Tetration below 1

[bo198214]

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?

Hmm, I don't know, whether I understand you correctly. If h=1 then all eigenvalues except the first are zero ( = [1,0,0,....]) and the result is always the same, independent of any power of log(h) since the "height" y of the tower occurs only as exponent of the eigenvalues....
Did I misread something obvious?

As said \( h \) is not 1 but \( e^{-1}<h<1 \).
\( f(x)=b^x \), \( A \) is the power derivation matrix of \( f \) (I think this is the transpose of your matrix \( B_b \)), and \( A^t \) is \( \exp(t\cdot \log A) \) (though we can also apply the powerseries \( (1+x)^t \) directly to \( A \)). Hence the power series \( f^{\circ t}(x) \) has as coefficients the first row of \( A^t \) (think transposed in your notation).
And now tetration is defined as \( {}^tb=f^{\circ t}(1) \). We set \( x=1 \) not \( h \).