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