Without further information about f, one can pretty much prescribe any value c>0 as function value of f at 0. Just define f as
f(c)=f(f(0)):=g(0)
f(f(f(0):=g(f(0)
...
f(f^n+1(0)):=g(f^n(0))
This works fine, since g is injectiv for positive arguments. Instead of starting with function value 0, we could take any other argument and prescribe any function value, as long as we avoid previously defined arguments. So unless we have other conditions for f, we can hardly calculate f(0).
Even if we assume that f can be expanded into a power series, it seems that there are quite many degrees of freedom. Maybe too many to fix f(0)?
Certainly quite a challenging Level2 puzzle ;)

Posted by JLo
on 20191223 07:59:55 