Change of base formula for Tetration
#29
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.
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Messages In This Thread
Change of base formula for Tetration - by jaydfox - 08/12/2007, 06:39 AM
RE: Parabolic Iteration - by jaydfox - 08/15/2007, 09:19 PM
RE: Parabolic Iteration - by bo198214 - 08/15/2007, 09:30 PM
RE: Parabolic Iteration - by jaydfox - 08/15/2007, 11:41 PM
RE: Parabolic Iteration - by bo198214 - 08/16/2007, 08:17 AM
RE: Parabolic Iteration - by jaydfox - 08/16/2007, 05:51 PM
RE: Parabolic Iteration - by bo198214 - 08/16/2007, 06:40 PM
RE: Parabolic Iteration - by jaydfox - 08/16/2007, 09:47 PM
RE: Parabolic Iteration - by bo198214 - 08/16/2007, 10:07 PM

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