f(2x) = ?

[bo198214]

consider f(f(x)) = exp(x)

with f(x) mapping R -> R and being strictly increasing.

you may choose your favorite solution f(x).

we know that exp(2x) = a(exp(x))

where a(x) = x^2.

the logical question becomes :

f(2x) = b(f(x))

b(x) = ???

...

=> f ( log(f(x)^2) ) = b(x)

Till here I could follow you but I dont know how you derive:

to get a different form :
use f(log(x)) = log(f(x)) = ...

Hm but let me see:
\( f(f(x)=\exp(x) \)
\( \exp(f(x))=f(f(f(x)))=f(\exp(x)) \)
\( f(\log(x))=\log(f(x)) \)

ok thats also correct

=> f( 2 x ) = log f ( f(f(x))^2 )

A NEW FUNCTIONAL EQUATION FOR f(x) ?!?

This functional equation can be simplified to:
\( f(2x)=\log f( \exp(2x)) \) and hence equivalent to
\( \exp(f(y) )=f(\exp(y)) \).

So it is a direct consequence from \( f(f(x))=\exp(x) \).

are they " equivalent " ? or are they " uniqueness criterions " ?

ill list them to be clear

f(x) with R -> R and strictly increasing.

1) f(f(x)) = exp(x)

2) log(f(x)) = f(log(x)) (*)

3) f( 2 x ) = log f ( f(f(x))^2 ) (*)

2) and 3) are consequences of 1)
1) does not follow from 2)
because also a function g, with g(g(g(x)))=exp(x) satisfies 2).
Surely 1) does not follow from 3).

So they are not equivalent, but also no uniqueness conditions.