Tetration extension for bases between 1 and eta

[Leo.W]

I just wonder if and when the quadratic approximation gives a different result than the linear approximation.

Im aware of the one period theta relating all solutions but that does not help at first sight.

maybe im asking a trivial thing ...

Well, if we're using Taylor expansion and dealing with infinitesimal very well we can have some kinda insight into it.
First we should know the basic rule that, suppose f is an analytic function, derivative (attractive) s at fixed point L \(f(z)=L+s(z-L)+O(z-L)^2\) and thus the Schroder function \(\sigma_f(z)=z-L+O(z-L)^2\) and \(\sigma_f^{-1}(z)=z-L+O(z-L)^2\) assuming \(\sigma'(L)=1\) and thus we get \(f^n(z)=L+s^n(z-L)+O(z-L)^2\).

And now let k be arbitrarily large number, we must have \(f^{-k}(f^t(f^k(z)))=f^t(z)\) even k extends to infinity in limit, or we say \(f^n(z)=\lim_{k\to\infty}{f^{-k}(f^n(f^k(z)))}\), and we assume the limit \(\lim_{k\to\infty}{f^k(z)}\) exist and equal to \(L\).
Recall \(f^n(f^k(z))=L+s^n(f^k(z)-L)+O(f^k(z)-L)^2=L+s^n(f^k(z)-L)+O(s^{2k}(z-L)^2)\) and thus \(f^{-k}(L+s^t(f^k(z)-L))=L+s^{-k}(s^t(f^k(z)-L))+O(s^{-2k}s^{2(k+t)}(z-L)^2)=L+s^t(z-L)+O(s^{2t}(z-L)^2)\), so as z approaches L, k goes to infinity, the limit \(\lim_{k\to\infty}{f^{-k}(L+s^t(f^k(z)-L))}\) exists.
Let's assume this limit equal to \(g(z,t)\). As t=0, we easily get by plugging t=0 in the limit: \(g(z,0)=z\).
And now we consider \(f^v(g(z,t))=\lim_{k\to\infty}{f^{-k+v}(L+s^t(f^k(z)-L))}=\)\(\lim_{k+v\to\infty}{f^{-k}(L+s^t(f^{k+v}(z)-L))}=\lim_{k\to\infty}{f^{-k}(L+s^t(f^k(f^v(z))-L))}=g(f^v(z),t)\)

Now we remind that we can rewrite the expansion as \(\lim_{z\to L}{\frac{f^t(z)-L}{s^t(z-L)}}=1\), and we can easily know the sequence \(\{z,f(z),f^2(z),f^3(z),\dots\}\) converge to some \(L\) (maybe infinity as well), so \(\lim_{k\to\infty}{\frac{f^t(f^k(z))-L}{s^t(f^k(z)-L)}}=1\)

Then we show directly by \(g(f^v(z),t)=\lim_{k\to\infty}{f^{-k}(L+s^t(f^{k+v}(z)-L))}=\)\(\lim_{k\to\infty}{f^{-k}(L+s^t(f^v(f^k(z))-L))}=\lim_{k\to\infty}{f^{-k}(L+s^t(s^v(f^k(z)-L)))}=g(z,t+s)\). So we arrive at \[f^t(z)=\lim_{k\to\infty}{f^{-k}(L+s^t(f^k(z)-L))}\]
This holds true for all such z that \(\{z,f(z),f^2(z),f^3(z),\dots\}\) converge to \(L\).

Now remind \(f^k(z)\) is very close to the fixed point \(L\), thus we rearrange the terms in the formula,
First we note and should denote that, \(\lim_L:=\lim_{k\to\infty}=\lim_{f^k(z)\to L}=\lim_{u=f^k(z),k\to\infty}\).
Then we have basically: \(\lim_L{f'(f^k(z))}=f'(L)=s\) and \(z\to L,f(z)-z\sim L+s(z-L)-z=(s-1)(z-L)\)
hence the formula reads \(\lim_L{f^{-k}(\frac{s^n(u-f(u))+f(u)-su}{1-s})}=\lim_L{f^{-k}(\frac{s^n(u-L)(1-s)+f(u)-su}{1-s})}\) \(=\lim_L{f^{-k}(\frac{s^n(u-L)(1-s)+L+s(u-L)-su}{1-s})}=\lim_L{f^{-k}(L+s^n(u-L))}=f^n(z)\)
Tah-dah!


btw there maybe many mistakes in this proof in limitation but the main idea is this