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 Principal Branch of the Super-logarithm tommy1729 Ultimate Fellow Posts: 1,372 Threads: 336 Joined: Feb 2009 06/20/2011, 09:32 PM (11/17/2007, 10:58 AM)andydude Wrote: $\text{slog}(x+\pi i) = \text{slog}(e^{x+\pi i}) - 1 = \text{slog}(-e^{x}) - 1$ i do not think that is correct ... it is tempting to substitute x => x + pi and similar tricks but in general $\text{slog}(x+f(x)) = \text{slog}(e^{x+f(x)}) - 1 = \text{slog}(e^{x}*e^{f(x)}) - 1$ is false. take for example x-> x^3 $\text{slog}(x^3) = \text{slog}(e^{x^3}) - 1$ is false because slog is not a "modified" inverse super of $e^{(x^{3})}$ ! so i think this is a mistake ... i would prefer a better explaination then my own very much perhaps this is the reason that we do not have many ( or many known ? ) calculus identities related to tetration ; substitution is troublesome , hence together with the growth speeds , integrals are harder or even impossible to express in closed form. i feel im missing something here and im still a bit confused about this... ( although i can unfortunately not ask for a concrete question - apart from the one below - , just an elaboration ) i must say i wonder about the existance of the following h and g do not commute : $h(g(x)) =/= g(h(x))$ $f(x) = f(g(x)) - 1$ derivate $f ' (x) = f ' (g(x)) * g ' (x)$ substitute $f ' (h(x)) * h'(x) = f ' (g(h(x))) * g ' (h(x)) * h'(x) (*)$ integrate $f(h(x)) = f(g(h(x))) + Constant$ ( * notice how h'(x) is on both sides as if it does not matter if we differentiate with respect to it , or just substitute directly , which mainly motives this question btw ) ??? *** it appears that " -1 " is a function of f and g in any $f(x) = f(g(x)) - 1$ , or at least my imagination tells me. so maybe we should define , or could define for some functions : $f(x) = f(g(x)) - dx$ or $f(x) = ( f(g(x)) - 1 ) dx$ or something similar ?? so another question becomes $f(x) = f(g(x)) - 1$ $f(h(x)) = f(g(h(x))) - h ' (x)$ regards tommy1729 « Next Oldest | Next Newest »

 Messages In This Thread Principal Branch of the Super-logarithm - by andydude - 11/17/2007, 10:58 AM RE: Principal Branch of the Super-logarithm - by jaydfox - 11/18/2007, 05:36 AM RE: Principal Branch of the Super-logarithm - by andydude - 11/21/2007, 10:00 PM RE: Principal Branch of the Super-logarithm - by andydude - 12/10/2007, 04:46 AM RE: Principal Branch of the Super-logarithm - by jaydfox - 12/11/2007, 03:00 AM RE: Principal Branch of the Super-logarithm - by andydude - 12/11/2007, 07:31 AM RE: Principal Branch of the Super-logarithm - by andydude - 01/10/2008, 08:19 AM RE: Principal Branch of the Super-logarithm - by tommy1729 - 06/20/2011, 09:32 PM

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