Tetration below 1

[bo198214]

Assume b=h^(1/h), or h=h(b) , where h() is the function described by Ioannis Galidakis, then the log of admissible h, hl=log(h) is -1< hl <1.
The eigenvalues of the operator for tetration are the consecutive powers of hl, so the diagonal contains a convergent sequence 1,hl,hl^2,hl^3,... if hl is in the admissible limit, and a divergent sequence if hl is outside.

Gottfried, this a wonderful assist. Before I could ask what the matrix operator method would answer for this case, you nearly gave the answer. So let me conclude.

If \( -1<\log(h)<0 \) then we have the negative Eigenvalues \( \log(h)^{2n+1} \) in the power derivation matrix A of \( f(x)=b^x \). Now we compute \( A^t \). It has the Eigenvalues \( (\log(h)^n)^t \). Take for example \( t=\frac{1}{2} \) then we see that \( A^t \) has also non-real Eigenvalues and hence has also non-real entries. Supposed \( f^{\circ \frac{1}{2}} \) had only real coefficients then \( A^{\frac{1}{2} \) would have only real coefficients. So it is clear that \( f^{\circ \frac{1}{2} \) must have some non-real coefficients and is a non-real function.

However perhaps it could be that \( {}^{\frac{1}{2}}b=f^{\circ \frac{1}{2}}(1) \) is real or generally that \( {}^tb=f^{\circ t}(1) \) is real, which I dont believe. Can someone just compute it?