As we now have the limit formula for the regular slog it is time to publish also the corresponding powerseries for the regular slog. Interestingly the computation is in a certain way similar to the computation of the natural Abel function and Andrew's slog.
This is the good thing about regular tetration that there is a well developed theory about direct compuation (limit formula) and about power series computation.
So let us start with the regular Schroeder function of a given powerseries \( f \) at 0 with fixed point at 0.
A Schroeder function \( \sigma \) satisfies
\( \sigma(f(x))=c\sigma(f(x)) \) (where \( c=f_1=f'(0)>0,\neq 1 \))
Let us write this with the Bell matrix (\( m \)th row contains the coefficients of the \( m \)-th power of the function) \( S \) for \( \sigma \) and \( F \) for \( f \). As \( f \) and \( \sigma \) dont have a constant/0th coefficient the matrix is correspondingly stripped:
\( FS=cS \)
\( (F-cI)S=0 \)
we only need to consider the first row of \( S \) which is \( \vec{\sigma} \):
\( (F-cI)\vec{\sigma}=0 \)
E.g., truncation to 4:
\(
\begin{pmatrix}
c-c & 0 & 0 & 0\\
f_2 & c^2-c & 0 & 0\\
f_3 & {f^2}_3 & c^3-c & 0\\
f_4 & {f^2}_4 & {f^3}_4 & c^4-c
\end{pmatrix}
\begin{pmatrix}
\sigma_1\\\sigma_2\\\sigma_3\\\sigma_2
\end{pmatrix}
=\begin{pmatrix}0\\0\\0\\0\end{pmatrix}
\)
We see that the first row is 0 and needs to be chopped this gives freedom up to a multiplicative constant for \( \sigma \) (which is anyway known for Schroeder functions) and we decide to choose \( \sigma_1=\pm 1 \) depending on whether \( c>1 \) or \( c<1 \) and from which side we approach the fixed point. This then leads to the equation with the matrix \( F' \), which is \( F \) with removed first row and column
\( F'(\sigma_2,\sigma_3,\dots)^T=\mp(f_2,f_3,\dots)^T \), eg.
\(
\begin{pmatrix}
c^2-c & 0 & 0\\
{f^2}_3 & c^3-c & 0\\
{f^2}_4 & {f^3}_4 & c^4-c
\end{pmatrix}
\begin{pmatrix}
\sigma_2\\\sigma_3\\\sigma_2
\end{pmatrix}
=-\begin{pmatrix}f_2\\f_3\\f_4\end{pmatrix}
\)
However we dont need an equation solver to solve this system, because we chopped off the first line and column and not the last line and the first column, as in Andrew's slog; we can solve it by hand:
\( \sigma_{k} = \left(\pm f_k + \sum_{i=2}^{k-1} {f^i}_k \sigma_i\right)/\left(c-c^k\right) \).
Also the solution of this equation system does not depend on the truncation size as it is with the slog. But of course this becomes relativated by needing a fixed point.
So we have a formula for the powerseries of the regular Schroeder function. Then the regular Abel function is just \( \alpha_f(x)=\log_c(\sigma_f(x)) \).
Let us apply this to \( b^x \). First we have to move the fixed point \( a \) to 0 by conjugation: \( f(x)=b^{x+a}-a=ab^x-a=ae^{x\ln(b)}-a \)
\( f \) has the coefficients:
\( f_k = a\frac{\ln(b)^k}{k!} \), \( f_0=0 \)
\( f^n(x)=a^n \sum_{m=0}^n (-1)^{n-m}\left(n\\m\right) e^{x\ln(b)m} \)
It has the coefficients
\( {f^n}_k = a^n \sum_{m=0}^n (-1)^{n-m}\left(n\\m\right)\frac{\ln(b)^k m^k}{k!}=a^n\frac{\ln(b)^k}{k!}\sum_{m=0}^n (-1)^{n-m}\left(n\\m\right)m^k \)
So
\( \sigma_k = \frac{\ln(b)^k}{k!\left(c-c^k\right)}\left(a + \sum_{n=2}^{k-1} a^n \sigma_n\sum_{m=0}^n (-1)^{n-m}\left(n\\m\right)m^k\right) \)
where \( c=f'(0)=a\ln(b)b^0=\ln(b^a)=\ln(a) \)
So the Abel function of \( f=\tau_a^{-1}\circ\exp_b\circ\tau_a \) is \( \alpha_f(x)=\log_{\ln(a)} (\sigma(x)) \) and so the
Abel function of \( b^x \) is \( \alpha(x)=\log_{\ln(a)}(\sigma(x-a)) \)
However it seems as if the convergence radius of \( \sigma \) is just \( a \). So you can not use this formula exclusively to plot for example the regular Abel function of \( \sqrt{2}^x \) in the range -1 to 1.9, it does not converge at 0. A numeric comparison with the natural slog will hopefully follow later.
This is the good thing about regular tetration that there is a well developed theory about direct compuation (limit formula) and about power series computation.
So let us start with the regular Schroeder function of a given powerseries \( f \) at 0 with fixed point at 0.
A Schroeder function \( \sigma \) satisfies
\( \sigma(f(x))=c\sigma(f(x)) \) (where \( c=f_1=f'(0)>0,\neq 1 \))
Let us write this with the Bell matrix (\( m \)th row contains the coefficients of the \( m \)-th power of the function) \( S \) for \( \sigma \) and \( F \) for \( f \). As \( f \) and \( \sigma \) dont have a constant/0th coefficient the matrix is correspondingly stripped:
\( FS=cS \)
\( (F-cI)S=0 \)
we only need to consider the first row of \( S \) which is \( \vec{\sigma} \):
\( (F-cI)\vec{\sigma}=0 \)
E.g., truncation to 4:
\(
\begin{pmatrix}
c-c & 0 & 0 & 0\\
f_2 & c^2-c & 0 & 0\\
f_3 & {f^2}_3 & c^3-c & 0\\
f_4 & {f^2}_4 & {f^3}_4 & c^4-c
\end{pmatrix}
\begin{pmatrix}
\sigma_1\\\sigma_2\\\sigma_3\\\sigma_2
\end{pmatrix}
=\begin{pmatrix}0\\0\\0\\0\end{pmatrix}
\)
We see that the first row is 0 and needs to be chopped this gives freedom up to a multiplicative constant for \( \sigma \) (which is anyway known for Schroeder functions) and we decide to choose \( \sigma_1=\pm 1 \) depending on whether \( c>1 \) or \( c<1 \) and from which side we approach the fixed point. This then leads to the equation with the matrix \( F' \), which is \( F \) with removed first row and column
\( F'(\sigma_2,\sigma_3,\dots)^T=\mp(f_2,f_3,\dots)^T \), eg.
\(
\begin{pmatrix}
c^2-c & 0 & 0\\
{f^2}_3 & c^3-c & 0\\
{f^2}_4 & {f^3}_4 & c^4-c
\end{pmatrix}
\begin{pmatrix}
\sigma_2\\\sigma_3\\\sigma_2
\end{pmatrix}
=-\begin{pmatrix}f_2\\f_3\\f_4\end{pmatrix}
\)
However we dont need an equation solver to solve this system, because we chopped off the first line and column and not the last line and the first column, as in Andrew's slog; we can solve it by hand:
\( \sigma_{k} = \left(\pm f_k + \sum_{i=2}^{k-1} {f^i}_k \sigma_i\right)/\left(c-c^k\right) \).
Also the solution of this equation system does not depend on the truncation size as it is with the slog. But of course this becomes relativated by needing a fixed point.
So we have a formula for the powerseries of the regular Schroeder function. Then the regular Abel function is just \( \alpha_f(x)=\log_c(\sigma_f(x)) \).
Let us apply this to \( b^x \). First we have to move the fixed point \( a \) to 0 by conjugation: \( f(x)=b^{x+a}-a=ab^x-a=ae^{x\ln(b)}-a \)
\( f \) has the coefficients:
\( f_k = a\frac{\ln(b)^k}{k!} \), \( f_0=0 \)
\( f^n(x)=a^n \sum_{m=0}^n (-1)^{n-m}\left(n\\m\right) e^{x\ln(b)m} \)
It has the coefficients
\( {f^n}_k = a^n \sum_{m=0}^n (-1)^{n-m}\left(n\\m\right)\frac{\ln(b)^k m^k}{k!}=a^n\frac{\ln(b)^k}{k!}\sum_{m=0}^n (-1)^{n-m}\left(n\\m\right)m^k \)
So
\( \sigma_k = \frac{\ln(b)^k}{k!\left(c-c^k\right)}\left(a + \sum_{n=2}^{k-1} a^n \sigma_n\sum_{m=0}^n (-1)^{n-m}\left(n\\m\right)m^k\right) \)
where \( c=f'(0)=a\ln(b)b^0=\ln(b^a)=\ln(a) \)
So the Abel function of \( f=\tau_a^{-1}\circ\exp_b\circ\tau_a \) is \( \alpha_f(x)=\log_{\ln(a)} (\sigma(x)) \) and so the
Abel function of \( b^x \) is \( \alpha(x)=\log_{\ln(a)}(\sigma(x-a)) \)
However it seems as if the convergence radius of \( \sigma \) is just \( a \). So you can not use this formula exclusively to plot for example the regular Abel function of \( \sqrt{2}^x \) in the range -1 to 1.9, it does not converge at 0. A numeric comparison with the natural slog will hopefully follow later.
