elementary superfunctions

[bo198214]

Next function \( f(x)=cx^a \), the natural numbered iterates are:
\( f^{[1]}(x)=cx^a \)
\( f^{[2]}(x)=c(cx^a)^a=c^{a+1} x^{a^2} \)
\( f^{[3]}(x)=c\left(c^{a+1}x^{a^2}\right)^a = c^{a^2+a+1} x^{a^3} \)
\( f^{[n]}(x)=c^{\sum_{k=0}^{n-1}a^k} x^{a^n}=c^{\frac{a^n-1}{a-1}} x^{a^n} \)

So a super-function would be
\( F(t)=c^{\frac{a^t-1}{a-1}}\exp(a^t)=\exp(a^t+\ln( c)\frac{a^t-1}{a-1}) \)
(*) \( F(t)=\exp\left(\left(1+\frac{\ln( c)}{a-1}\right)a^t -\frac{\ln( c)}{a-1}\right) \)

For the regular iteration we need to find a fixed point \( \lambda \)
\( c\lambda^a = \lambda \)
\( c\lambda^{a-1}=1 \) or \( \lambda=0 \)
\( \lambda=c^{-\frac{1}{a-1}}=c^{\frac{1}{1-a}} \), \( a\neq 1 \).

Then (*) looks like:
\( F(x) = (e /\lambda)^{a^x} \lambda \)

If we translate \( F \) along the x-axis we can get
\( F(x)=\exp(a^x)\lambda \)
Check: \( cF(x)^a=\exp(a^x)^a c\lambda^a=\exp(a^{x+1})\lambda=F(x+1) \)
Edit: It is regular at \( \lambda \).

Summary: \( F(x)=\exp(a^x)c^{\frac{1}{1-a}} \) is the at \( \lambda=c^{\frac{1}{1-a}} \) regular superfunction of \( f(x)=cx^a \), for \( a\neq 1 \).