Hey, Tommy. I looked at this before; and found a numerical procedure that looks similar to this.
We start by using differential-bullet notation. So we start by setting:
\[
\int_a^x g(s,z)\,ds\bullet z = h(x,z)\\
\]
Then \(h\) is the unique solution to the equation:
\[
\begin{align}
h'(x,z) &= g(x,h(x,z))\\
h(a,z) &= z\\
\end{align}
\]
When taking the half iterate \(f(f(x)) = \exp(x)\); we know that:
\[
f'(x) = \frac{\exp(x)}{f'(f(x))}\\
\]
Let's try to sequentially solve this equation for \(|x| < \delta\)--by writing:
\[
f_{n+1}'(x) = \frac{\exp(x)}{f_n'(f_{n+1}(x))}\\
\]
This would mean that:
\[
f_{n+1}(x) = \int_0^x \frac{\exp(s)}{f_{n}'(z)}\,ds\bullet z\\
\]
If you choose \(|x| < \delta\)--this becomes very close to a linear operation; as:
\[
f_{n+1}(x) = z + \int_0^x \frac{\exp(s)}{f_{n}'(z)}\,ds + O(x^2)\\
\]
Which satisfies the proper initial condition:
\[
\begin{align}
f_{n+1}(0) &= z\\
f_{n+1}'(0) &= \frac{1}{f_n'(z)}\\
\end{align}
\]
I would always start this iteration with \(f_0(x) = z+x\).
I could never prove it converges; but it did look like it was converging for certain values of \(z\). But it was fairly uneconomical-and I never wrote good code for it. What this would converge to is a function:
\[
\begin{align}
f(0) = z\\
f(f(x)) = \exp(x)\\
\end{align}
\]
So what you choose here is the appropriate value of \(z\), which is somewhere in \((-\infty,0)\). And it would only converge for specific values of \(z\); which I could never figure out... Plus my hacked together mock proof of the convergence of this sequence could only get smooth convergence; never holomorphy.
For example, if you take \(z = \text{tet}_K(\frac{1}{2} + \text{slog}_K(0)) = \text{tet}_K(-\frac{1}{2})\), for Kneser's tetration, it did look like it was was converging. I wasn't detailed enough to tell if it actually converged to Kneser though...
I eventually abandoned this approach when I started with the beta method--as that seemed more promising. But maybe it's time to crack this open again
.
EDIT: I should add the description of differential bullet product:
\[
\int_a^b g(s,z)\,ds\bullet z = \Omega_{j=1}^M \left(z+ g(s_j,z)\Delta s_j \right)\bullet z\\
\]
Where omega is composition in z; and \(s_j\) are the sample points of \(s_j \in [a,b]\) and \(\Delta s_j\) are the difference in samples.... This is just a fancier, more modern way of writing Euler's method for first order differential equations. Or, Euler's method.
We start by using differential-bullet notation. So we start by setting:
\[
\int_a^x g(s,z)\,ds\bullet z = h(x,z)\\
\]
Then \(h\) is the unique solution to the equation:
\[
\begin{align}
h'(x,z) &= g(x,h(x,z))\\
h(a,z) &= z\\
\end{align}
\]
When taking the half iterate \(f(f(x)) = \exp(x)\); we know that:
\[
f'(x) = \frac{\exp(x)}{f'(f(x))}\\
\]
Let's try to sequentially solve this equation for \(|x| < \delta\)--by writing:
\[
f_{n+1}'(x) = \frac{\exp(x)}{f_n'(f_{n+1}(x))}\\
\]
This would mean that:
\[
f_{n+1}(x) = \int_0^x \frac{\exp(s)}{f_{n}'(z)}\,ds\bullet z\\
\]
If you choose \(|x| < \delta\)--this becomes very close to a linear operation; as:
\[
f_{n+1}(x) = z + \int_0^x \frac{\exp(s)}{f_{n}'(z)}\,ds + O(x^2)\\
\]
Which satisfies the proper initial condition:
\[
\begin{align}
f_{n+1}(0) &= z\\
f_{n+1}'(0) &= \frac{1}{f_n'(z)}\\
\end{align}
\]
I would always start this iteration with \(f_0(x) = z+x\).
I could never prove it converges; but it did look like it was converging for certain values of \(z\). But it was fairly uneconomical-and I never wrote good code for it. What this would converge to is a function:
\[
\begin{align}
f(0) = z\\
f(f(x)) = \exp(x)\\
\end{align}
\]
So what you choose here is the appropriate value of \(z\), which is somewhere in \((-\infty,0)\). And it would only converge for specific values of \(z\); which I could never figure out... Plus my hacked together mock proof of the convergence of this sequence could only get smooth convergence; never holomorphy.
For example, if you take \(z = \text{tet}_K(\frac{1}{2} + \text{slog}_K(0)) = \text{tet}_K(-\frac{1}{2})\), for Kneser's tetration, it did look like it was was converging. I wasn't detailed enough to tell if it actually converged to Kneser though...
I eventually abandoned this approach when I started with the beta method--as that seemed more promising. But maybe it's time to crack this open again
.EDIT: I should add the description of differential bullet product:
\[
\int_a^b g(s,z)\,ds\bullet z = \Omega_{j=1}^M \left(z+ g(s_j,z)\Delta s_j \right)\bullet z\\
\]
Where omega is composition in z; and \(s_j\) are the sample points of \(s_j \in [a,b]\) and \(\Delta s_j\) are the difference in samples.... This is just a fancier, more modern way of writing Euler's method for first order differential equations. Or, Euler's method.