Change of base formula for Tetration

[jaydfox]

Sorry Jay we dont speak about the same thing.
You ask me why I assume \( a<b \).
I answered: because your formula does not converge for some x if we allow \( a>b \).
You answered: that I didnt pay attention to \( \mu_b(a) \).
I answered: there are despite values of x for which it doesnt converge.

And I can not relate your current answer to this problem.
To express it in formulas and numbers:
Your definition: For a given tetration \( {}^xb \) for base \( b \) compute another tetration for base \( a \) by the formula
\( {}^xa = \lim_{n\to\infty} \log_a^{\circ n}({}^{x+b+\mu_b(a)}b) \)
where \( \mu_b(a) \) is chosen accordingly that \( {}^1a=a \).

This formula is equivalent to
\( {}^xa = \lim_{n\to\infty} \log_a^{\circ n}(\exp_b^{\circ n}({}^{x+\mu_b(a)}b) \)

By definition \( {}^0 b=1 \), so if I chose \( x=-\mu_b(a) \)
then \( {}^{x+\mu_b(a)}b=1 \).
If I now compute the sequence
\( t_n = \lim_{n\to\infty} \log_a^{\circ n}(\exp_b^{\circ n}(1.0)) \)
say for bases \( a=3.0>\eta \) and \( b=1.5>\eta \) I get for \( n=1,2,3 \):
\( .3690702464, -.5382273450, -.9460372733+2.859600867*I \)

For me this means your formula gives no result in this case.

Well, yeah, if mu_b(a) >= 2, then you can't solve for a^^(-mu_b(a)). I still don't see why this is a problem.

For a=3.0, b=1.5, the constant mu_b(a) is going to be something in the neighborhood of 10. So, yeah, if you try to set x=-10 or thereabouts, it's not going to converge properly.