We have investigated the coefficients of the super-logarithm for quite a long time now, and so far, all my attempts have been met with approximations of approximations. Finally, I may have found a super-logarithm with exact coefficients. You be the judge.

First lets start with some recent realizations about Abel functions, Julia functions, and topological conjugacy. So, I'll use the \( \text{dxp}_b(x) = b^x - 1 \) notation, and thus the topological conjugacy between "exp" and "dxp" can be expressed as:

\( \exp_{(b^{1/b})}^{\circ t}(x) = b(\text{dxp}_b^{\circ t}(x/b-1)+1) \)

which can also be found in this thread. The Abel functions and Julia functions can be related as:

\( \mathcal{J}[f](x) = \frac{1}{\frac{\partial}{\partial x} \mathcal{A}[f](x)} \)

which, as they apply to exp/dxp, can be proven (and is proven later) to imply:

\( \mathcal{A}[\exp_{(b^{1/b})}](x) = \mathcal{A}[\text{dxp}_b](x/b-1) \)

\( \mathcal{J}[\exp_{(b^{1/b})}](x) = b\mathcal{J}[\text{dxp}_b](x/b-1) \)

Second, I think the process of finding iterated-dxp is well understood by now, so I'll start with that. One recent observation in this thread has been that Szekeres' Julia functions and Jabotinsky's L-functions (or iterative logarithm) are actually the same functions, which has opened my eyes to a whole new approach to iteration. With this in mind, not only can we express the Abel function as:

\( \mathcal{A}[f](x) = \lim_{n\to\infty}(\log_a(f^{\circ n}(x)) - n) \)

or the logarithm of the Schroeder function, but we can also express it as:

\( \mathcal{A}[f](x) = \int\frac{dx}{\left[\frac{\partial}{\partial t} f^{\circ t}(x)\right]_{t=0}} \)

which also serves to emphasize the fact that Abel functions are only determined up to a constant, and that a solution \( \alpha(x) \) (to the Abel functional equation) can always be generalized to \( \alpha(x) + C \), which is also true of integrals.

Iterated-dxp can be expressed as the hyperbolic iteration of \( e^{ax}-1 \) as:

\( \text{dxp}_{(e^a)}^{\circ t}(x) = a^tx + \frac{a^{t+1}(a^t-1)x^2}{2(a-1)}

+ \frac{a^{t+2}(a^t-1)(-1-2a+a^t(a+2))x^3}{6(a-1)^2(a+1)} \)

and since the Julia function of dxp is also the iterative logarithm of dxp:

\( \left[\frac{\partial}{\partial t}\text{dxp}_{(e^a)}^{\circ t}(x)\right]_{t=0} = \mathcal{J}[\text{dxp}_{(e^a)}](x) \)

which evaluates to the power series:

\( \mathcal{J}[\text{dxp}_{(e^a)}](x) = \ln(a)\left(

x + \frac{ax^2}{2(a-1)} + \frac{a^2x^3}{6(1-a^2)}

- \frac{a^3(1+2a)x^4}{24(-1-a+a^3+a^4)} + \cdots

\right) \)

We can now find the Abel function of dxp by integrating its Julia function, but we need to find the reciprocal first. Finding the reciprocal of a power series can be a tedious task, so I've done the work for you:

\( \frac{1}{f'(x)} = \frac{1}{f'(0)} - \frac{f''(0)}{f'(0)^2}x + \frac{2f''(0)^2-f'(0)f'''(0)}{2f'(0)^3}x^2 + \cdots \)

and since \( \mathcal{J}[f](x) = \frac{1}{\frac{\partial}{\partial x}\mathcal{A}[f](x)} \), we solve for \( f^{(k)}(0) \) by equating the coefficients of x, and the solution to these equations is:

\( \mathcal{A}[\text{dxp}_{(e^a)}](x) = \frac{1}{\ln(a)}\left(

x + \frac{ax^2}{4(1-a)} + \frac{a^2(1+5a)x^3}{36(a-1)^2(a+1)}

- \frac{a^4(2+a+3a^2)x^4}{32(a-1)^3(a+1)(1+a+a^2)} + \cdots

\right) \)

Lastly, we can relate these findings back to the super-logarithm.

Let \( \alpha(x) = \mathcal{A}[\text{dxp}_{b}](x) \) be the Abel function of dxp.

Let \( \beta(x) = \alpha(x/b - 1) \).

Then \( \beta(x) \) is an Abel function of exp.

Proof.

\( \alpha(b^x-1) = \alpha(x) + 1 \)

\( \alpha(b^{(x/b - 1)}-1) = \alpha(x/b - 1) + 1 \)

\( \alpha(b^{(x/b)}/b-1) = \alpha(x/b - 1) + 1 \)

\( \alpha((b^{1/b})^x/b-1) = \alpha(x/b - 1) + 1 \)

\( \beta((b^{1/b})^x) = \beta(x) + 1 \). []

This means that \( \beta(x) = \text{slog}_{(b^{1/b})}(x) \) which relates back to the super-logarithm as follows:

\( \text{slog}_{(e^{a(e^{-a})})}(x) = C + \frac{1}{\ln(a)}\left(

(x(e^{-a})-1) + \frac{a(x(e^{-a})-1)^2}{4(1-a)} + \frac{a^2(1+5a)(x(e^{-a})-1)^3}{36(a-1)^2(a+1)}

- \frac{a^4(2+a+3a^2)(x(e^{-a})-1)^4}{32(a-1)^3(a+1)(1+a+a^2)} + \cdots

\right) \)

Andrew Robbins

First lets start with some recent realizations about Abel functions, Julia functions, and topological conjugacy. So, I'll use the \( \text{dxp}_b(x) = b^x - 1 \) notation, and thus the topological conjugacy between "exp" and "dxp" can be expressed as:

\( \exp_{(b^{1/b})}^{\circ t}(x) = b(\text{dxp}_b^{\circ t}(x/b-1)+1) \)

which can also be found in this thread. The Abel functions and Julia functions can be related as:

\( \mathcal{J}[f](x) = \frac{1}{\frac{\partial}{\partial x} \mathcal{A}[f](x)} \)

which, as they apply to exp/dxp, can be proven (and is proven later) to imply:

\( \mathcal{A}[\exp_{(b^{1/b})}](x) = \mathcal{A}[\text{dxp}_b](x/b-1) \)

\( \mathcal{J}[\exp_{(b^{1/b})}](x) = b\mathcal{J}[\text{dxp}_b](x/b-1) \)

Second, I think the process of finding iterated-dxp is well understood by now, so I'll start with that. One recent observation in this thread has been that Szekeres' Julia functions and Jabotinsky's L-functions (or iterative logarithm) are actually the same functions, which has opened my eyes to a whole new approach to iteration. With this in mind, not only can we express the Abel function as:

\( \mathcal{A}[f](x) = \lim_{n\to\infty}(\log_a(f^{\circ n}(x)) - n) \)

or the logarithm of the Schroeder function, but we can also express it as:

\( \mathcal{A}[f](x) = \int\frac{dx}{\left[\frac{\partial}{\partial t} f^{\circ t}(x)\right]_{t=0}} \)

which also serves to emphasize the fact that Abel functions are only determined up to a constant, and that a solution \( \alpha(x) \) (to the Abel functional equation) can always be generalized to \( \alpha(x) + C \), which is also true of integrals.

Iterated-dxp can be expressed as the hyperbolic iteration of \( e^{ax}-1 \) as:

\( \text{dxp}_{(e^a)}^{\circ t}(x) = a^tx + \frac{a^{t+1}(a^t-1)x^2}{2(a-1)}

+ \frac{a^{t+2}(a^t-1)(-1-2a+a^t(a+2))x^3}{6(a-1)^2(a+1)} \)

and since the Julia function of dxp is also the iterative logarithm of dxp:

\( \left[\frac{\partial}{\partial t}\text{dxp}_{(e^a)}^{\circ t}(x)\right]_{t=0} = \mathcal{J}[\text{dxp}_{(e^a)}](x) \)

which evaluates to the power series:

\( \mathcal{J}[\text{dxp}_{(e^a)}](x) = \ln(a)\left(

x + \frac{ax^2}{2(a-1)} + \frac{a^2x^3}{6(1-a^2)}

- \frac{a^3(1+2a)x^4}{24(-1-a+a^3+a^4)} + \cdots

\right) \)

We can now find the Abel function of dxp by integrating its Julia function, but we need to find the reciprocal first. Finding the reciprocal of a power series can be a tedious task, so I've done the work for you:

\( \frac{1}{f'(x)} = \frac{1}{f'(0)} - \frac{f''(0)}{f'(0)^2}x + \frac{2f''(0)^2-f'(0)f'''(0)}{2f'(0)^3}x^2 + \cdots \)

and since \( \mathcal{J}[f](x) = \frac{1}{\frac{\partial}{\partial x}\mathcal{A}[f](x)} \), we solve for \( f^{(k)}(0) \) by equating the coefficients of x, and the solution to these equations is:

\( \mathcal{A}[\text{dxp}_{(e^a)}](x) = \frac{1}{\ln(a)}\left(

x + \frac{ax^2}{4(1-a)} + \frac{a^2(1+5a)x^3}{36(a-1)^2(a+1)}

- \frac{a^4(2+a+3a^2)x^4}{32(a-1)^3(a+1)(1+a+a^2)} + \cdots

\right) \)

Lastly, we can relate these findings back to the super-logarithm.

Let \( \alpha(x) = \mathcal{A}[\text{dxp}_{b}](x) \) be the Abel function of dxp.

Let \( \beta(x) = \alpha(x/b - 1) \).

Then \( \beta(x) \) is an Abel function of exp.

Proof.

\( \alpha(b^x-1) = \alpha(x) + 1 \)

\( \alpha(b^{(x/b - 1)}-1) = \alpha(x/b - 1) + 1 \)

\( \alpha(b^{(x/b)}/b-1) = \alpha(x/b - 1) + 1 \)

\( \alpha((b^{1/b})^x/b-1) = \alpha(x/b - 1) + 1 \)

\( \beta((b^{1/b})^x) = \beta(x) + 1 \). []

This means that \( \beta(x) = \text{slog}_{(b^{1/b})}(x) \) which relates back to the super-logarithm as follows:

\( \text{slog}_{(e^{a(e^{-a})})}(x) = C + \frac{1}{\ln(a)}\left(

(x(e^{-a})-1) + \frac{a(x(e^{-a})-1)^2}{4(1-a)} + \frac{a^2(1+5a)(x(e^{-a})-1)^3}{36(a-1)^2(a+1)}

- \frac{a^4(2+a+3a^2)(x(e^{-a})-1)^4}{32(a-1)^3(a+1)(1+a+a^2)} + \cdots

\right) \)

Andrew Robbins