Take $f(x)$ and do a "double-call":
$f^2(x)=f(f(x))$
I use this notation here to explain my problem.
This can be easy calculated for any function. Also $f^{100}(x)$ is not really a problem. This means this operation works well for natural numbers.
Then it is already possible to define some easy rules:
- $f^1(x)=f(x)$
- $f^{n+1}(x)=f(f^n(x))$
- $f^a(f^b(x))=f^{a+b}(x)$
The next question is: Can this be extended to all integers? Yes. For this we only require the inverse function $f^{-1}(x)$, because we know $f^{-1}(f(x))=x$. Or more detailed: $f^{-1}(f^n(x))=f^{n-1}(x)$.
This results in some more rules:
- $f^0(x)=x$
- $f^{-1}(f(x))=x$
- $f^{n-1}(x)=f^{-1}(f^n(x))$
Of course it is not that easy to calculate $f^{-1}(x)$ for many functions, but this is ok.
One may ask if this can be extended to rational numbers?
E.g. $f^{0.5}(x)$ would mean that $f^{0.5}(f^{0.5}(x))=f(x)$. It sounds reasonable, but this seems to be very hard to calculate for many function and the result does not seem to be unique. Take $f(x)=x$ then $f^{0.5}(x)$ could be $x$ or also $-x$. Maybe there are much more other solutions?
Or is it also possible to extend this thing to the real numbers? Or even complex numbers? If it can be done: how? And if not: why?
Thank you very much
Kevin
