Bifurcation of tetration below E^-E

[bo198214]

Finally, my formula for the perimeter of the yellow zone is correct,

yup thats a nice formula and one wouldnt have guessed that \( b^{b^x}=x \) is solvable for \( b \) with help of the plog/Lambert W.

What somebody could argue about is that: yes, y = b[4]oo has two branches (two values) and only the lower one can be reached.

What you are probably referring to is
\( H_b(x)=\lim_{n\to\infty} \exp_b^{\circ n}(x) \) which is more general then just \( b[4]\infty \), i.e. \( b[4]\infty = H_b(1) \).

However the behaviour of this function for \( 1< b\le e^{1/e} \) with real fixed points at \( 1<a_1\le a_2 \) is as follows:

1. \( H_b(x)=\infty \) for \( x>a_2 \)
   
2. \( H_b(x)=a_2 \) for \( x=a_2 \)
   
3. \( H_b(x)=a_1 \) for \( 0<x<a_2 \)
   

This is because \( a_1 \) is an attracting fixed point, while \( a_2 \) is a repelling fixed point. I.e. only if we give exactly \( a_2 \) to \( H_b \) it stays at \( a_2 \) otherwise the limit is scared away from \( a_2 \) (either to \( a_1 \) or to \( \infty \)).

For \( e^{-e}\le b\le 1 \) there is only one real fixed point \( a\le 1 \), which is attracting, for \( b^x \). The behaviour hence is \( H_b(x)=a \) for each \( 0<x<1 \).

For \( 0<b<e^{-e} \) the limit \( b[4]\infty=H_b(1) \) is not defined. However we have two accumulation points \( y_1<y_2 \). They form a 2-cycle as the dynamics people call it. This is because \( b^{y_1}=y_2 \) and \( b^{y_2}=y_1 \). So here would be the right place to introduce multi numbers. One could say that \( \exp_b^{\circ n}(1) \) converges to the set \( \{y_1,y_2\} \). And if you consider convergence of filters in general topology they indeed converge to sets.

The upper accumulation point \( y_2 \) can be described as
\( \lim_{n\to\infty} f_b^{\circ n}(1) \) where \( f_b(x)=b^{b^x} \). When we have a look at the graph of \( b^{b^x}-x \) we see that there are 3 fixed points \( a_1<a_2<a_3 \) of \( f_b(x)=b^{b^x} \):
   

The middle one is repelling, the lower and the upper are attracting. So we can say that for \( 0<b<e^{-e} \)
\( f_b^{\infty}(x)=a_3 \) for \( x>a_2 \)
\( f_b^{\infty}(x)=a_2 \) for \( x=a_2 \)
\( f_b^{\infty}(x)=a_1 \) for \( 0<x<a_2 \)

So we can complete our case distinction regarding the behaviour of \( H_b \) with:

For \( 0<b<e^{-e} \) where \( a_1<a_2<a_3 \) are the 3 real fixed points of \( b^{b^x} \):

1. \( H_b(x)=\{a_1,a_3\} \) for \( x\neq a_2 \)
   
2. \( H_b(x)=a_2 \) for \( x=a_2 \)
   

The middle fixed point \( a_2 \) is the fixed point of \( b^x \) which is however now repelling for the chosen \( b \).

and of course for \( b>e^{1/e} \) always \( H_b(x)=\infty \).

Therefore, it would be "cleaner" to start from the lower "civilized rational" branch of y = b[4]oo

As we just have seen there are no "branches" but one value for \( b[4]\infty=H_b(1) \) for \( e^{-e}\le b\le e^{1/e} \).