I still can not completely follow your argumentation, you write:
Ok, going with the base on the upper halfplane from e to sqrt(2).
As long as we are not landed back on the real axis,
the fixpoints are not conjugate, so the tetration value will not be real.
But this is nothing new.
Now when on the real axis: the Kneser method is not applicable to two real fixpoints.
So we must take the value there as limit from above.
The kneser tetration \( b\mapsto b[4]p \) is real on the real axis \( b>\eta \), which implies that \( \overline{b} [4] p = \overline{b[4]p} \) (conjugation).
So approaching from above or below is just conjugate to each other.
So what one need to show imho is that the imaginary part will not tend to zero when approaching the real axis at \( b<\eta \).
I wonder whether Sheldon could supply some pictures of that, with his fourier algorithm.
I dont think it is 2 at \( +i\infty \) and 4 at \( -i\infty \). As mentioned before, the limits must be conjugate for the kneser tetration.
I guess there is a problem in the assumption that the Kneser tetration behaves at \( +i\infty \) like the regular iteration of the upper fixpoint in the case where the twe developing fixpoints are not conjugate.
I still agree with you that everything looks like there is a branchpoint, but I can not follow your particular arguments in the moment.
(06/02/2011, 01:55 AM)mike3 Wrote: Already we see that these are not conjugate, thus the tetrational will not be conjugate-symmetric and hence cannot be real at real heights. This contrasts with the behavior at \( b = e \). If a complex analytic function is real along some part of the real axis, but not another, there must be a branchpoint (not sure what the proof is, though.).
Ok, going with the base on the upper halfplane from e to sqrt(2).
As long as we are not landed back on the real axis,
the fixpoints are not conjugate, so the tetration value will not be real.
But this is nothing new.
Now when on the real axis: the Kneser method is not applicable to two real fixpoints.
So we must take the value there as limit from above.
The kneser tetration \( b\mapsto b[4]p \) is real on the real axis \( b>\eta \), which implies that \( \overline{b} [4] p = \overline{b[4]p} \) (conjugation).
So approaching from above or below is just conjugate to each other.
So what one need to show imho is that the imaginary part will not tend to zero when approaching the real axis at \( b<\eta \).
I wonder whether Sheldon could supply some pictures of that, with his fourier algorithm.
Quote:But if we do go around through the lower half-plane, we get \( L_1 = 4 \) and \( L_2 = 2 \). Note that this may at first be thought to be the same since these are the same two fixed points as before, it is NOT, since now the \( +i\infty \) is 2, not 4, and the \( -i\infty \) is 4, not 2: they have been swapped! Then, we see that this function cannot be the same.
I dont think it is 2 at \( +i\infty \) and 4 at \( -i\infty \). As mentioned before, the limits must be conjugate for the kneser tetration.
I guess there is a problem in the assumption that the Kneser tetration behaves at \( +i\infty \) like the regular iteration of the upper fixpoint in the case where the twe developing fixpoints are not conjugate.
I still agree with you that everything looks like there is a branchpoint, but I can not follow your particular arguments in the moment.