New answers tagged tetration
Extending my earlier comment. First we write $f(x)=x^x$ and $f(x,h)=f(f(x,h-1))$ , $f(x,0)=x$ and $f(x,1)=f(x)$ where $h$ indicates the "iteration-height". Then we consider the function $g(x,h)=f(x+1,h)-1$ and get a nicely iterable power series for $g(x)$ . The coefficients of $g(x,h)$ depend on $h$ and are in fact polynomials in $h$ of increasing order. The ...
Hint: Other super roots at Tetration wikipedia article.
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