Functional power - Printable Version +- Tetration Forum (https://math.eretrandre.org/tetrationforum) +-- Forum: Tetration and Related Topics (https://math.eretrandre.org/tetrationforum/forumdisplay.php?fid=1) +--- Forum: Mathematical and General Discussion (https://math.eretrandre.org/tetrationforum/forumdisplay.php?fid=3) +--- Thread: Functional power (/showthread.php?tid=1158) Functional power - Xorter - 03/11/2017 Let f and g be total functions (so e. g. C -> C) and N and M be complexes. Then (f o g)(x) and f o a = f(a) are so-called functional multiplications. But the interesting thing is the following: functional power: $f^{oN} = f o f o ... o f (N-times)$ When N is an integer, it is trivial, just look: $f^{o0} = x$ $f^{o1} = f$ $f^{o2} = f o f$ $f^{o3} = f o f o f$ ... $f^{o-1} = f^{-1}$ We have rules for it, like these ones: $(f^{oN}) o (f^{oM}) = f^{o N+ M}$ $(f^{oN})^{oM} = f^{o N M}$ $f o (f^{oN}) = (f^{oN}) o f = f^{o N+1}$ But for instance: $(f^{oN}) o (g)^{oN} != (f o g)^{oN}$ (Also functional tetration exists.) My theory is that if we can get an explicit formula for $f^{oN}$ with x and N, then N is extendable to any total function. For example: $(2x)^{oN} = 2^N x N := log_2 (x) (2x)^{o log_2 (x)} = x^2$ And in the same way, theoritacelly you could do the same with all the functions. But how? My concept is that by Carleman matrices.