jaydfox Wrote:You're not factoring in the superlogarithmic constant. You tetrate base b n+mu times, then take the logarithm base a only n times. If base b is greater than base a, then mu will be negative.
You gave the formula \( \log_a^{\circ n}(\exp_b^{\circ n}({}^{x+\mu_b(a)}b)) \) for arbitrary \( x \). So if \( \mu_{b}(a) \) is negative I simply choose \( x:=-\mu_{b}(a) \) such that for example \( y:={}^{x+\mu_{b}(a)}b={}^{0}b=1 \) (actually I tried \( y=1.0 \) and \( y=2.0 \)). Otherwise you have to specify for which x your formula is valid (which makes it clear again that the convergence of your formula wasnt at all "obvious"). Normally the tetration shall be defined for exponents \( x>-2 \) I guess.
However if we can show analyticity it suffices to define it in a small vicinity of some point \( x_0 \) because from there we can analytically extend it. However the question then is what is \( x_0 \) and what \( \varepsilon>0 \) has the vicinity.