Change of base formula for Tetration
#19
jaydfox Wrote:Really, I thought it obvious it converges. As n goes to infinity, the epsilon goes to 0 (and quite rapidly at that), making the formula exact in the limiting case.
For publishing this argumentation would not suffice Wink

Quote:The proof relies on bases greater than eta, but if one starts with one of the bases greater than eta (e seems the obvious choice), and uses iterated logarithms for the other base (see Andrew Robbins's notation), then one can even get solutions for bases between 1 and eta, though I'm not sure if there is a defensible "unique" definition of the zeroeth iterate. (more on that when I get to bases less than eta).
That would be the interesting case, we want to transfrom from a base \( b\le\eta \) to a base \( a\gt\eta \). For soundness it should also be verified that if \( a,b<\eta \) (for this case we have a unique solution demanding differentiability at the smaller fixed point of \( x\mapsto{}^{\text{slog}_b(x)+t}b \)) the change of base yields the other unique solution for base \( a \).
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Messages In This Thread
Change of base formula for Tetration - by jaydfox - 08/12/2007, 06:39 AM
RE: Change of base formula for Tetration - by bo198214 - 08/15/2007, 09:36 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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