08/15/2007, 11:41 PM
bo198214 Wrote:jaydfox Wrote:How do you figure that it only works for a<b? It works for any two bases greater than eta, regardless of relative size.Did you try it?
The sequence is decreasing for \( b<a \) (you can even prove this, similar to my proof that it increases for \( a<b \)) and after some steps it becomes negative and after that the logarithm yields complex numbers. Its quite obvious also from the experimentation.
And I would expect real numbers from a change of base, as the tetration is real for real arguments.
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.
So, for example, if mu_b(a) is -1, and we use n=20, then you'll tetrate base b 19 times, then take 20 logarithms base a, and you'll be back at 1 (and some small epsilon value that is essentially nil).
Tetrate base b 20 times and take 20 logarithms base a, and you'll be back at a. Tetrate base b 21 times and take 20 logarithms base a, and you'll be at a^a. Tetrate base b 22 times, then 20 log_a's, and you'll be at a^^3. And so on. Going the other way, tetrate base b 18 times, then 20 log_a's, and you'll arrive at 0.
~ Jay Daniel Fox
