uniqueness

[tommy1729]

NO

of course not.

for example : the first case :

exp(x) = f ' (f(x)) * f ' (x)

now consider a solution that satisfies f(f(x)) = exp(x)

and let assume f ' (x) = 0 has a finite real zero at x = r1.

thus f ' (r1) = 0

exp( r1 ) = 0 * f ' (f(r1)) => ??????

you see , contradiction.

That just shows that there is no halfiterate with f'(x)=0 for some x.
The equation \( \exp(x)=f'(f(x))*f'(x) \) is just a consequence of \( \exp(x)=f(f(x)) \), so it is valid for *every* halfiterate f (which of course must be differentiable).

yes.

so basicly its about f(x) being Coo or at least C4.

( notice bo didnt first notice , or at least didnt mention , " must be differentiable " )

i used to identies of the OP because those equations show restrictions ...

so lets restate :

conjecture

if f(f(x)) = exp(x)


f(x) is real -> real and > x

and f(x) is Coo

then f(x) is the unique solution to the above.


???