (x) ) = integral f^[3](x) dx + C

I feel like Im getting closer to a solution but Im stuck now.

How about subtitution f^[-1](z) = x ?

Anyways if we plug in f(x) = a x^b we get a b x^(b-1) = a (a x^b)^b = a a^b x^(b^2) = a^(b+1) x^(b^2)

Hence we get the system a b = a^(b+1) and b-1 = b^2.

b = (-1)^(1/3) or -(-1)^(2/3) ( yes can be simplified )
ab = a^(b+1) => b = a^b => a = b^(1/b) ( typical tetration )

however what about other solutions ??

regards

tommy1729

[tommy1729]

Notice how the superfunction of a x^b involves the generalizations.

e.g. f ' (x) = f^[2,5](x)

regards

tommy1729