Solving tetration for base 0 < b < e^-e
#2
Gottfried considered this case already here. He provided pictures of the regular iteration (base 0.04) at the repelling fixed point. If we however rather want to consider iteration at the attracting 2-cycle, the following comes to my mind:

At this base range the limits \( \lim_{n\to\infty} f^{\circ 2n}(x_0) = p_1 \) and \( \lim_{n\to\infty} f^{\circ 2n+1}(x_0)= p_2 = f(p_1) \) exist (\( f(x)=b^x \)).

So I would consider the regular iteration of \( g(x)=f^{\circ 2}(x)=b^{b^x} \) and then just always take the half of the iteration number \( f^{\circ t}(x)=g^{\circ t/2}(x) \).

I will carry that out perhaps in the next post.
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RE: Solving tetration for base 0 < b < e^-e - by bo198214 - 09/12/2009, 06:56 AM

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