base holomorphic tetration

[bo198214]

That's correct. Though as I've mentioned, there is as of yet no rigorous proof that the series converges even though the first 25 terms given do, hence why I mentioned it suggests, not proves, the hypothesis.

That it takes on a real value here suggests this continuation cannot be interpreted as regular iteration at these bases, which would take on complex values.

No, Mike, thats no hint at all. You can take any function real analytic at some point, close to some real singularity. The truncated powerseries always yields real values outside the convergence radius. Better would be a graph of the root test, or at least a graph that shows some convergence of the series.

Not analytic at \( e^{1/e} \)? That's strange. Not analytic at \( e \) would make more sense.

yes, of course you are right. I correct it in my original post.

Can you tell me what this series formula is, by the way?

Have a look at the tetration method draft formula 2.5. This is regular iteration at fixed point 0.
You get the formula by coefficient comparison of \( f^{\circ t} \circ f = f\circ f^{\circ t} \) and \( {f^{\circ t}}_1={f_1}^t \), where the index indicates the corresponding coefficient of the powerseries at 0.
For fixed points \( a \) different from 0 you consider the function \( h(x)=f(x+a)-a \) which has the fixed point at 0. Then \( f^{\circ t}(x)=h^{\circ t}(x-a)+a \).

I.e. in our case we have \( h_b(x)=b^{x+a}-a=a(b^x-1)=a(e^{\ln(b)x}-1)=\frac{\ln(a)}{\ln(b)}(e^{\ln(b)x}-1) \).

To work with a simpler formula we consider another conjugation \( h(x)=\frac{1}{\ln(b)}(g(\ln(b) x)) \).
\( g(x)=\ln(a) (e^x -1 ) \), \( \tau(x)=\frac{x}{\ln(b)}+a \), \( \tau^{-1}(x)=\ln(b)(x - a) \), \( g = \tau^{-1}\circ f\circ \tau \).

\( g(x)=\ln(a)x + \frac{\ln(a)}{2} x^2 + \frac{\ln(a)}{6} x^3 + \dots \)

This gives the following unsimplified coefficients of \( g^{\circ t} \):
\(
{g^{\circ t}}_1 = \mbox{lna}^{t}\\
{g^{\circ t}}_2 = \left(\frac{1}{2} \, \frac{{({(\mbox{lna})}^{t}^{2} - \mbox{lna}^{t})}}{{(\mbox{lna}^{2} - \mbox{lna})}}\right) \mbox{lna}\\
{g^{\circ t}}_3 = \left(\frac{1}{2} \, \frac{{(\frac{{({(\mbox{lna})}^{t}^{2} - \mbox{lna}^{t})} \mbox{lna}^{t}}{{(\mbox{lna}^{2} - \mbox{lna})}} - \frac{{({(\mbox{lna})}^{t}^{2} - \mbox{lna}^{t})} \mbox{lna}}{{(\mbox{lna}^{2} - \mbox{lna})}})}}{{(\mbox{lna}^{3} - \mbox{lna})}}\right) \mbox{lna}^{2} + \left(\frac{1}{6} \, \frac{{({(\mbox{lna})}^{t}^{3} - \mbox{lna}^{t})}}{{(\mbox{lna}^{3} - \mbox{lna})}}\right) \mbox{lna}
\)

One can see that the coefficients are polynomials in \( \ln(a)^t \) with rational coefficients in \( \ln(a) \).
One needs to investigate whether \( f^{\circ t}(1)=\tau\circ g^{\circ t}\circ\tau^{-1}(1)=a+\frac{1}{\ln(b)}\sum_{n=1}^\infty {g^{\circ t}}_n (\ln(b)(1 - a))^n \) is analytic in \( b=e^{1/e} \) with \( a=\exp(-W(-\ln(b))) \).