Triggered by the interesting finding of Andrew I open this thread for the further investigation of elementary superfunctions, i.e. functions \( F \) that are expressible with elementary functions and operations such that
\( F(x+1)=f(F(x)) \)
for a given elementary function \( f \).
Our first example is:
\( f(x)=2x^2-1 \) with a superfunction \( F(x)=\cos(2^x) \).
Now the \( \cosh \) has the same property \( \cosh(2x)=2\cosh(x)^2 -1 \) as the \( \cos \). Hence \( F(x)=\cosh(2^x) \) is another superfunction of \( 2x^2-1 \).
Indeed \( f^{[t]}(x)=F(t+F^{-1}(x)) \) exists and is differentiable at \( x=1 \).
But it does not exist at the other fixed point \( -\frac{1}{2} \), because \( \operatorname{arccosh}\left(-\frac{1}{2}\right) \) is not defined.
Edit: both are regular super-functions at fixed point 1. \( \lim_{x\to-\infty} F(x)=1 \).
So if we are at polynomials \( f \), we can also give an elementary superfunction for \( f(x)=x^a \), i.e. \( F(x)=c^{a^x} \).
Because \( F(x+1)=c^{a^xa}=F(x)^a \).
Edit: these are the regular super-exponentials at 1. \( \lim_{x\to-\infty} F(x)=1 \).
Generally for Chebyshev polynomials, these are the polynomials \( T_n \) such that \( \cos(nx)=T_n(\cos(x)) \) - for example above we used \( T_2(x)=2x^2-1 \) -, we know already two elementary superfunctions of \( f(x)=T_n(x) \), these are \( F(x)=cos(n^x) \) and \( F(x)=\cosh(n^x) \).
\( F(x+1)=f(F(x)) \)
for a given elementary function \( f \).
Our first example is:
\( f(x)=2x^2-1 \) with a superfunction \( F(x)=\cos(2^x) \).
Now the \( \cosh \) has the same property \( \cosh(2x)=2\cosh(x)^2 -1 \) as the \( \cos \). Hence \( F(x)=\cosh(2^x) \) is another superfunction of \( 2x^2-1 \).
Indeed \( f^{[t]}(x)=F(t+F^{-1}(x)) \) exists and is differentiable at \( x=1 \).
But it does not exist at the other fixed point \( -\frac{1}{2} \), because \( \operatorname{arccosh}\left(-\frac{1}{2}\right) \) is not defined.
Edit: both are regular super-functions at fixed point 1. \( \lim_{x\to-\infty} F(x)=1 \).
So if we are at polynomials \( f \), we can also give an elementary superfunction for \( f(x)=x^a \), i.e. \( F(x)=c^{a^x} \).
Because \( F(x+1)=c^{a^xa}=F(x)^a \).
Edit: these are the regular super-exponentials at 1. \( \lim_{x\to-\infty} F(x)=1 \).
Generally for Chebyshev polynomials, these are the polynomials \( T_n \) such that \( \cos(nx)=T_n(\cos(x)) \) - for example above we used \( T_2(x)=2x^2-1 \) -, we know already two elementary superfunctions of \( f(x)=T_n(x) \), these are \( F(x)=cos(n^x) \) and \( F(x)=\cosh(n^x) \).