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 = b^x - 1)
notation, and thus the topological conjugacy between "exp" and "dxp" can be expressed as:
}^{\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:
 = \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:
}](x) = \mathcal{A}[\text{dxp}_b](x/b-1))
}](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:
 = \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:
 = \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)
(to the Abel functional equation) can always be generalized to  + C)
, which is also true of integrals.
Iterated-dxp can be expressed as the hyperbolic iteration of
as:
}^{\circ t}(x) = a^tx + \frac{a^{t+1}(a^t-1)x^2}{2(a-1)}<br />
+ \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:
}^{\circ t}(x)\right]_{t=0} = \mathcal{J}[\text{dxp}_{(e^a)}](x))
which evaluates to the power series:
}](x) = \ln(a)\left(<br />
x + \frac{ax^2}{2(a-1)} + \frac{a^2x^3}{6(1-a^2)}<br />
- \frac{a^3(1+2a)x^4}{24(-1-a+a^3+a^4)} + \cdots<br />
\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'(0)} - \frac{f''(0)}{f'(0)^2}x + \frac{2f''(0)^2-f'(0)f'''(0)}{2f'(0)^3}x^2 + \cdots)
and since = \frac{1}{\frac{\partial}{\partial x}\mathcal{A}[f](x)})
, we solve for }(0))
by equating the coefficients of x, and the solution to these equations is:
}](x) = \frac{1}{\ln(a)}\left(<br />
x + \frac{ax^2}{4(1-a)} + \frac{a^2(1+5a)x^3}{36(a-1)^2(a+1)}<br />
- \frac{a^4(2+a+3a^2)x^4}{32(a-1)^3(a+1)(1+a+a^2)} + \cdots<br />
\right))
Lastly, we can relate these findings back to the super-logarithm.
Let = \mathcal{A}[\text{dxp}_{b}](x))
be the Abel function of dxp.
Let = \alpha(x/b - 1))
.
Then)
is an Abel function of exp.
Proof.
 = \alpha(x) + 1)
}-1) = \alpha(x/b - 1) + 1)
}/b-1) = \alpha(x/b - 1) + 1)
^x/b-1) = \alpha(x/b - 1) + 1)
^x) = \beta(x) + 1)
. []
This means that = \text{slog}_{(b^{1/b})}(x))
which relates back to the super-logarithm as follows:
})}(x) = C + \frac{1}{\ln(a)}\left(<br />
(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)}<br />
- \frac{a^4(2+a+3a^2)(x(e^{-a})-1)^4}{32(a-1)^3(a+1)(1+a+a^2)} + \cdots<br />
\right))
Andrew Robbins
First lets start with some recent realizations about Abel functions, Julia functions, and topological conjugacy. So, I'll use the
which can also be found in this thread. The Abel functions and Julia functions can be related as:
which, as they apply to exp/dxp, can be proven (and is proven later) to imply:
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:
or the logarithm of the Schroeder function, but we can also express it as:
which also serves to emphasize the fact that Abel functions are only determined up to a constant, and that a solution
Iterated-dxp can be expressed as the hyperbolic iteration of
and since the Julia function of dxp is also the iterative logarithm of dxp:
which evaluates to the power series:
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:
and since
Lastly, we can relate these findings back to the super-logarithm.
Let
Let
Then
Proof.
This means that
Andrew Robbins
