06/20/2008, 01:26 PM
There is an error in your calculation of the Abel function of dxp:
The regular Abel function (for a function with fixed point at 0) always has a singularity at 0. You can not expand it into a powerseries at 0.
In the hyperbolic case the Abel function is the logarithm of the Schroeder function (as Gottfried also pointed out in his post), here you can also see that it is not developable at 0, because log is not.
What you however can do is to express the Abel function as \( c\log(z)+M(z) \) where M is a function, such that \( z^m M(z) \) is holomorphic and so developable at at 0. This is an extension to holomorphic functions (so called meromorphic functions) which also allow a finite number of negative powers in the power series development, i.e.
\( M(z) = \sum_{n=-m}^\infty M_n z^n \).
I explained here how this comes.
To compute the inverse of a powerseries \( f \) - which is a meromorphic function - we determine the first index \( m \) such that \( f_m\neq 0 \) then we divide \( f \) by \( z^m \) and get a powerseries with \( g \) with \( g_0\neq 0 \). We can then compute the reciprocal of this powerseries and the inverse of \( f \) is then
\( \frac{1}{f(z)} = z^{-m} \frac{1}{f(z)/z^m} \)
is a powerseries with \( m \) negative powers.
In our case \( f(z)=\mathcal{J}[f](z) \), \( f_0=0 \) and \( m=1 \).
\( \frac{1}{f(z)}=z^{-1}\frac{1}{f(z)/z} = \frac{1}{\log(a)} z^{-1} + \frac{-{a}^{2} }{{2 \left( {a}^{2} - a \right)}}z^0 - \frac{\frac{{-{a}^{5} }}{{2 \left( {a}^{2} - a \right)}} + \frac{{{a}^{4} }}{{2 \left( {a}^{2} - a \right)}} + \frac{{{a}^{3} }}{3}}{{\left( {a}^{3} - a \right) }} z + \dots \)
If we now integrate this:
\( \alpha(z)=\mathcal{A}[\text{dxp}_{e^a}](z)=\int{\frac{dz}{f(z)}=C+\log_a(z)+\frac{-{a}^{2} }{{2 \left( {a}^{2} - a \right)}}z^1 - \frac{\frac{{-{a}^{5} }}{{2 \left( {a}^{2} - a \right)}} + \frac{{{a}^{4} }}{{2 \left( {a}^{2} - a \right)}} + \frac{{{a}^{3} }}{3}}{{\left( {a}^{3} - a \right) }} \frac{z^2}{2} + \dots \)
Yes this is correct and this is equal to the rslog because of the following:
You put the formula
\( \left(\exp_{p^{1/p}}\right)^{\circ t}=\mu_p\circ \tau_1\circ \text{dxp}_p^{\circ t} \circ \tau_1^{-1}\circ \mu_p^{-1} \)
where \( p \) is the lower fixed point, \( \mu_c(z)=cz \) and \( \tau_c(z)=z+c \).
If \( \alpha \) is the (principal) Abel function of \( \text{dxp}_p \) we write this is as:
\( \left(\exp_{p^{1/p}}\right)^{\circ t}=\mu_p\circ \tau_1\circ \alpha^{-1}\circ \tau_t \circ \alpha \circ \tau_1^{-1}\circ \mu_p^{-1}=\beta^{-1}\circ \tau_t \circ \beta \) with \( \beta = \alpha\circ \tau_1^{-1}\circ \mu_p^{-1} \) as you already pointed out with \( \beta(z)=\alpha(z/p-1) \).
Now the rslog computation is slightly different, we start with
\( h=\tau_p^{-1}\circ \exp_{b}\circ \tau_p=\mu_p\circ \text{dxp}_b \)
i.e. \( \exp_b^{\circ t} = \tau_p \circ h^{\circ t} \circ \tau_p^{-1} = \tau_p \circ \left(\mu_p\circ \text{dxp}_b\right)^{\circ t} \circ \tau_p^{-1} \)
to show that both Abel functions of \( \exp_b \) are identical we just need to show that
\( \mu_p\circ \tau_1\circ \text{dxp}_p^{\circ t} \circ \tau_1^{-1}\circ \mu_p^{-1} = \tau_p \circ \left(\mu_p\circ \text{dxp}_b\right)^{\circ t} \circ \tau_p^{-1} \)
This is be done by the following equivalences:
\(
\begin{align*}
\tau_p\circ\mu_p \circ \text{dxp}_p^{\circ t} \circ \mu_p^{-1}\circ \tau_p^{-1} &= \tau_p \circ \left(\mu_p\circ \text{dxp}_b\right)^{\circ t} \circ \tau_p^{-1}\\
\mu_p \circ \text{dxp}_p^{\circ t} \circ \mu_p^{-1} &= \left(\mu_p\circ \text{dxp}_b\right)^{\circ t}=\left(\mu_p \circ \text{dxp}_p \circ \mu_p^{-1}\right)^{\circ t}
\end{align*}
\)
And the last line is automatically satisfied by the regular iteration.
If you now redevelop
\( \beta(z)=\alpha(z/b - 1) \) with the correct \( \alpha \) - note that the logarithmic term \( \log_a(z/b-1) \) in \( \alpha(z/b-1) \) is now developable at 0 - you see that the coefficients become infinite sums, which is again no finite description (except you find some closed expression).
andydude Wrote: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:
\( \mathcal{A}[\text{dxp}_{(e^a)}](x) = \frac{1}{\ln(a)}\left(
x + \frac{ax^2}{4(1-a)} + \frac{a^2(1+5a)x^3}{36(a-1)^2(a+1)}
- \frac{a^4(2+a+3a^2)x^4}{32(a-1)^3(a+1)(1+a+a^2)} + \cdots
\right) \)
The regular Abel function (for a function with fixed point at 0) always has a singularity at 0. You can not expand it into a powerseries at 0.
In the hyperbolic case the Abel function is the logarithm of the Schroeder function (as Gottfried also pointed out in his post), here you can also see that it is not developable at 0, because log is not.
What you however can do is to express the Abel function as \( c\log(z)+M(z) \) where M is a function, such that \( z^m M(z) \) is holomorphic and so developable at at 0. This is an extension to holomorphic functions (so called meromorphic functions) which also allow a finite number of negative powers in the power series development, i.e.
\( M(z) = \sum_{n=-m}^\infty M_n z^n \).
I explained here how this comes.
To compute the inverse of a powerseries \( f \) - which is a meromorphic function - we determine the first index \( m \) such that \( f_m\neq 0 \) then we divide \( f \) by \( z^m \) and get a powerseries with \( g \) with \( g_0\neq 0 \). We can then compute the reciprocal of this powerseries and the inverse of \( f \) is then
\( \frac{1}{f(z)} = z^{-m} \frac{1}{f(z)/z^m} \)
is a powerseries with \( m \) negative powers.
In our case \( f(z)=\mathcal{J}[f](z) \), \( f_0=0 \) and \( m=1 \).
\( \frac{1}{f(z)}=z^{-1}\frac{1}{f(z)/z} = \frac{1}{\log(a)} z^{-1} + \frac{-{a}^{2} }{{2 \left( {a}^{2} - a \right)}}z^0 - \frac{\frac{{-{a}^{5} }}{{2 \left( {a}^{2} - a \right)}} + \frac{{{a}^{4} }}{{2 \left( {a}^{2} - a \right)}} + \frac{{{a}^{3} }}{3}}{{\left( {a}^{3} - a \right) }} z + \dots \)
If we now integrate this:
\( \alpha(z)=\mathcal{A}[\text{dxp}_{e^a}](z)=\int{\frac{dz}{f(z)}=C+\log_a(z)+\frac{-{a}^{2} }{{2 \left( {a}^{2} - a \right)}}z^1 - \frac{\frac{{-{a}^{5} }}{{2 \left( {a}^{2} - a \right)}} + \frac{{{a}^{4} }}{{2 \left( {a}^{2} - a \right)}} + \frac{{{a}^{3} }}{3}}{{\left( {a}^{3} - a \right) }} \frac{z^2}{2} + \dots \)
Quote: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.
\( \alpha(b^x-1) = \alpha(x) + 1 \)
\( \alpha(b^{(x/b - 1)}-1) = \alpha(x/b - 1) + 1 \)
\( \alpha(b^{(x/b)}/b-1) = \alpha(x/b - 1) + 1 \)
\( \alpha((b^{1/b})^x/b-1) = \alpha(x/b - 1) + 1 \)
\( \beta((b^{1/b})^x) = \beta(x) + 1 \). []
Yes this is correct and this is equal to the rslog because of the following:
You put the formula
\( \left(\exp_{p^{1/p}}\right)^{\circ t}=\mu_p\circ \tau_1\circ \text{dxp}_p^{\circ t} \circ \tau_1^{-1}\circ \mu_p^{-1} \)
where \( p \) is the lower fixed point, \( \mu_c(z)=cz \) and \( \tau_c(z)=z+c \).
If \( \alpha \) is the (principal) Abel function of \( \text{dxp}_p \) we write this is as:
\( \left(\exp_{p^{1/p}}\right)^{\circ t}=\mu_p\circ \tau_1\circ \alpha^{-1}\circ \tau_t \circ \alpha \circ \tau_1^{-1}\circ \mu_p^{-1}=\beta^{-1}\circ \tau_t \circ \beta \) with \( \beta = \alpha\circ \tau_1^{-1}\circ \mu_p^{-1} \) as you already pointed out with \( \beta(z)=\alpha(z/p-1) \).
Now the rslog computation is slightly different, we start with
\( h=\tau_p^{-1}\circ \exp_{b}\circ \tau_p=\mu_p\circ \text{dxp}_b \)
i.e. \( \exp_b^{\circ t} = \tau_p \circ h^{\circ t} \circ \tau_p^{-1} = \tau_p \circ \left(\mu_p\circ \text{dxp}_b\right)^{\circ t} \circ \tau_p^{-1} \)
to show that both Abel functions of \( \exp_b \) are identical we just need to show that
\( \mu_p\circ \tau_1\circ \text{dxp}_p^{\circ t} \circ \tau_1^{-1}\circ \mu_p^{-1} = \tau_p \circ \left(\mu_p\circ \text{dxp}_b\right)^{\circ t} \circ \tau_p^{-1} \)
This is be done by the following equivalences:
\(
\begin{align*}
\tau_p\circ\mu_p \circ \text{dxp}_p^{\circ t} \circ \mu_p^{-1}\circ \tau_p^{-1} &= \tau_p \circ \left(\mu_p\circ \text{dxp}_b\right)^{\circ t} \circ \tau_p^{-1}\\
\mu_p \circ \text{dxp}_p^{\circ t} \circ \mu_p^{-1} &= \left(\mu_p\circ \text{dxp}_b\right)^{\circ t}=\left(\mu_p \circ \text{dxp}_p \circ \mu_p^{-1}\right)^{\circ t}
\end{align*}
\)
And the last line is automatically satisfied by the regular iteration.
Quote:This means that \( \beta(x) = \dots \)
If you now redevelop
\( \beta(z)=\alpha(z/b - 1) \) with the correct \( \alpha \) - note that the logarithmic term \( \log_a(z/b-1) \) in \( \alpha(z/b-1) \) is now developable at 0 - you see that the coefficients become infinite sums, which is again no finite description (except you find some closed expression).